dbms theory

DATABASE MANAGEMENT SYSTEMS LAB MANUAL

DEPARTMENT OF COMPUTER APPLICATIONS i

INDEX

CONTENTS PAGE NO STRUCTURED QUERY LANGUAGE (SQL) 1

BASIC SQL OPERATION 2

OPERATORS IN SQL 11

FUNCTIONS IN SQL 13

Single-Row Functions 13

Group-Value Functions 19

SET OPERATORS 20

JOIN OPERATION 21

Inner join 21

Outer joins 21

SUBQUERIES 24

VIEWS 25

Creating a read only view 26

Dropping a VIEW 26

INTEGRITY CONSTRAINTS IN CREATE TABLE COMMAND 26

PL/SQL 28

Anonymous block 29

Named block 29

Features of PL/SQL 29

Structure of a PL/SQL block 29

Conditional Logic 32

Cursor 34

Ref Cursors 38

Varying Arrays 38

Named Program Units 40

PROCEDURES 40

Parameter modes 40

Calling a Procedure 41

Dropping procedure 41

FUNCTIONS 41

PACKAGES 42

Create package Syntax 42

Package specification 43

Package body 43

TRIGGERS 44

Types of Triggers 44


DEPARTMENT OF COMPUTER APPLICATIONS ii

Enabling and Disabling Triggers 45

Dropping Triggers 45

TABLESPACE 46

Creating tablespace 46

VISUAL BASIC 48

The Visual Basic Environment 48

Properties Window 49

The tool box 49

Data source name 50

Creating a dsn 50

Open DataBase Connectivity 51

SQL SOLVED PROBLEMS 52

PL/SQL SOLVED PROBLEMS 89

DATABASE CONNECTIVITY SOLVED PROBLEMS 107


DEPARTMENT OF COMPUTER APPLICATIONS 1

INTRODUCTION

A database management system (DBMS) is a collection of interrelated data and a set

of programs to access those data .The primary goal of DBMS is to provide a way to

store and retrieve database information that is both convenient and efficient.

The DBMS a general purpose software system that facilitates the process of

i) defining,

ii) constructing and

iii) manipulating database for various application. NEED FOR DATABASE SYSTEMS

DBMS has been developed to overcome the disadvantages with the conventional

systems like file-processing system, viz.

1. Data redundancy and inconsistency

2. Difficulty in accessing data

3. Data isolation

4. Integrity problems

5. Atomicity problems

6. Concurrent-access anomalies

7. Security problems THE RELATIONAL DATABASE: A relational database uses tables to store information. DATABASE OBJECTS: A database object is anything defined and stored in a

database.

Some of the database objects are:-tables, views, indexes etc

TABLES: Table is the basic unit of data storage in a relational database management

system. Every table has a table name and a set of columns and rows in which the data

is stored. Each column is given a column name, a datatype or domain type and a

width.

STRUCTURED QUERY LANGUAGE (SQL)

SQL is the standard relational database language. It includes features for defining the

structure of the data, for modifying data in the database and for specifying security



constraints. DDL and DML form part of SQL. Data Definition Language (DDL) provides

commands for defining the relation schemas. Data Manipulation Language (DML)

includes commands to insert tuples , delete tuples, and modify tuples in the database.

SQL language has the following parts:

* Data Definition Language (DDL)

* Data Manipulation Language (DML)

* Transaction Control Language (TCL)

* Data Control Language (DCL)

BASIC SQL OPERATION

SQL statements can be grouped into four general categories: Data definition language

(DDL), Data manipulation language (DML), Transaction control language (TCL), Data

control language (DCL).

1. DATA DEFINITION LANGUAGE (DDL)

The overall design of a database is called the database schema. A database schema is

specified by a set of definitions that are expressed using a data definition language. It

is used for structuring the database.

We use the data definition language statements CREATE, ALTER, DROP, TRUNCATE to

create new objects, alter the structure of existing objects, completely remove objects

from system, or delete all rows permanently from the table leaving the structure of the

table.

DDL COMMANDS

1.1 CREATE TABLE COMMAND

While naming a table,

i) First letter should be an alphabet ii) Reserved words cannot be used to name a table

iii) Maximum length for a table name is 30 character

iv) Table name should also be unique

v) Underscore, numerals and letters are allowed but not blank space and

single quotes Syntax:

Create table<table name> (attribute-1 domaintype-1,

attribute-2 domaintype-2…..



attribute-n domaintype-n,

integrity_CONSTRAINT<constraint_name-

1><constraints>,

CONSTRAINT<constraint_name-2><constraints>,

……,

CONSTRAINT<constraint name n><constraints>);

Example:

Create table manager (mnane varchar2(15), eno int, mid varchar2(10) primary key); Output:

Table created

COMMAND FOR A TABLE FROM AN EXISTING TABLE

A new table is created with attributes whose domain type and contents are same as

the existing table’s.

Syntax:

Create table <new table name> (<attribute1….attribute n>) as select attribute-1,

attribute-2 … attribute-n from <existing table name>

Example-1:

Create table branch_new(branchname, branchno) as select bname,bno from

branch_exist;

Output:

Table created

Example-2:

Create table basset(bname,asset) as select bname,assetfrom branch ;

Output:

Table created

1.2. DESC COMMAND

Command to see the structure of a table

Syntax:

desc<table name>

Example:

desc branch;



Output:

Name Null? Type

-------------------- ------------ ---------------

BNAME CHAR(15)

BNO NOT NULL VARCHAR2(15)

MANAGER VARCHAR2(15)

ASSET NUMBER

1.3. ALTER TABLE COMMAND

Command for modifying the fields in a table

This command is used to

1. Add a new column

2. Change the width of a data type or the data type itself

3. Include or drop integrity constraints

Syntax: - to add a new field into a table

alter table<table name> add(attribute-1 domintype, attribute-2 domaintype

…attribute-n domaintype-n);

Example:

alter table branch add (city varchar(15), state char(15));

Output:

Table altered.

Syntax: - To modify the domain type

alter table<table name>modify(attribute newdomaintype1, attribute2

newdomaintype2)

Example:

alter table branch modify (state varchar(15),city char(15));

Output:

Table altered

Syntax: - to add integrity constraint

alter table add integrity constraint (attribute)

Example:

alter table add primary key (bno);



alter table add unique (name, manager);

Output:

Table altered

Syntax: - to drop integrity constraint

alter table<table name>drop integrity constraints;

Example:

alter table breanch drop primary key;

alter table branch drop unique (name,manager);

Output:

Table altered

1.4. DROP TABLE COMMAND

Command to delete the entire schema of the table (table with its content is deleted)

Syntax:

drop table<table name>;

Example:

drop table basset;

Output:

Table dropped.

1.5. TRUNCATE TABLE COMMAND

This command is to delete the records in the table, while retaining the structure of the

table.

Syntax:

truncate table <table name> ;

Example:

truncate table basset;

Output:

Table truncated

2. DATA MANIPULATION LANGUAGE (DML)

These commands are used to query and manipulate existing objects like tables.



DML commands are: INSERT, SELECT, UPDATE, DELETE

DML COMMANDS

2.1 INSERT STATEMENT - To insert values into a table

To insert a single record

Syntax:

Insert into <table name> (attribute-1,attribute-2,…attribute-n) values (value-1,value-

2,…value-n);

Example:

insert into branch (bname,bno,manager,asset) values (‘tvmbranch’,’b1’,’Smith’,1000);

Output:

1 row created.

Mass insertion

Syntax:

insert into <table name> (attribute-1,attribute-2,…,attribute-n) values(‘&attribute-

1’,’&attribute-2’,…..’&attribute-n’);

Example:

Insert into branch(bname,bno,manager) values (‘&bname’,’&bno’,’&manager’);

Output:

Enter value for name: klmbranch

Enter value for bno: b2

Enter value for manager: Jones

Old 1:insert into branch(bname,bno,manager)

Values(‘&bname’,’&bno’,’&manager’)

New 1:insert into branch(bname,bno,manager)

Values(‘klmbranch’,’b2’,’Jones’)

1 row created

2.2 SELECT STATEMENT

Select command is used to retrieve information stored in the table.

Syntax:

select <column name> from <table name>;



Example:

Select * from branch; // Select all the columns of the table

VARIATIONS OF SELECT STATEMENT

� To see the values of specific attributes

Syntax:

Select attribute-1,attribute-2,….attribute-n from<table name>;

Example:

Select bname,bno from branch;

� Selecting distinct rows

Command to list the tuples in a table without duplication

Syntax:

select distinct attribute-1,attribute-2,…..from<table name> where<condition>;

Example:

Select distinct bname from branch;

� Command to list the tuples in a table in a sorted order based on some

attributes

Syntax:

Select attribute-1,attribute-2,……attribute-n from<table name> order by attribute-1

asc;

Select attribute-1,attribute-2,……attribute-n from<table name> order by attribute-1

desc;

Example:

Select * from branch order by asset asc;

Select * from branch order by asset desc;

� Select command with ‘where’ clause

To select specific rows or records which satisfy some specific conditions, we can use

‘where’ clause in the select command.

Syntax:

select <attribute> from <table name> where <conditions>;

Example:

Select bname from branch where bno=123;



� Select command to create a table

User can include a ‘select’ clause in a create table command to create table and copy

the records into it.

Syntax:

create table <table name> as select <column names> from <existing table name>;

Example:

create table branch_new as select * from branch;

‘where’ clause can be included to create table with those records which satisfy the

‘where’ condition

Example:

create table branch_new as select * from branch where bno>200;

� Select command to insert records

This command either create a table or consider an existing table prior to insertion, with

the same structure as the other table and then performs the insertion command.

Syntax:

insert in to <table name> (select column_name from <existing table name>);

Example:

insert into branch select * from branch_usa;

insert into branch(branchname,branchno) select bname,bno from branch_usa;

2.3 UPDATE STATEMENT

command to modify the value of records in a table

Syntax:

update<table-name>set <change-value-of-attribute> where<condition>;

Example:

update brach set city=’tvm’ where bno=’b1’;

Output:

1 row updated

update comfort set midnight = (select temperature from weather where city =

'manchester') where city = 'walpole' and sampledate = to_date('22-dec-1999','dd-

mon-yyyy');



1 row updated.

2.4 DELETE STATEMENT

Command for deletion of rows

� Deletion of specified rows

Syntax:

Delete from<table name>where<condition>;

Example:

delete from branch1 where asset>2000;

Output:

0 rows deleted

� Deletion of entire rows

Syntax:

delete from<table name>;

Example:

Delete from branch1;

Output:

4 rows deleted

3. TRANSACTION CONTROL LANGUAGE (TCL)

The various commands in TCL are

� COMMIT

� SAVEPOINT

� ROLLBACK

3.1 COMMIT COMMAND

Transaction changes can be made permanent to the database by commit command.

Syntax:

commit work;

or

commit;

user can set auto commit option by using the command ‘set autocommit on’



3.2 SAVEPOINT COMMAND

These are markers to divide a very lengthy transaction into smaller ones. They are

used to identify a point in a transaction to which we can later roll back. This is used

along with roll back.

Syntax:

savepoint savepoint_id;

3.3 ROLLBACK

It is used to undo the work done in the current transaction. We can either rollback the

entire transaction or to a savepoint so that statements after the savepoint are rolled

back.

Syntax:

rollback work;

or

rollback;

Syntax to rollback to a particular stage in a transaction:

rollback to savepoint save_pt;

where save_pt is the savepoint.

4. DATA CONTROL LANGUAGE (DCL)

Data Control Language is used to manage user access to a database. DCL statements

such as GRANT, REVOKE control access to database and affirm or revoke database

transactions. It provides the user with privilege commands. The owner of the database

can grant privileges or withdraw (revoke) privilege to other database.

4.1 GRANT

An object privilege specifies what a user can do with a database object such as a table,

a sequence or a view. The privileges are granted using GRANT statement.

Syntax:

grant privileges on <object_name> to <username>;

Example:

grant select, update on branch to new_branch;



grant all on customers to ashraf; // Grants all permissions on the table customers to

the user who logs in as 'ashraf'.

grant select on customers to sunil; // Grants SELECT permission on the table

customers to the user 'sunil'. User 'sunil' does not have permission to insert, update,

delete or perform any other operation on customers table.

grant select on customers to sunil with grant option; // Enables user 'sunil' to give

SELECT permission on customers table to other users

4.2 REVOKE

The privileges, once granted, can be taken away. REVOKE statement takes privileges

not only from the grantee but also from users who are granted privileges by the

grantee.

Syntax:

revoke privileges on <object_name> from <username>;

Example:

revoke select, update on branch from mca;

revoke delete on customers from ashraf

OPERATORS IN SQL

The operators supported by SQL are

1. Arithmetic operators

2. Comparison operators

3. Logical operators

4. Pattern matching operators 1. Arithmetic operators

The various arithmetic operators are

Function Definition

value1 + value2 Addition

value1 - value2 Subtraction

value1 * value2 Multiplication

value1 / value2 Division



Example:

select * from pnum;

NAME NO1 NO2

---------- ---------- ----------

os 60 70

nm 70 40

ds 60 90

select name, no1 + no2 as Additon, no1 - no2 as Substraction, no1 / no2 as Division,

no1 * no2 as Multiplication from pnum;

Output:

NAME ADDITON SUBSTRACTION DIVISION MULTIPLICATION

---------- ---------- ------------------ -------------- -------------------

os 130 -10 .857142857 4200

nm 110 30 1.75 2800

ds 150 -30 .666666667 5400

2. Comparison operators

The various comparison operators are:

=, !=, <, >, <=, >=,

between ( to check between any two values)

in (to match with any of the values in the list) //not in – opposite of the in function

result.

like (to match a character pattern)

is null (to check whether it is Null) // is not null - opposite of the is null function

result.

Example:

Select bname from branch where bno > 100;

Select bname from branch where bno is null;

3. Logical Operators

These are used to combine the results of two conditions to produce a single result. The

logical operators are:

AND, NOT, OR

Example:

select regno from car where colour = 'blue' and regno = 1230



4. Pattern matching operators

Pattern matching operators are like, %, _ (under score)

LIKE is useful for slightly more complex queries, in which a non-exact match is

required. it can be used to select data in which the column entry matches a pattern

containing the wildcards % or _

� % matches 1 or more characters of any type � _ matches any single character

Example:

Select bname,asset from branch where bname like ‘tv%’;

Select bname from branch where bname like ‘tv_%’;

FUNCTIONS IN SQL

These are built-in functions for performing operations using the Data Manipulation

commands.

Single-Row Functions

Single-row functions return a single result row for every row of a queried table or view.

These functions can appear in select lists, WHERE clauses, START WITH and CONNECT

BY clauses, and HAVING clauses.

Group Functions (Aggregate functions)

Group Functions Returns a result based on a group of rows.

Single-Row functions

1. Numeric function

2. Character function

3. Date function

4. Conversion function

1. Numeric Functions

Numeric functions accept numeric input and return numeric values. Most numeric

functions that return NUMBER values that are accurate to 38 decimal digits.

Function Definition

ABS(value) ABSolute value



ACOS(value) Arc COSine of value, in radians

ASIN(value) Arc SINe of value, in radians

ATAN(value) Arc TANgent of value, in radians

CEIL(value) Numeric CEILing: the smallest integer larger than

or equal to value

COS(value) COSine of value

COSH(value) Hyperbolic COSine of value

EXP(value) e raised to value EXPonent

FLOOR(value) Largest integer smaller than or equal to value

LN(value) Natural Logarithm of value

LOG(value) Base 10 LOGarithm of value

MOD(value, divisor) MODulus

NVL(value, substitute) substitute for value if value is NULL

POWER(value, exponent) value raised to an exponent POWER

ROUND(value, precision) ROUNDing of value to precision

SIGN(value) 1 if value is positive, –1 if negative, 0 if zero

SIN(value) SINe of value

SINH(value) Hyperbolic SINe of value

Example:

select * from emp;

ENO SAL COMM

---------- ---------- ----------

101 4000 2.2

102 8000 1.76

select sal + nvl(comm,0) from emp;

SAL+NVL(COMM,0)

---------------------

4002.2

8001.76

8000

0

select abs(-123) from dual;

ABS(-123)

------------

123

select ceil(1.9),ceil(2),ceil(1.3),floor(1.4),floor(1.9),floor(2) from dual;

CEIL(1.9) CEIL(2) CEIL(1.3) FLOOR(1.4) FLOOR(1.9) FLOOR(2)

---------- --------- ---------- ------------- ------------- -----------

2 2 2 1 1 2



SQL> select mod(6,5),mode(6,2) from dual;

MOD(6,5) MOD(6,2)

---------- ----------

1 0

select power(3,2),sqrt(64),round(1.2),round(1.9) from dual;

POWER(3,2) SQRT(64) ROUND(1.2) ROUND(1.9)

------------- ---------- -------------- --------------

9 8 1 2

2. Character Functions

These are functions operated on characters. Input will be character and output will be

either character or number values. Character functions that return character values

return values of the same datatype as the input argument. The length of the value

returned by the function is limited by the maximum length of the datatype returned.

Function Name Use

|| Glues or concatenates two strings together. The | symbol is called

a vertical bar or pipe.

CONCAT CONCATenates two strings together (same as | |).

INITCAP INITial CAPital. Capitalizes the first letter of a word or series of

words.

INSTR Finds the location of a character IN a STRing.

LENGTH Tells the LENGTH of a string.

LOWER Converts every letter in a string to LOWERcase.

LPAD Left PAD. Makes a string a certain length by adding a certain set

of characters to the left.

RPAD Right PAD. Makes a string a certain length by adding a certain set

of characters to the right.

RTRIM Right TRIM. Trims all the occurrences of any one of a set of

characters off of the right side of a string.

SOUNDEX Finds words that SOUND like the EXample specified.

SUBSTR SUBSTRing. Clips out a piece of a string.

UPPER Converts every letter in a string into UPPERcase.

Examples

select * from location;



CITY COUNTRY

--------------- -----------

athens greece

trivandrum inda

southwales Australia

select upper(city),initcap(country),length(city),city||country from location;

UPPER(CITY) INITCAP(COUNTRY) LENGTH(CITY) CITY||COUNTRY

--------------- ---------------------- ---------------- -----------------------

ATHENS Greece 6 athensgreece

TRIVANDRUM Inda 10 trivandruminda

SOUTHWALES Australia 10 southwalesaustralia

select lower(city) from location;

LOWER(CITY)

---------------

athens

trivandrum

southwales

select concat(city,country) from location;

CONCAT(CITY,COUNTRY)

-----------------------------

athensgreece

trivandruminda

southwalesaustralia

select rpad(city,10,'*') from location;

RPAD(CITY,10,'*')

---------------------

athens****

trivandrum

southwales

select lpad(city,15,'*') from location;

LPAD(CITY,15,'*')

---------------------

*********athens

*****trivandrum

*****southwales

select substr(city,2,4) from location;

SUBSTR(CITY,2,4)

---------------------

then

riva

outh



select city,instr(city,'t') from location;

CITY INSTR(CITY,'T')

--------------- ------------------

athens 2

trivandrum 1

southwales 4

select * from location where soundex(city)=soundex('athans');

CITY COUNTRY

--------------- ---------------

athens Greece

soundex – has the unusual ability to find words that sound like other

3. Date Functions

Date functions operate on date (DATE), and interval (INTERVAL YEAR TO MONTH)

values.

Function Use

ADD_MONTHS(date,count) Adds count months to date.

GREATEST(date1,date2,date3,...) Picks latest date from list of dates.

LEAST(date1,date2,date3,...) Picks earliest date from list of dates.

LAST_DAY(date) Gives date of last day of month that date is

in.

MONTHS_BETWEEN(date2,date1) Gives date2−date1 in months

(can be fractional months).

NEXT_DAY(date,’day’) Gives date of next day after date, where

‘day’is ‘Monday’, ‘Tuesday’, and so on.

Example:

select add_months(sysdate,3) from dual;

ADD_MONTH

--------------

27-JUL-08

select sysdate,last_day(sysdate) "last",last_day(sysdate) - sysdate "days left" from

dual;

SYSDATE Last Days Left

------------- ------------- ----------

30-MAY-01 31-MAY-01 1



select next_day('02-feb-2001','tuesday') "next day" from dual;

NEXT DAY

-----------

06-FEB-2001

select months_between (t_date,n_date ) from ex;

select greatest('10-mar-2008','2-apr-2008') from dual;

GREATEST('

--------------

2-apr-2008

SQL> select least('10-mar-2008','2-apr-2008') from dual;

LEAST('10-M

---------------

10-mar-2008

Note: SYSDATE

SYSDATE returns the current date and time set for the operating system on which the

database resides.

select sysdate from dual;

SYSDATE

-----------

27-APR-08

4. Conversion Functions

Conversion functions convert a value from one datatype to another. Generally, the

form of the function names follows the convention datatype TO datatype. The first

datatype is the input datatype. The second datatype is the output datatype.

The SQL conversion functions are:

Function Use

TO_CHAR transforms a DATE or NUMBER into a character string.

TO_DATE transforms a NUMBER, CHAR, or VARCHAR2 into a DATE.

TO_NUMBER transforms a CHAR or VARCHAR2 into a NUMBER.

Example:

select to_char(sysdate,'yy>>MM>>dd') from dual;



TO_CHAR(SY

---------------

08>>03>>07

select to_char(sysdate,'dd month year') from dual;

TO_CHAR(SYSDATE,'DDMONTHYEAR')

-------------------------------------------

07 march two thousand eight

select to_date('01032008','mm dd yyyy') from dual;

TO_DATE('

------------

03-JAN-08

Group-Value Functions

Group value functions return a single result row based on groups of rows, rather than

on single rows. Group-Value functions can appear in select lists and in ORDER BY and

HAVING clauses. They are commonly used with the GROUP BY clause in a SELECT

statement.

Function Definition

AVG(value) Average of value for group of rows

COUNT(value) COUNT of rows for column

GROUPING(expression) Used in conjunction with ROLLUP and CUBE functions to

detect NULLs

MAX(value) MAXimum of all values for group of rows

MIN(value) MINimum of all values for group of rows

STDDEV(value) STanDard DEViation of all values for group of rows

SUM(value) SUM of all values for group of rows

VARIANCE(value) VARIANCE of all values for group of rows

Example:

select avg(salary) from employees;

select min(salary) from employees;

select max(salary) from employees;

select sum(salary) from employees;

select count(name) from employees;

select count(*) from emp;

select count(empno) from emp;



Count(distinct attribute) – count all rows except null valued rows but eliminates

duplicates while counting.

select count(distinct empno) from emp;

Group By

GROUP BY is used in conjunction with aggregating functions to group the results by the

unaggregated columns.

Example:

select deptno,sum(sal) from emp group by deptno;

DEPTNO SUM(SAL)

----------- ---------------

1 11000

2 5000

select deptno,sum(sal) from emp group by deptno order by deptno desc;

DEPTNO SUM(SAL)

------------ ---------------

2 5000

1 11000

Group By With Having

HAVING is used to perform an action on groups created by GROUP BY similar to that of

the WHERE clause on rows in a basic SQL statement. The WHERE clause limits the

rows evaluated. The HAVING clause limits the grouped rows returned.

Example:

select deptno,sum(sal) from emp group by deptno having sum(sal) > 9000;

DEPTNO SUM(SAL)

---------- -----------

1 11000

SET OPERATORS

The four set operators union, union all, intersects and minus allows to serially combine

more than one select statements. Although more than one select statement will then

be present, only one result set is then returned.

Operators Use

union all selects all rows from all select statements

union union all is very similar to union, however, it dismisses duplicate

rows found across different select statements



intersect returns the rows that are found in all select statements

minus minus returns all rows from the first select statements except

those who are duplicated in a following select statement

Example:

Select ename, empno from emp Union all Select mname, eno from manager;

Select ename,empno from emp Union Select mname,eno from manager;

Select ename,empno from emp Intersect Select mname,eno from manager;

Select ename,empno from emp Minus Select mname,eno from manager;

JOIN OPERATION

An SQL JOIN clause combines records from two tables in a relational database,

resulting in a new, temporary table, sometimes called a "joined table".

Inner join

� Equi-join

� Natural join

� Cross join

Outer joins

� Left outer join

� Right outer join

� Full outer join

An inner join requires each record in the two joined tables to have a matching record.

An inner join essentially combines the records from two tables (A and B) based on a

given join-predicate.

Example of an explicit inner join

select * from employee inner join department on employee.departmentid =

department.departmentid

Example of an implicit inner join

select * from employee, department where employee.departmentid =




Types of inner joins

Equi-join

An equi-join is a specific type of comparator-based join, or theta join, that uses only

equality comparisons in the join-predicate. Using other comparison operators (such as

<) disqualifies a join as an equi-join.

The query shown above has already provided an example of an equi-join

select * from employee inner join department on employee.departmentid =


Natural join

A natural join offers a further specialization of equi-joins. The join predicate arises

implicitly by comparing all columns in both tables that have the same column-name in

the joined tables. The resulting joined table contains only one column for each pair of

equally-named columns.

The above sample query for inner joins can be expressed as a natural join in the

following way:

select * from employee natural join department

Cross join

A cross join, cartesian join or product provides the foundation upon which all types

of inner joins operate. A cross join returns the cartesian product of the sets of records

from the two joined tables. Thus, it equates to an inner join where the join-condition

always evaluates to True or join-condition is absent in statement.

If A and B are two sets, then cross join = A × B.

Example of an explicit cross join

select * from employee cross join department

Example of an implicit cross join:

select * from employee, department;

Outer joins

An outer join does not require each record in the two joined tables to have a matching

record. The joined table retains each record—even if no other matching record exists.



Outer joins subdivide further into left outer joins, right outer joins, and full outer joins,

depending on which table(s) one retains the rows from (left, right, or both).

Left outer join

The result of a left outer join for tables A and B always contains all records of the

"left" table (A), even if the join-condition does not find any matching record in the

"right" table (B). This means that if the ON clause matches 0 (zero) records in B, the

join will still return a row in the result—but with NULL in each column from B. This

means that a left outer join returns all the values from the left table, plus matched

values from the right table (or NULL in case of no matching join predicate).

Example:

select * from employee left outer join department on employee.departmentid =


Right outer join

A right outer join closely resembles a left outer join, except with the tables reversed.

Every record from the "right" table (B) will appear in the joined table at least once. If

no matching row from the "left" table (A) exists, NULL will appear in columns from A

for those records that have no match in A.

A right outer join returns all the values from the right table and matched values from

the left table (NULL in case of no matching join predicate).

Example:

select * from employee right outer join department on employee.departmentid =


Full outer join

A full outer join combines the results of both left and right outer joins. The joined

table will contain all records from both tables, and fill in NULLs for missing matches on

either side.

Example:

select * from employee full outer join department on employee.departmentid =




SUBQUERIES

Nesting of queries, one within another is termed as a subquery. Subqueries allow to

realize complex queries which would normally require several queries with storage of

the intermediate results.

Example: Determine the name, sex and age of the oldest student.

select nometu, cdsexe, ( current_date-dtnaiss)/365 as age from etudiant

where (current_date-dtnaiss) /365 = ( select max(( current_date-dtnaiss) /365) from

etudiant );

Predicates - IN, ANY, ALL, EXISTS

A subquery can return a subset of zero to n values. According to the conditions which

one wants to express, one can use the predicates IN, ANY, ALL or EXISTS.

Predicate Meaning

IN The comparison operatror is the equality and the logical

operation between values is OR.

ANY Allows to check if at least a value of the list satisfies condition.

ALL Allows to check if condition is realized for all the values of the list.

EXISTS If the subquery returns a result, the value returned is True otherwise the value returned is False.

Example: Display the marks of the student number 1 which are equal to the marks of

the student number 2.

select note from notes where numetu=1 and note in ( select note from notes where

numetu=2);

Example: Display the marks of the student number 1 which are greater than the

marks of the student number 2.

select note from notes where numetu=1 and note > any ( select note from notes

where numetu=2);

Example: Display the marks of the student number 1 which are lower than all the

marks of the student number 9.

select note from notes where numetu=1 and note < all ( select note from notes where

numetu=9);



Example: Display all information on the students who do not have any mark.

select * from etudiant e where not exists ( select numetu from notes where

numetu=e.numetu);

VIEWS

A view is a customized presentation of the data from one or more tables. Views derive

their data from tables on which they are based, which are known as base tables. All

operations performed on a view actually affect the base tables of the view.

Views can be used for several purposes:

� To give you an additional level of table security by restricting access to a

predetermined set of table rows and columns.

� To hide data complexity. Oracle 8 databases usually include many tables, and

by creating a view combining information from two or more tables,you make it

easier for other users to access information from your database.

� To present the data in a different perspective from that of the base table.Views

provide a means to rename columns without affecting the base table.

� To store complex queries.

Syntax:

create [or replace] view <view-name> [(<column(s)>)] as <select-statement> [with

check option [constraint <name>]][with read only];

Example:

create view DEPT20 (ENAME, JOB, ANNUAL SALARY) as select ENAME, JOB, SAL * 12

from EMP where DEPTNO = 20;

OUTPUT:

View created.

To list the view:

Select * from DEPT20;

ENAME JOB SALARY

------------ ------------- ----------

SMITH 111 2000

HAYES 123 7000

JOHNES 221 8000

TURNE 241 1000



Creating a read only view

To prevent modifications to the base tables via view, user can use with read only

clause of the create view command.

Create or replace view RAIN as select * from TROUBLE with read only;

Dropping a VIEW

Syntax:

drop view view_name;

example:

drop view rain;

INTEGRITY CONSTRAINTS IN CREATE TABLE COMMAND

A constraint is a rule that restricts the values for one or more columns in a table.

Constraints are of two types. Table constraints and Column constrains.

Some of the constraints are:

1. Primary key: It is the minimal combination of attribute in the table, which is used

to uniquely identify the rows of the table.

2. Foreign key: This is a key to join a table to another table. Foreign in a table is an

attribute, which is a primary key of another table. If an attribute A1 in table R1 is a

primary key in another table R2, the A1 is the Foreign key in R1 and A1 refers R2.

3. Unique: If some attribute (column) combination is said to be unique, then their

combination cannot have duplicate values in the table.

4. Not Null: If a column is assigned to be not null then it cannot have null values.

5. Check: This constraint is used to specify some condition for the columns.

6. Default: This constraint is used to specify a default value for a specific column.

Examples:

primary key

create table emp(empno number(3 ) primary key, ename varchar2(20));

create table emp(empno number(3),ename varchar2(20), primary key(empno));



NOT NULL

create table emp(empno number(3), ename vrchar2(15) not null);

unique

create table emp(empno number(3) unique,ename varchar(20))

create table emp (empno number(3),ename varchar(20),unique(empno)));

check

create table emp(empno number(3),sal number(8) check(sal>0), ename varchar(20));

create table emp5(empno number(3),sal number(8), ename

varchar(20),check(sal>0));

foreign key

cretate table maneger table(mname varchar(15), eno int references emp(empno),mid

varchar(10)primary key);

default

create table worker(wname varchar(15) default (‘not entered’), dob date,salary

numeric(6,2) default(3000));

USAGE OF CONSTRIANT keyword for implementing integrity constraints in a

CREATE TABLE command

Syntax:

Create table<table name>(<attribute-1 domain type-1,….attribute-n domain type-n,

CONSTRAINT<constaintname1>,constraints>,….CONSTRAINT<constraintname2><con

straints>, …CONSTRAINT<constaintnamren><constraints>);

Example:

Create table borrower(loanno varchar(10),amount number(10), CONSTRAINT chk1

check(amount>=0), CONSTRIAINT conpri primary key(loanno));

Note: Here the name of constraint is chk1

To remove the constraint chk1 command is:- alter table borrower drop constraint

chk1;



PL/SQL

(PROCEDURAL LANGUAGE/ STRUCTURED QUERY LANGUAGE)



PL/SQL stands for Procedural Language/SQL. PL/SQL is super set of the Structured

Query Language. Using PL/SQL user can do things like codify business rules through

the creation of stored procedures and packages, trigger database events to occur, or

add programming logic to the execution of SQL commands.

PL/SQL code is grouped into structures called blocks.

Anonymous block - If the block of PL/SQL code is not given a name then it is called

anonymous block.

Named block - block with a code with a name. Procedures and Functions are

examples.

Features of PL/SQL

� Block(modular) structures

� Flow-control statements and loops

� Variables, constants and types

� Structured data

� Customized error handling

� Allows to store compiled code directly in the database

PL/SQL datatypes include all of the valid SQL datatypes as well as complex datatypes

based on query structures.

A block of PL/SQL contains three sections (for anonymous PL/SQL block)

Section Description

Declarations Defines and initializes the variable and cursors used in the

block

Executable commands Uses flow-control commands (such as if commands and

loops) to execute the commands and assign values to the

declared variables

Exception Handling Provides customized handling of error conditions

Structure of a PL/SQL block

declare

<declarations section>

begin

<executable commands>

exception

<exception handling>

end;



Declarations Section

The Declarations section begins a PL/SQL block. The Declarations section starts with

the declare keyword, followed by a list of variable and cursor definitions.

User can define variables to have constant values, and variables can inherit datatypes

from existing columns and query results.

Example:

declare

pi constant NUMBER(9,7) := 3.1415926;

radius INTEGER(5);

area NUMBER(14,2);

begin

<executable commands>

end;

/

%TYPE and %ROWTYPE

Both %TYPE and %ROWTYPE are used to define variables in PL/SQL as it is defined

within the database. If the datatype or precision of a column changes, the program

automically picks up the new definition from the database without having to make any

code changes. The %TYPE and %ROWTYPE constructs allows programs to adapt as the

database changes to meet new business needs.

%TYPE

%TYPE is used to declare a field with the same type as that of a specified table's

column.

Example:

DECLARE

v_EmpName emp.ename%TYPE;

BEGIN

SELECT ename INTO v_EmpName FROM emp WHERE ROWNUM = 1;

DBMS_OUTPUT.PUT_LINE('Name = ' || v_EmpName);

END;

/

%ROWTYPE

%ROWTYPE is used to declare a record with the same types as found in the specified

database table, view or cursor. Examples:



DECLARE

v_emp emp%ROWTYPE;

BEGIN

v_emp.empno := 10;

v_emp.ename := 'JOY';

END;

/

Executable Commands Section

In the Executable Commands section, user can manipulate the variables and cursors

declared in the Declarations section of your PL/SQL block. The Executable Commands

section always starts with the keyword begin.

Example:

declare


radius INTEGER(5);

area NUMBER(14,2);

begin

radius := 3;

area := pi*power(radius,2);

insert into AREAS values (radius, area);

end;

/

Exception Handling Section

When user-defined or system-related exceptions (errors) are encountered, the control

of the PL/SQL block shifts to the Exception Handling section. Within the Exception

Handling section, the when clause is used to evaluate which exception is to be

“raised”—that is, executed.

If an exception is raised within the Executable Commands section of your PL/SQL

block, the flow of commands immediately leaves the Executable Commands section

and searches the Exception Handling section for an exception matching the error

encountered. PL/SQL provides a set of system-defined exceptions and allows you to

add your own exceptions.

Every exception in PL/SQL has an error number and error message; some exceptions

also have names.

Declaring Exceptions

DECLARE

exception_name EXCEPTION;



User can associate an error number with a declared exception with the PRAGMA

EXCEPTION_INIT statement:

DECLARE

exception_name EXCEPTION;

PRAGMA EXCEPTION_INIT (exception_name, error_number);

where error_number is a literal value (variable references are not allowed). This

number can be an Oracle error, such as -1855, or an error in the user-definable -

20000 to -20999 range.

Raising Exceptions

An exception can be raised in three ways:

� By the PL/SQL runtime engine

� By an explicit RAISE statement in your code

� By a call to the built-in function RAISE_APPLICATION_ERROR

The syntax for the RAISE statement is:

RAISE exception_name;

Example:

Raise_application_error

It is used to create user defined error messages.

Syntax:

raise_application_error( error number, error message);

where , error number – between –20,000 and –20,999.

Error message – text associated with this error.

Example:

exception lo_bal then

raise_application_error(-20001, ‘ balance is low’);

--

end;

Conditional Logic

Within PL/SQL, user can use if, else, and elsif commands to control the flow of

commands within the Executable Commands section

Syntax:

if <some condition>

then <some command>

elsif <some condition>

then <some command>



else <some command>

end if;

Example:

if area >30

then

insert into AREAS values (rad_val.radius, area);

end if;

Loops

User can use loops to process multiple records within a single PL/SQL block. PL/SQL

supports three types of loops

Simple loops

A loop that keeps repeating until an exit or exit when statement is reached within the

loop. The loop is started by the loop keyword, and the exit when clause determines

when the loop should be exited. An end loop clause signals the end of the loop.

Example:

declare


radius INTEGER(5);

area NUMBER(14,2);

begin

radius := 3;

loop



radius := radius+1;

exit when area >100;

end loop;

end;

/

FOR Loops

A loop that repeats a specified number of times. The FOR loop’s start is indicated by

the keyword for, followed by the criteria used to determine when the processing

should exit the loop. Since the number of times the loop is executed is set when the

loop is begun, an exit command isn’t needed within the loop.

Example:

declare


radius INTEGER(5);

area NUMBER(14,2);

begin

for radius in 1..7 loop





end loop;

end;

/

WHILE Loops

In a WHILE loop, the loop is processed until an exit condition is met. Instead of

specifying the exit condition via an exit command within the loop, the exit condition is

specified in the while command that initiates the loop.

Example:

declare


radius INTEGER(5);

area NUMBER(14,2);

begin

radius := 3;

while radius<=7

loop



radius := radius+1;

end loop;

end;

/

Cursor

A cursor is handle or pointer to the context area. The PL/SQL program can control the

context area by using the cursor. There can be either static cursors, whose SQL

statement is determined at compile time, or dynamic cursors, whose SQL statement is

determined at runtime.

Types of Cursors

PL/SQL uses two types of cursors: explicit and implicit

Explicit Cursors

Explicit cursors are SELECT statements that are DECLAREd explicitly in the declaration

section of the current block or in a package specification. Use OPEN, FETCH, and

CLOSE in the execution or exception sections of your programs.

Declaring explicit cursors

To use an explicit cursor, first declare it in the declaration section of a block or

package. There are three types of explicit cursor declarations:

A cursor without parameters



cursor company_cur is select company_id from company;

A cursor that accepts arguments through a parameter list

cursor company_cur (id_in in number) is select name from company where

company_id = id_in;

A cursor header that contains a RETURN clause in place of the SELECT statement:

cursor company_cur (id_in in number) return company%rowtype is select * from

company;

Opening explicit cursors

To open a cursor, use the following syntax:

OPEN cursor_name [(argument [,argument ...])];

where cursor_name is the name of the cursor as declared in the declaration section.

The arguments are required if the definition of the cursor contains a parameter list.

DECLARE

CURSOR c1 IS SELECT ename, job FROM employee WHERE sal < 7000;

...

BEGIN

OPEN c1;

...

END;

Fetching from explicit cursors

The FETCH statement places the contents of the current row into local variables. To

retrieve all rows in a result set, each row needs to be fetched.

The syntax for a FETCH statement is:

FETCH cursor_name INTO record_or_variable_list;

where cursor_name is the name of the cursor as declared and opened. Closing explicit

cursors

The syntax of the CLOSE statement is:

CLOSE cursor_name;

where cursor_name is the name of the cursor declared and opened.



Explicit cursor attributes

Attribute Description

%ISOPEN

TRUE if cursor is open. FALSE if cursor is not open.

%FOUND

Returns true if the last fetch returned a row else a false

%NOTFOUND

Returns true if the last fetch did not return a row else a false

%ROWCOUNT

The number of rows fetched from the cursor.

INVALID_CURSOR if cursor has been CLOSEd.

Example:

declare


area NUMBER(14,2);

cursor rad_cursor is

select * from RADIUS_VALS;

rad_val rad_cursor%ROWTYPE;

begin

open rad_cursor;

loop

fetch rad_cursor into rad_val;

exit when rad_cursor%NOTFOUND;

area := pi*power(rad_val.radius,2);


end loop;

close rad_cursor;

end;

/

Implicit Cursors

Whenever a SQL statement is directly in the execution or exception section of a PL/SQL

block, you are working with implicit cursors. These statements include INSERT,

UPDATE, DELETE, and SELECT INTO statements. Unlike explicit cursors, implicit

cursors do not need to be declared, OPENed, FETCHed, or CLOSEd.

SELECT statements handle the %FOUND and %NOTFOUND attributes differently from

explicit cursors. When an implicit SELECT statement does not return any rows, PL/SQL

immediately raises the NO_DATA_FOUND exception and control passes to the

exception section. When an implicit SELECT returns more than one row, PL/SQL



immediately raises the TOO_MANY_ROWS exception and control passes to the

exception section.

implicit cursor attributes

Attribute Description

%ISOPEN

Always false as it is closed immediately after executing its

associated SQL statement

%FOUND

Returns true if the last DML statement returned a row else

a false

%NOTFOUND

Returns true if the last DML statement did not return any

row else a false

%ROWCOUNT

Returns the total number of rows returned.

cursor for loop

In a Cursor FOR loop, the results of a query are used to dynamically determine the

number of times the loop is executed. In a Cursor FOR loop, the opening, fetching, and

closing of cursors is performed implicitly.

Example:

declare


area NUMBER(14,2);

cursor rad_cursor is

select * from RADIUS_VALS;

begin

for rad_val in rad_cursor

loop

area := pi*power(rad_val.radius,2);


end loop;

end;

/



Ref Cursors

To create cursor variables, you take two steps. First, you define a REF CURSOR type,

then declare cursor variables of that type. You can define REF CURSOR types in any

PL/SQL block, subprogram, or package using the syntax

TYPE ref_type_name IS REF CURSOR RETURN return_type;

Declaring Cursor Variables

Once you define a REF CURSOR type, you can declare cursor variables of that type in

any PL/SQL block or subprogram. In the following example, you declare the cursor

variable dept_cv:

DECLARE

TYPE DeptCurTyp IS REF CURSOR RETURN dept%ROWTYPE;

dept_cv DeptCurTyp; -- declare cursor variable

Opening a Cursor Variable

The OPEN-FOR statement associates a cursor variable with a multi-row query,

executes the query, and identifies the result set. Here is the syntax:

OPEN {cursor_variable_name | :host_cursor_variable_name}

FOR select_statement;

where host_cursor_variable_name identifies a cursor variable declared in a PL/SQL

host environment such as an OCI or Pro*C program.

Closing a Cursor Variable

The CLOSE statement disables a cursor variable. After that, the associated result set is

undefined. Here is the syntax:

CLOSE {cursor_variable_name | :host_cursor_variable_name);

Varying Arrays

A varying array allows you to store repeating attributes of a record in a single row.

Creating a Varying Array

SQL> create or replace type mark as varray(5) of number;

/

Type created.

SQL> create table varr(no number,marks mark);

Table created.

SQL> desc varr;



Name Null? Type

----------------------------------------- -------- -----------------

NO NUMBER

MARKS MARK

SQL> select * from varr;

NO

----------

MARKS

-------------------------

101

MARK(33, 44, 55, 66)

Example:

declare

cursor borrower_cursor is

select * from varr;

begin

for borrower_rec in borrower_cursor

loop

for i in 1..borrower_rec.marks.count

loop

dbms_output.put_line(borrower_rec.marks(i));

end loop;

end loop;

end;

/

33

44

55

66

PL/SQL procedure successfully completed.



Named Program Units

The PL/SQL programming language allows you to create a variety of named program

units they include:

Procedure - A program that executes one or more statements

Function - A program that returns a value

Package - A container for procedures, functions, and data structures

Triggers - Programs that execute in response to database changes

PROCEDURES

Procedures are program units that execute one or more statements and can receive or

return zero or more values through their parameter lists.

The syntax of a procedure is:

CREATE [OR REPLACE] PROCEDURE procedure_name

[ (parameter [,parameter]) ]

IS

[declaration_section]

BEGIN

executable_section

[EXCEPTION

exception_section]

END [procedure_name];

Example:

create procedure NEW_WORKER (Person_Name IN varchar2)

AS

BEGIN

insert into WORKER

(Name, Age, Lodging)

values

(Person_Name, null, null);

END;

/

In the above example the NEW_WORKER procedure will accept a person’s name as its

input. It can be called from any application. It inserts a record into the WORKER table,

with NULL values for the Age and Lodging columns.

Parameter modes

When you create a procedure or function, you may define parameters. There are three

types of parameters that can be declared:



IN The parameter can be referenced by the procedure or function. The

value of the parameter can not be overwritten by the procedure or

function.

OUT The parameter can not be referenced by the procedure or function, but

the value of the parameter can be overwritten by the procedure or

function.

IN OUT The parameter can be referenced by the procedure or function and the

value of the parameter can be overwritten by the procedure or function.

Calling a Procedure

A call to the procedure is made through an executable PL/SQL statement. The

procedure has the following syntax:

Procedure_Name [(parameters)];

The procedure can be executed from SQL environment with the execute command as

follows:

EXECUTE Procedure_Name [ (parameters) ];

Dropping procedure

To drop a procedure, use the drop procedure command, as follows:

drop procedure NEW_WORKER;

FUNCTIONS

Unlike procedures, functions can return a value to the caller (procedures cannot return

values). This value is returned through the use of the return keyword within the

function.

A function is characterized as follows:

• A function can be with one, more or no parameters.

• A function must have an explicit RETURN statement in the executable section to

return a value.

• The data type of the return value must be declared in the function’s header.

• A function cannot be executed as a standalone program.

The syntax for a function is:

CREATE [OR REPLACE] FUNCTION function_name

[ (parameter [,parameter]) ]

RETURN return_datatype



IS | AS

[declaration_section]

BEGIN

executable_section

[EXCEPTION

exception_section]

END [function_name];

Example:

create function BALANCE_CHECK (Person_Name IN varchar2)

RETURN NUMBER

IS

balance NUMBER(10,2);

BEGIN

select SUM(DECODE(Action,'BOUGHT',Amount,0))

- SUM(DECODE(Action,'SOLD',Amount,0))

INTO balance

from LEDGER

where Person = Person_Name;

RETURN(balance);

END;

/

Dropping function

To drop a function, use the drop function command, as follows:

drop function BALANCE_CHECK;

PACKAGES

A package is a database object, which is an encapsulation of related PL/SQL types,

subprograms, cursors, exceptions, variables and constants.

Packages have

package specification: PL/SQL types, subprograms, cursors, exceptions, variables

and constants are declared here.

package body: implements cursors, subprograms defined in package specification.

Note:

packages can not be called

we can not pass parameters to packages

nesting of packages is not possible

Create package Syntax

When creating packages, the package specification and the package body are created

separately. Thus, there are two commands to use: create package for the package

specification, and create package body for the package body.



Syntax for creating package specifications:

create [or replace] package [user.] package

{IS | AS}

[AUTHID {DEFINER | CURRENT_USER} ]

package specification;

A package specification consists of the list of functions, procedures, variables,

constants, cursors, and exceptions that will be available to users of the package.

creating package bodies is as follows:

create [or replace] package body [user.] package body

{IS | AS}

package body;

The name of the package body should be the same as the name of the package

specification.

Example:

Package specification:

create or replace package LEDGER_PACKAGE

AS

function BALANCE_CHECK(Person_Name VARCHAR2) return NUMBER;

procedure NEW_WORKER(Person_Name IN VARCHAR2);

end LEDGER_PACKAGE;

/

Package body:

create package body LEDGER_PACKAGE

AS

function BALANCE_CHECK (Person_Name IN varchar2)

RETURN NUMBER

IS

balance NUMBER(10,2);

BEGIN

select SUM(DECODE(Action,'BOUGHT',Amount,0)) -

SUM(DECODE(Action,'SOLD',Amount,0))

INTO balance from LEDGER where Person = Person_Name;

RETURN(balance);

EXCEPTION

WHEN NO_DATA_FOUND THEN

RAISE_APPLICATION_ERROR (-20100, 'No BOUGHT or SOLD entries for that Person.');

END BALANCE_CHECK;

procedure NEW_WORKER (Person_Name IN varchar2)

AS

BEGIN

insert into WORKER (Name, Age, Lodging) values (Person_Name, null, null);

END NEW_WORKER;

END LEDGER_PACKAGE;

/



Executing the function

execute LEDGER_PACKAGE.BALANCE_CHECK('ADAH TALBOT');

Dropping Packages

To drop a package, use the drop package command, as follows:

drop package LEDGER_PACKAGE;

TRIGGERS

Trigger defines an action the database should take when some database-related event

occurs. Triggers are executed by the database when specific types of data

manipulation commands are performed on specific tables. Such commands may

include inserts, updates, and deletes. Updates of specific columns may also be used

as triggering events.

Types of Triggers

A trigger’s type is defined by the type of triggering transaction and by the level at

which the trigger is executed.

Row-Level Triggers

Row-level triggers execute once for each row in a transaction.

Row-level triggers are created using the for each row clause in the create trigger

command.

Statement-Level Triggers

Statement-level triggers execute once for each transaction. Statement-level triggers

are the default type of trigger created via the create trigger command.

The general syntax of the trigger is:

Create [or replace] trigger Trigger_name

[Before / after] [Insert/update/delete]

On Table_name

[For each row]

[When condition]

Declare

Declaration statements

Begin

Executable statements

Exception

Exception handling statements

End;



The before and after keywords indicate whether the trigger should be executed

before or after the triggering transaction.

The delete, insert, and update keywords (the last of which may include a column

list) indicate the type of data manipulation that will constitute a triggering event. When

referring to the old and new values of columns, you can use the defaults (“old” and

“new”) or you can use the referencing clause to specify other names.

When the for each row clause is used, the trigger will be a row-level trigger;

otherwise, it will be a statement-level trigger.

The when clause is used to further restrict when the trigger is executed. The

restrictions enforced in the when clause may include checks of old and new data

values.

Example:

create trigger ledger_bef_upd_row

before update on LEDGER

for each row

when (new.Amount/old.Amount>1.1)

begin

insert into LEDGER_AUDIT values (:old.ActionDate, :old.Action, :old.Item,

:old.Quantity, :old.QuantityType, :old.Rate, :old.Amount, :old.Person);

end;

Enabling and Disabling Triggers

A trigger can be either enabled or disabled. By default, triggers are enabled To enable

a trigger, use the alter trigger command with the enable keyword

Syntax for disabling a trigger:

alter trigger <Trigger_name> disable;

Syntax for disabling all triggers:

alter table <Table_name> disable all triggers;

Syntax for enabling a disabled trigger:

alter table <Table_name> enable <Trigger_name>;

Syntax for enabling all the disabled triggers:

alter table <Table_name> enable all Triggers;

Dropping Triggers

Triggers may be dropped via the drop trigger command.



Syntax for dropping triggers:

Drop Trigger <Trigger_name>;

TABLESPACE

A tablespace is an area of disk consisting of one or more disk files. A tablespace can

contain many tables, indexes, or clusters. Because a tablespace has a fixed size, it can

get full as rows are added to its tables.

Creating tablespace

The create tablespace command allows one or more files to be assigned immediately to

the tablespace. It also specifies a default space for any tables created without an

explicit storage clause mentioned in the create table statement.

SQL> create tablespace TALBOT datafile '/oraclexe/oradata/talbot.dbf' size 5000k

default storage (initial 1M next 1M minextents 1 maxextents 100 pctincrease 0)

permanent;

Tablespace created.

The permanent keyword in the create tablespace command tells Oracle that you

will be storing permanent objects (such as tables) in the tablespace. If the tablespace

is only used for temporary segments, then you can specify the temporary keyword

instead. The default value is permanent.

The initial default extent is 1M bytes (not blocks) and the next (incremental) extent is

1M bytes.

minextents allows you to set aside additional extents beyond the first at the time a

table is created. These additional extents will not necessarily be contiguous (physically

adjacent) with the initial extent, or with each other, but the space will at least be

reserved.

maxextents is the limit of additional extents allowed. You can specify

maxextents unlimited, in which case there is no limit to the number of extents

allowed for the table or index.

pctincrease is a growth factor for extents. When set to a non-zero value, each

incremental extent will be the specified percentage larger than the one before it. This



has the effect of reducing the number of extents, and noncontiguous space, used by a

table that grows large. However, it causes the space allocated to the table to grow

exponentially. If the data volume in the table grows at a constant rate, you should set

pctincrease to 0.

SQL> create table emp(name varchar2(10),age number) tablespace talbot;

Table created.



VISUAL BASIC is a high level programming language which was evolved from the

earlier DOS version called BASIC. VISUAL BASIC is a VISUAL and events driven

Programming Language. In VISUAL BASIC, programming is done in a graphical

environment.

a VISUAL BASIC Program is made up of many subprograms, each has its own program

codes, and each can be executed independently and at the same time each can be

linked together in one way or another.

The Visual Basic Environment

The above diagram shows the development environment with all the important points

labeled. You can choose to start a new project, open an existing project or select a list

of recently opened programs.

A project is a collection of files that make up your application. There are various types

of applications we could create.

The Project explorer window gives you a tree-structured view of all the files inserted

into the application. The project explorer window displays forms, modules or other

separators which are supported by the visual basic like class'es and Advanced Modules.



Properties Window

By default, the properties window displays its properties alphabetically (with the

exception of the name value) when you click on the categorized button the window

changes to left picture.

The tool box

Example:

Vertical Scroll bar Horizontal Scroll bar



Data source name

A data source name (DSN) is a data structure that contains the information about a

specific database that an Open Database Connectivity (ODBC) driver needs in order to

connect to it. Included in the DSN, which resides either in the registry or as a separate

text file, is information such as the name, directory and driver of the database, and,

depending on the type of DSN, the ID and password of the user.

User and system DSNs are specific to a particular computer, and store DSN

information in the registry. A user DSN allows database access for a single user on a

single computer, and a system DSN for any user of a particular computer. A file DSN

contains the relevant information within a text file with a .DSN file extension, and can

be shared by users of different computers who have the same drivers installed.

Creating a dsn

Control panel -> Administrative tools -> Data Sources (ODBC)

Click add

Choose the driver ‘Oracle in XE’ (for Oracle 10g)



Give the data source name and click ok, as show in the above picture dsn ‘rvs’ will be

created.

Open DataBase Connectivity

Open DataBase Connectivity, a standard database access method developed by the

SQL Access group in 1992. The goal of ODBC is to make it possible to access any data

from any application, regardless of which database management system (DBMS) is

handling the data. ODBC manages this by inserting a middle layer, called a database

driver, between an application and the DBMS. The purpose of this layer is to translate

the application's data queries into commands that the DBMS understands.



SQL

SOLVED PROBLEMS



Exercise no 1

1. Write an SQL statement to create an employee table with following details:

a. empno number(4)

b. empname varchar2(30)

c. doj date

d. manager number(4)

e. salary number(7,2)

f. commission number(5,2)

g. deptnumber number(2)

2. Write an SQL statement to create a department with following details.

a. deptnumber number(2)

b. deptname varchar2(20)

c. location varchar2(20)

3. Write an SQL statement to insert records into employee table.

4. Insert two rows into employee table for all columns except manager and

commission.

1. Write an SQL statement to insert values into the department table dynamically

at runtime.

2. Write an SQL statement to copy all the values from already existing employee

table to new table.

3. Write an SQL statement to insert two records into the employee table by giving

null values for doj and deptnumber column.

4. Write an SQL statement to insert some records with doj varying 1st jan 1980 to

31st dec 2010

5. Write an SQL statement to insert 5 records with salary and commission column

Varying from 1500 to 15000 and 150 to 3000 respectively

6. Write an SQL statement to alter the employee table by adding a column ‘sex’

with datatype char(1).



Oueries and Outputs:

Ans Ans Ans Ans 1111 SQL> create table emp (

empno number(4), ename varchar2(30), doj date, mgrssn number(4), sal number(7,2), comm number(5,2), dno number(2) );

Table created. Ans Ans Ans Ans 2222 SQL> create table dept ( dno number(2), dname varchar2(15), loc varchar2(10) ); Table created. Ans Ans Ans Ans 3333 SQL> insert into emp values(1101,'Smith','1-jan-84','1120',14000,200,20); 1 row created. SQL> insert into emp values(1102,'john','10-march-95','1121',8000,100,15); 1 row created. SQL> insert into emp values(1120,'william','20-may-2005',1122,15000,500,10); 1 row created. Ans Ans Ans Ans 4444 SQL> insert into emp values('1122','james','20-july-1985',null,20000,null,10); 1 row created. SQL> insert into emp values('1121','johnson','20-july-1990',null,20000,null,20); 1 row created. Ans Ans Ans Ans 5555 SQL> insert into dept values(&dno,'&dname','&loc'); Enter value for dno: 20 Enter value for dname: research Enter value for loc: california



old 1: insert into dept values(&dno,'&dname','&loc'); new 1: insert into dept values(20,'research','california'); 1 row created. SQL> / Enter value for dno: 10 Enter value for dname: sales Enter value for loc: washington old 1: insert into dept values(&dno,'&dname','&loc'); new 1: insert into dept values(10,'sales','washington'); 1 row created. SQL> / Enter value for dno: 15 Enter value for dname: marketing Enter value for loc: newyork old 1: insert into dept values(&dno,'&dname','&loc'); new 1: insert into dept values(15,'marketing','newyork'); 1 row created. AnsAnsAnsAns 6666 SQL> create table newemp as (select * from emp); Table created. AnsAnsAnsAns 7777 SQL> insert into emp values(1126,'james',null,1121,10000,95,null); 1 row created. SQL> insert into emp values(1127,'zam dem',null,1122,10000,85,null); 1 row created. Ans 8Ans 8Ans 8Ans 8 SQL> insert into emp values(1129,'mone jo','20-april-85',1122,15000,500,10); 1 row created. SQL> insert into emp values(1130,'xelestina','20-april-95',1121,16000,200,15); 1 row created. Ans 9Ans 9Ans 9Ans 9 SQL> insert into emp values(&empno,'&ename','&date','&mgrssn',&sal,&comm,&dno); Enter value for empno: 1140 Enter value for ename: satheesh Enter value for date: 12-june-1999 Enter value for mgrssn: 1120 Enter value for sal: 2500 Enter value for comm: 150 Enter value for dno: 20 old 1: insert into emp values(&empno,'&ename','&date','&mgrssn',&sal,&comm,&dn o) new 1: insert into emp values(1140,'satheesh','12-june-1999' ,'1120',2500,150,20) 1 row created.



SQL> / Enter value for empno: 1141 Enter value for ename: santhosh Enter value for date: 12-march-2002 Enter value for mgrssn: 1120 Enter value for sal: 6000 Enter value for comm: 160 Enter value for dno: 20 old 1: insert into emp values(&empno,'&ename','&date','&mgrssn',&sal,&comm,&dn o) new 1: insert into emp values(1141,'santhosh','12-march-2002' ,'1120',6000,160,20) 1 row created. SQL> / Enter value for empno: 1142 Enter value for ename: jonathen Enter value for date: 12-april-1998 Enter value for mgrssn: 1120 Enter value for sal: 8000 Enter value for comm: 190 Enter value for dno: 20 old 1: insert into emp values(&empno,'&ename','&date','&mgrssn',&sal,&comm,&dn o) new 1: insert into emp values(1142,'jonathen','12-april-1998','1120',8000,190,20) 1 row created. SQL> / Enter value for empno: 1143 Enter value for ename: sam Enter value for date: 18-march-2007 Enter value for mgrssn: 1120 Enter value for sal: 8000 Enter value for comm: 200 Enter value for dno: 20 old 1: insert into emp values(&empno,'&ename','&date','&mgrssn',&sal,&comm,&dn o) new 1: insert into emp values(1143,'sam','18-march-2007', '1120',8000,200,20) 1 row created. SQL> / Enter value for empno: 1144 Enter value for ename: husain Enter value for date: 17-dec-2005 Enter value for mgrssn: 1120 Enter value for sal: 9000 Enter value for comm: 300 Enter value for dno: 20 old 1: insert into emp values(&empno,'&ename','&date','&mgrssn',&sal,&comm,&dn o) new 1: insert into emp values(1144,'husain','17-dec-2005', '1120',9000,300,20) 1 row created.



Ans 10Ans 10Ans 10Ans 10 SQL> alter table emp add sex char(1); Table altered. SQL> select * from emp; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------- --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 14 rows selected.



Exercise no 2

1. Write an SQL statement to display all the records from an employee table

2. Write an SQL statement to display all the records from department table

3. Write an SQL statement to display empno,empname,salary,deptno from

employee table

4. Write an SQL statement to display deptno, deptname from department

table.

5. Write an SQL statement display the empname who are getting salary more

than 5000

6. Write an SQL statement to display the employees who are getting salary=0

7. Write an SQL statement to display the employee who are getting null

commission

8. Write an SQL statement to display employee whose name starts with ‘s’

9. Write an SQL statement to display employee details who are getting salary

between 5000 & 15000

10. Write an SQL statement to display employee who join organization

between 01-jan-1980 and 31-mar-2006.

11. Write an SQL statement to display employees who are getting commission

less than 1000 only.

12. Write an SQL statement to display the employee working in department

number 12 & 30.

13. Write an SQL statement to display employees working in all department

except deptno 20 and 30

14. Write an SQL statement to display employees who are getting salaries not

equal to 1000

15. Write an SQL statement to display the employee details who are getting

salary more than 1000(1000 inclusive).



Oueries and Outputs: Ans 1Ans 1Ans 1Ans 1 SQL> column ename format a12; SQL> select * from emp; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 14 rows selected. Ans 2Ans 2Ans 2Ans 2 SQL> select * from dept; DNO DNAME LOC ---------- --------------- ---------- 20 research california 10 sales washington 15 marketing newyork Ans 3Ans 3Ans 3Ans 3 SQL> select empno,ename,sal,dno from emp; EMPNO ENAME SAL DNO ---------- ------------ ---------- ---------- 1101 Smith 14000 20 1102 john 8000 15



1120 william 15000 10 1122 james 20000 10 1121 johnson 20000 20 1126 james 10000 1127 zam dem 10000 1129 mone jo 15000 10 1130 xelestina 16000 15 1140 satheesh 2500 20 1141 santhosh 6000 20 EMPNO ENAME SAL DNO ---------- ------------ ---------- ---------- 1142 jonathen 8000 20 1143 sam 8000 20 1144 husain 9000 20 14 rows selected. Ans 4Ans 4Ans 4Ans 4 SQL> select dno,dname from dept; DNO DNAME ---------- --------------- 20 research 10 sales 15 marketing Ans 5Ans 5Ans 5Ans 5 SQL> select ename from emp where sal>5000; ENAME ------------ Smith john william james johnson james zam dem mone jo xelestina santhosh jonathen ENAME ------------ sam husain 13 rows selected. AnsAnsAnsAns 6666 SQL> select * from emp where sal=0; no rows selected AnsAnsAnsAns 7777 SQL> select * from emp where comm is null;



EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 Ans Ans Ans Ans 8888 SQL> select * from emp where ename like 's%'; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 1143 sam 18-MAR-07 1120 8000 200 20 Ans Ans Ans Ans 9999 SQL> select * from emp where sal between 5000 and 15000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1141 santhosh 12-MAR-02 1120 6000 160 20 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 10 rows selected. Ans: 10Ans: 10Ans: 10Ans: 10 SQL> select ename from emp where doj between '01-jan-1980' and '31-march-2006'; ENAME ------------ Smith john william



james johnson mone jo xelestina satheesh santhosh jonathen husain 11 rows selected. Ans:Ans:Ans:Ans: 11111111 SQL> select * from emp where nvl(comm,0)<1000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 14 rows selected. Ans:Ans:Ans:Ans: 12121212 SQL> select * from emp where dno in(12,30); no rows selected Ans:Ans:Ans:Ans: 13131313 SQL> select * from emp where dno not in(20,30); EMPNO ENAME DOJ MGRSSN SAL COMM DNO S



---------- ------------ --------- ---------- ---------- ---------- ---------- - 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 Ans :14Ans :14Ans :14Ans :14 SQL> select * from emp where sal!=1000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 14 rows selected.



Ans :15Ans :15Ans :15Ans :15 SQL> select * from emp where sal>=1000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 14 rows selected.



Exercise no 3

1. Write an SQL statement to display the employee details who are working in

deptno 10 and 30, who are getting salary more than 3000

2. Write an SQL statement to display the employee details who are working in

deptno 10 and 30 or who are drawing salary more than 3000

3. Write an SQL statement to display the employee details who joined before 01-

jan-2000 and salary <5000

4. Write an SQL statement to display the employee whose names sounds like

‘zam’

5. Write an SQL statement to sort the employee details in an alphabetical order

name wise

6. Write an SQL statement to sort the employee name department wise and each

department alphabetical wise

7. Write an SQL statement to give the employee details in descending order of

employee names.

8. Write an SQL statement to give the employee details in ascending order of

department number wise and in each department descending order of salary

wise

9. Write an SQL statement to give the employee details and arrange them inorder

based on position.

10. Write an SQL statement to display empno, name, salary and deptno and display

the records in the ascending order based on position.



AnsAnsAnsAns:::: 1 1 1 1 SQL> select * from emp where dno in(10,30) and sal>3000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1129 mone jo 20-APR-85 1122 15000 500 10 AnsAnsAnsAns:::: 2 2 2 2 SQL> select * from emp where dno in(10,30) or sal>3000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1102 john 10-MAR-95 1121 8000 100 15 1120 william 20-MAY-05 1122 15000 500 10 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1141 santhosh 12-MAR-02 1120 6000 160 20 1142 jonathen 12-APR-98 1120 8000 190 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1143 sam 18-MAR-07 1120 8000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 13 rows selected. AnsAnsAnsAns:::: 3 3 3 3 SQL> select * from emp where doj < '1-jan-2000' and sal < 5000; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1140 satheesh 12-JUN-99 1120 2500 150 20



AnsAnsAnsAns:::: 4 4 4 4 SQL> select * from emp where soundex(ename)=soundex('zam'); no rows selected SQL> select * from emp where soundex(ename)=soundex('sam'); EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1143 sam 18-MAR-07 1120 8000 200 20 AnsAnsAnsAns:::: 5 5 5 5 SQL> select * from emp order by ename asc; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1101 Smith 01-JAN-84 1120 14000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 1122 james 20-JUL-85 20000 10 1126 james 1121 10000 95 1102 john 10-MAR-95 1121 8000 100 15 1121 johnson 20-JUL-90 20000 20 1142 jonathen 12-APR-98 1120 8000 190 20 1129 mone jo 20-APR-85 1122 15000 500 10 1143 sam 18-MAR-07 1120 8000 200 20 1141 santhosh 12-MAR-02 1120 6000 160 20 1140 satheesh 12-JUN-99 1120 2500 150 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1120 william 20-MAY-05 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1127 zam dem 1122 10000 85 14 rows selected. AnsAnsAnsAns:::: 6 6 6 6 SQL> select ename,dno from emp order by dno asc,ename asc; ENAME DNO ------------------------------ ---------- james 10 mone jo 10 william 10



john 15 xelestina 15 Smith 20 husain 20 johnson 20 jonathen 20 sam 20 santhosh 20 ENAME DNO ------------------------------ ---------- satheesh 20 james zam dem 14 rows selected. AnsAnsAnsAns:::: 7 7 7 7 SQL> select * from emp order by ename desc; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1127 zam dem 1122 10000 85 1130 xelestina 20-APR-95 1121 16000 200 15 1120 william 20-MAY-05 1122 15000 500 10 1140 satheesh 12-JUN-99 1120 2500 150 20 1141 santhosh 12-MAR-02 1120 6000 160 20 1143 sam 18-MAR-07 1120 8000 200 20 1129 mone jo 20-APR-85 1122 15000 500 10 1142 jonathen 12-APR-98 1120 8000 190 20 1121 johnson 20-JUL-90 20000 20 1102 john 10-MAR-95 1121 8000 100 15 1126 james 1121 10000 95 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1122 james 20-JUL-85 20000 10 1144 husain 17-DEC-05 1120 9000 300 20 1101 Smith 01-JAN-84 1120 14000 200 20 14 rows selected. AnsAnsAnsAns:::: 8 8 8 8 SQL> select * from emp order by dno asc, sal desc; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S



---------- ------------ --------- ---------- ---------- ---------- ---------- - 1122 james 20-JUL-85 20000 10 1120 william 20-MAY-05 1122 15000 500 10 1129 mone jo 20-APR-85 1122 15000 500 10 1130 xelestina 20-APR-95 1121 16000 200 15 1102 john 10-MAR-95 1121 8000 100 15 1121 johnson 20-JUL-90 20000 20 1101 Smith 01-JAN-84 1120 14000 200 20 1144 husain 17-DEC-05 1120 9000 300 20 1142 jonathen 12-APR-98 1120 8000 190 20 1143 sam 18-MAR-07 1120 8000 200 20 1141 santhosh 12-MAR-02 1120 6000 160 20 EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1140 satheesh 12-JUN-99 1120 2500 150 20 1126 james 1121 10000 95 1127 zam dem 1122 10000 85 14 rows selected. AnsAnsAnsAns:::: 9 9 9 9 SQL> select * from emp order by sal desc,mgrssn desc; EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 1130 xelestina 20-APR-95 1121 16000 200 15 1120 william 20-MAY-05 1122 15000 500 10 1129 mone jo 20-APR-85 1122 15000 500 10 1101 Smith 01-JAN-84 1120 14000 200 20 1127 zam dem 1122 10000 85 1126 james 1121 10000 95 1144 husain 17-DEC-05 1120 9000 300 20 1102 john 10-MAR-95 1121 8000 100 15 1142 jonathen 12-APR-98 1120 8000 190 20



EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1143 sam 18-MAR-07 1120 8000 200 20 1141 santhosh 12-MAR-02 1120 6000 160 20 1140 satheesh 12-JUN-99 1120 2500 150 20 14 rows selected. AnsAnsAnsAns:::: 10 10 10 10 SQL> select empno, ename, sal,dno from emp order by sal desc,mgrssn asc; EMPNO ENAME SAL DNO ---------- ------------ ---------- ---------- 1122 james 20000 10 1121 johnson 20000 20 1130 xelestina 16000 15 1120 william 15000 10 1129 mone jo 15000 10 1101 Smith 14000 20 1126 james 10000 1127 zam dem 10000 1144 husain 9000 20 1142 jonathen 8000 20 1143 sam 8000 20 EMPNO ENAME SAL DNO ---------- ------------ ---------- ---------- 1102 john 8000 15 1141 santhosh 6000 20 1140 satheesh 2500 20 14 rows selected.



Exercise no 4

1. Write an SQL statement to display empno, empname, salary and deptnumber

with appropriate headings.

2. Write an SQL statement to display empname, salary and commission without

any duplicates

3. Write an SQL statement to display empno, empname and total salary. Total

salary is calculated by adding salary and commission.

4. Write an SQL statement to display empno, name, their total salary and total

annual salary of all employees.

5. Write an SQL statement to display the empno, empname and their

experience. Experience is calculated based upon the ((todays date-doj)/12).

6. Write an SQL statement to display empno, empname, salary, 10% of salary,

50% of salary, 30% of commission, 110% of commission and commission of

all employees.

7. Write an SQL statement to display the empno, empname, total salary, 10% of

total salary, 50% of total salary of all the employees.

8. Write an SQL statement to display the employee details who is getting the

highest salary.

9. Write an SQL statement to display the employee details with lowest salary

10. Write an SQL statement to display empno and empname in 1st column and

total salary in another column.



Queries and Outputs: AnsAnsAnsAns:::: 1 1 1 1 SQL> select ename as employee_name,empno as dept_no,sal as payment, dno as depa rtment_no from emp; EMPLOYEE_NAME DEPT_NO PAYMENT DEPARTMENT_NO ------------------------------ ---------- ---------- ------------- Smith 1101 14000 20 john 1102 8000 15 william 1120 15000 10 james 1122 20000 10 johnson 1121 20000 20 james 1126 10000 zam dem 1127 10000 mone jo 1129 15000 10 xelestina 1130 16000 15 satheesh 1140 2500 20 santhosh 1141 6000 20 EMPLOYEE_NAME DEPT_NO PAYMENT DEPARTMENT_NO ------------------------------ ---------- ---------- ------------- jonathen 1142 8000 20 sam 1143 8000 20 husain 1144 9000 20 14 rows selected. AnsAnsAnsAns:::: 2 2 2 2 SQL> select distinct ename,sal,comm from emp; ENAME SAL COMM ------------ ---------- ---------- Smith 14000 200 james 20000 xelestina 16000 200 john 8000 100 satheesh 2500 150 sam 8000 200 zam dem 10000 85 william 15000 500 james 10000 95 mone jo 15000 500 santhosh 6000 160 ENAME SAL COMM ------------ ---------- ---------- husain 9000 300 johnson 20000 jonathen 8000 190 14 rows selected. AnsAnsAnsAns:::: 3 3 3 3 SQL> select empno,ename,(sal+comm) as total_salary from emp; EMPNO ENAME TOTAL_SALARY ---------- ------------ ------------ 1101 Smith 14200



1102 john 8100 1120 william 15500 1122 james 1121 johnson 1126 james 10095 1127 zam dem 10085 1129 mone jo 15500 1130 xelestina 16200 1140 satheesh 2650 1141 santhosh 6160 EMPNO ENAME TOTAL_SALARY ---------- ------------ ------------ 1142 jonathen 8190 1143 sam 8200 1144 husain 9300 14 rows selected. AnsAnsAnsAns:::: 4 4 4 4 SQL> select empno,ename,((sal+comm)*12) as total_annual_salary from emp; EMPNO ENAME TOTAL_ANNUAL_SALARY ---------- ------------ ------------------- 1101 Smith 170400 1102 john 97200 1120 william 186000 1122 james 1121 johnson 1126 james 121140 1127 zam dem 121020 1129 mone jo 186000 1130 xelestina 194400 1140 satheesh 31800 1141 santhosh 73920 EMPNO ENAME TOTAL_ANNUAL_SALARY ---------- ------------ ------------------- 1142 jonathen 98280 1143 sam 98400 1144 husain 111600 14 rows selected. AnsAnsAnsAns:::: 5 5 5 5 SQL> select empno,ename,((sysdate-doj)/12) as experience from emp order by exper ience desc; EMPNO ENAME EXPERIENCE ---------- ------------ ---------- 1127 zam dem 1126 james 1101 Smith 707.788011 1129 mone jo 668.204678 1122 james 660.621345 1121 johnson 508.454678 1102 john 367.288011 1130 xelestina 363.871345 1142 jonathen 273.204678 1140 satheesh 237.704678 1141 santhosh 154.038011



EMPNO ENAME EXPERIENCE ---------- ------------ ---------- 1120 william 56.9546779 1144 husain 39.3713445 1143 sam 1.37134452 14 rows selected. AnsAnsAnsAns:::: 6 6 6 6 SQL> select empno, ename, sal,(sal*10/100) as salper10,(sal*50/100) as salper50 ,(comm*30/100) as commper30,(comm*110/100)as commper110 from emp; EMPNO ENAME SAL SALPER10 SALPER50 COMMPER30 COMMPER110 ---------- ------------ ---------- ---------- ---------- ---------- ---------- 1101 Smith 14000 1400 7000 60 220 1102 john 8000 800 4000 30 110 1120 william 15000 1500 7500 150 550 1122 james 20000 2000 10000 1121 johnson 20000 2000 10000 1126 james 10000 1000 5000 28.5 104.5 1127 zam dem 10000 1000 5000 25.5 93.5 1129 mone jo 15000 1500 7500 150 550 1130 xelestina 16000 1600 8000 60 220 1140 satheesh 2500 250 1250 45 165 1141 santhosh 6000 600 3000 48 176 EMPNO ENAME SAL SALPER10 SALPER50 COMMPER30 COMMPER110 ---------- ------------ ---------- ---------- ---------- ---------- ---------- 1142 jonathen 8000 800 4000 57 209 1143 sam 8000 800 4000 60 220 1144 husain 9000 900 4500 90 330 14 rows selected. AnsAnsAnsAns:::: 7 7 7 7 SQL> select empno,ename,(sal+comm) as total_salary,((sal+comm)*10/100) as total_ sal10,((sal+comm)*50/100) as total_sal50 from emp; EMPNO ENAME TOTAL_SALARY TOTAL_SAL10 TOTAL_SAL50 ---------- ------------ ------------ ----------- ----------- 1101 Smith 14200 1420 7100 1102 john 8100 810 4050 1120 william 15500 1550 7750 1122 james 1121 johnson 1126 james 10095 1009.5 5047.5



1127 zam dem 10085 1008.5 5042.5 1129 mone jo 15500 1550 7750 1130 xelestina 16200 1620 8100 1140 satheesh 2650 265 1325 1141 santhosh 6160 616 3080 EMPNO ENAME TOTAL_SALARY TOTAL_SAL10 TOTAL_SAL50 ---------- ------------ ------------ ----------- ----------- 1142 jonathen 8190 819 4095 1143 sam 8200 820 4100 1144 husain 9300 930 4650 14 rows selected. AnsAnsAnsAns:::: 8 8 8 8

SQL> select * from emp where sal in (select max(sal) from emp); EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1122 james 20-JUL-85 20000 10 1121 johnson 20-JUL-90 20000 20 AnsAnsAnsAns:::: 9999 SQL> select * from emp where sal in (select min(sal) from emp); EMPNO ENAME DOJ MGRSSN SAL COMM DNO S ---------- ------------ --------- ---------- ---------- ---------- ---------- - 1140 satheesh 12-JUN-99 1120 2500 150 20

AnsAnsAnsAns:::: 10 10 10 10

SQL> column no_name format a16; SQL> select empno||ename as no_name,(sal+comm) as total from emp; NO_NAME TOTAL ---------------- ---------- 1101Smith 14200 1102john 8100 1120william 15500 1122james 1121johnson 1126james 10095 1127zam dem 10085 1129mone jo 15500 1130xelestina 16200 1140satheesh 2650 1141santhosh 6160 NO_NAME TOTAL ---------------- ---------- 1142jonathen 8190 1143sam 8200 1144husain 9300 14 rows selected.



Exercise no 5

1. Write an SQL plus statement to create a new employee table with following

details

Empno number(5) primary key

ename varchar2(15) default 'Not Null'

doj date

bp number default 5000

desg varchar2(15) default 'Not Entered'.

2. Display the experience of each employee in the descending order of

experience (in years)

3. Display empno, empname, basic pay, da, hra and netsalary of all employees

with suitable heading as DA, HRA,…etc.DA=7.5% of basic, HRA=9% of basic,

netsal=basic+HRA+DA.

4. Insert some values of empno, doj, empname.

5. Display the total basic salary given to employees to each designation which is

greater than 10000.

6. Display the empno, name and basic of all employees for either manager or

clerk and basic salary in the range of 6000 and 11000

7. Display the empno, empname, designation and salary of employee with

largest salary

8. Alter the table employee by dropping the primary key

9. Alter the table employee by adding a new field ssn as primary key.

10. Create a table emp_sal with fields empno ,empname, bp, da using new

employee table.



Ans:Ans:Ans:Ans: 1 1 1 1 SQL> create table emp1 2 ( 3 empno number(5) primary key, 4 ename varchar2(15) default 'Not Null', 5 doj date , 6 bp number default 5000, 7 desg varchar2(15) default 'Not Entered' 8 ); Table created. SQL> insert into emp1 values(&empno,'&ename','&doj',&bp,'&desg'); Enter value for empno: 1101 Enter value for ename: smith Enter value for doj: 12-mar-2000 Enter value for bp: 5000 Enter value for desg: manager old 1: insert into emp1 values(&empno,'&ename','&doj',&bp,'&desg') new 1: insert into emp1 values(1101,'smith','12-mar-2000',5000,'manager') 1 row created. SQL> / Enter value for empno: 1102 Enter value for ename: john Enter value for doj: 12-april-1985 Enter value for bp: 3000 Enter value for desg: clerk old 1: insert into emp1 values(&empno,'&ename','&doj',&bp,'&desg') new 1: insert into emp1 values(1102,'john','12-april-1985',3000,'clerk') 1 row created. SQL> / Enter value for empno: 1103 Enter value for ename: willian Enter value for doj: 12-dec-1999 Enter value for bp: 8000 Enter value for desg: MD old 1: insert into emp1 values(&empno,'&ename','&doj',&bp,'&desg') new 1: insert into emp1 values(1103,'willian','12-dec-1999',8000,'MD') 1 row created. SQL> / Enter value for empno: 1104 Enter value for ename: james Enter value for doj: 14-feb-1988 Enter value for bp: 11000 Enter value for desg: director old 1: insert into emp1 values(&empno,'&ename','&doj',&bp,'&desg') new 1: insert into emp1 values(1104,'james','14-feb-1988',11000,'director') 1 row created. AnsAnsAnsAns:::: 2 2 2 2 SQL> select ename,round(months_between(sysdate,doj)/12) as exp from

emp1 order by exp desc;



ENAME EXP ------------ ---------- john 22 james 19 willian 7 smith 7 Ans: 3Ans: 3Ans: 3Ans: 3 SQL> select empno,bp,ename,bp,0.075*bp as DA,0.09*bp as

HRA,bp+(0.075*bp)+(0.09*bp)as NET from emp1;

EMPNO BP ENAME BP DA HRA NET ---------- ---------- ------------ ---------- ---------- ---------- ---------- 1101 5000 smith 5000 375 450 5825 1102 3000 john 3000 225 270 3495 1103 8000 willian 8000 600 720 9320 1104 11000 james 11000 825 990 12815 Ans: 4Ans: 4Ans: 4Ans: 4 SQL> insert into emp1(empno,ename,doj) values (1110,'mathew','02-dec-

2006');

1 row created. Ans: 5Ans: 5Ans: 5Ans: 5 SQL> select desg,sum(bp) as NET_SALARY from emp1 group by desg having

sum(bp)> 10000;

DESG NET_SALARY --------------- ---------- director 11000 Ans: 6Ans: 6Ans: 6Ans: 6 SQL> select desg,bp from emp1 where desg in('manager','clerk') and bp

between 6000 and 11000;

no rows selected Ans: 7Ans: 7Ans: 7Ans: 7 SQL> select empno,ename,bp from emp1 where bp=(select max(bp) from

emp1);

EMPNO ENAME BP ---------- ------------ ----------



1104 james 11000 Ans: 8Ans: 8Ans: 8Ans: 8 SQL> alter table emp1 drop primary key; Table altered. SQL> select * from emp1; EMPNO ENAME DOJ BP DESG ---------- ------------ --------- ---------- --------------- 1101 smith 12-MAR-00 5000 manager 1102 john 12-APR-85 3000 clerk 1103 willian 12-DEC-99 8000 MD 1104 james 14-FEB-88 11000 director 1110 mathew 02-DEC-06 5000 Not Entered SQL> desc emp1; Name Null? Type ----------------------------------------- -------- ----------------------- EMPNO NUMBER(5) ENAME VARCHAR2(15) DOJ DATE BP NUMBER DESG VARCHAR2(15) Ans: 9Ans: 9Ans: 9Ans: 9 SQL> alter table emp1 add ssn number(5); Table altered. SQL> update emp1 set ssn=empno; 5 rows updated. SQL> alter table emp1 add primary key(ssn); Table altered. SQL> select * from emp1; EMPNO ENAME DOJ BP DESG SSN ---------- ------------ --------- ---------- --------------- ---------- 1101 smith 12-MAR-00 5000 manager 1101 1102 john 12-APR-85 3000 clerk 1102 1103 willian 12-DEC-99 8000 MD 1103 1104 james 14-FEB-88 11000 director 1104 1110 mathew 02-DEC-06 5000 Not Entered 1110 SQL> desc emp1; Name Null? Type ----------------------------------------- -------- ----------------------------



EMPNO NUMBER(5) ENAME VARCHAR2(15) DOJ DATE BP NUMBER DESG VARCHAR2(15) SSN NOT NULL NUMBER(5) Ans: 1Ans: 1Ans: 1Ans: 10000 SQL> create table emp_salary (empno,ename,bp,da) as (select

empno,ename,bp,bp*7/100 from emp1);

Table created. SQL> select * from emp_salary; EMPNO ENAME BP DA ---------- ------------ ---------- ---------- 1101 smith 5000 350 1102 john 3000 210 1103 willian 8000 560 1104 james 11000 770 1110 mathew 5000 350 SQL> desc emp_salary; Name Null? Type ---------------------------- ------------------ -------------- EMPNO NUMBER(5) ENAME VARCHAR2(15) BP NUMBER DA NUMBER



Exercise no 6

1. Write an SQL statement to display the employee details who are working for

accounts department.

2. Write an SQL statement to display the department details for who are getting

salary >10000.

3. Write an SQL statement to display, who is reporting to whom (ie manager_no,

name, worker_no, name of all employees.

4. Write an SQL statement to display all employees who are not associated with

any department.

5. Write an SQL statement to display all departments without any employees.

6. Write an SQL statement to display employee details who are working for

department sales and not getting any commission.

7. Display the employee details who are working for the marketing department

and salary > 5000.

8. Display the employees who are working for accounts department getting salary

between 5000 and 10000.

9. Display the employees who are working in loation 'Dallas’ and getting salary

between 5000 and 10000.

10. Display the employees who are getting commission as null and working for the

sales department in location ‘Texas’.



SQL> create table emp ( EMPNO NUMBER(3) primary key, ENAME VARCHAR2(30) not null, JOB VARCHAR2(20), MGR NUMBER(3), DOJ DATE, SAL NUMBER(7,2), COMM NUMBER(5,2), DEPTNO NUMBER(2) ); Table created. SQL> create table dept ( DEPTNO NUMBER(2) primary key, DNAME VARCHAR2(20), LOC VARCHAR2(20) ); Table created.

THE TABLE CONTENTS AFTER INSERTION OPERATIONS ---------------------------------------------

EMPNO ENAME JOB MGR DOJ SAL COMM DEPTNO ---------- ---------- ----------------- ---------- --------- ---------- ---------- ---------- 222 mahesh analyst 333 31-JAN-84 5000 500 10 333 john project_leader 444 31-JAN-83 7000 20 444 satheesh manager 555 31-JAN-82 10000 30 555 mathew managing director 26-JAN-82 15000 30 112 ajay operator 222 22-FEB-84 2000 111 suresh programmer 222 22-FEB-84 3000 300 10 6 rows selected. SQL> select * from dept; DEPTNO DNAME LOC ---------- -------------------- -------------------- 10 accounts texas 20 sales dallas 30 marketing california 40 production miniapolis Ans:Ans:Ans:Ans: 1111 SQL> select emp.ename,emp.empno,emp.sal,emp.doj,emp.job from emp,dept

where emp.deptno=dept.deptno and dept.dname='accounts';

ENAME EMPNO SAL DOJ JOB ---------- ---------- ---------- --------- -------------------- mahesh 222 5000 31-JAN-84 analyst suresh 111 3000 22-FEB-84 programmer



AnAnAnAnssss:::: 2 2 2 2 SQL> select distinct dept.deptno,dept.dname,dept.loc from dept ,emp

where emp.sal>=10000 and emp.deptno=dept.deptno;

DEPTNO DNAME LOC ---------- -------------------- -------------------- 30 marketing california Ans: 3Ans: 3Ans: 3Ans: 3 SQL> select e1.ename as "employee",e2.ename as "manager" from emp e1,emp e2 wher e e1.mgr=e2.empno; employee manager ------------------------------ ------------------------------ mahesh john john satheesh satheesh mathew ajay mahesh suresh mahesh Ans: 4Ans: 4Ans: 4Ans: 4 SQL> (select emp.ename from emp,dept where emp.deptno=dept.deptno(+))

minus (select emp.ename from emp,dept where emp.deptno=dept.deptno);

ENAME ---------- ajay Ans: 5Ans: 5Ans: 5Ans: 5 SQL> (select dept.dname from emp,dept where emp.deptno(+)=dept.deptno)

minus (select dept.dname from emp,dept where emp.deptno=dept.deptno);

DNAME -------------------- production AnAnAnAnssss:::: 6 6 6 6 SQL> select emp.ename from emp,dept where emp.deptno=dept.deptno and

dept.dname='sales' and emp.comm is null;

ENAME ---------- john SQL> Ans: 7Ans: 7Ans: 7Ans: 7 SQL> select emp.empno,emp.ename,emp.job from emp,dept where

emp.deptno=dept.deptno and dept.dname='marketing' and emp.sal>5000;

EMPNO ENAME JOB



---------- ---------- -------------------- 444 satheesh manager 555 mathew managing director AnsAnsAnsAns:::: 8 8 8 8 SQL> select emp.ename from emp,dept where emp.deptno=dept.deptno and

dept.dname='accounts' and emp.sal between 5000 and 10000;

ENAME ---------- mahesh AnsAnsAnsAns:::: 9 9 9 9 SQL> select emp.ename from emp,dept where emp.deptno=dept.deptno and

dept.loc='dallas' and emp.sal between 5000 and 10000;

ENAME ---------- john AnsAnsAnsAns: 10: 10: 10: 10 SQL> select emp.ename from emp,dept where emp.deptno=dept.deptno and

dept.loc='texas' and emp.sal between 5000 and 10000;

ENAME ---------- mahesh



Exercise no 7

1. Write an SQL statement to create a view to display employee number,

employee name, job, salary, department number based on the employee table

2. Write an SQL statement to create view to display employee number, employee

name, job, salary, department number based on employee table who are

getting a salary greater than 5000

3. Write an SQL statement to create a view to display employee number,

employee name, department no, department name from employee table and

department table of all employees.

4. Write an SQL statement to insert records into the view created on the basis of

question 1 and 2.

THE TABLE CONTENTS -------------------

EMPNO ENAME JOB MGR DOJ SAL COMM DEPTNO ---------- ---------- ----------------- ---------- --------- ---------- ---------- ---------- 222 mahesh analyst 333 31-JAN-84 5000 500 10 333 john project_leader 444 31-JAN-83 7000 20 444 satheesh manager 555 31-JAN-82 10000 30 555 mathew managing director 26-JAN-82 15000 30 112 ajay operator 222 22-FEB-84 2000 111 suresh programmer 222 22-FEB-84 3000 300 10 6 rows selected. SQL> select * from dept; DEPTNO DNAME LOC ---------- -------------------- -------------------- 10 accounts texas 20 sales dallas 30 marketing california 40 production miniapolis Ans: 1Ans: 1Ans: 1Ans: 1 SQL> create or replace view empview as select empno, ename, job, sal,

deptno from emp;

View created. SQL> SELECT * FROM EMPVIEW; EMPNO ENAME JOB SAL DEPTNO



---------- ---------- -------------------- ---------- ---------- 222 mahesh analyst 5000 10 333 john project_leader 7000 20 444 satheesh manager 10000 30 555 mathew managing director 15000 30 112 ajay operator 2000 111 suresh programmer 3000 10 6 rows selected. AnsAnsAnsAns: 2: 2: 2: 2 SQL> create or replace view empview2 as select empno, ename, job,

sal, deptno from emp where sal>5000;

View created. SQL> SELECT * FROM EMPVIEW2; EMPNO ENAME JOB SAL DEPTNO ---------- ---------- -------------------- ---------- ---------- 333 john project_leader 7000 20 444 satheesh manager 10000 30 555 mathew managing director 15000 30 Ans: Ans: Ans: Ans: 3333 SQL> create or replace view empdeptview as select emp.empno,

emp.ename,emp.deptno,dept.dname from emp,dept where

emp.deptno=dept.deptno;

View created. SQL> SELECT * FROM EMPDEPTVIEW; EMPNO ENAME DEPTNO DNAME ---------- ---------- ---------- -------------------- 222 mahesh 10 accounts 333 john 20 sales 444 satheesh 30 marketing 555 mathew 30 marketing 111 suresh 10 accounts Ans: 4Ans: 4Ans: 4Ans: 4 SQL> DESC EMPVIEW; Name Null? Type ----------------------------------------- -------- ---------------------------- EMPNO NOT NULL NUMBER(3) ENAME NOT NULL VARCHAR2(30) JOB VARCHAR2(20) SAL NUMBER(7,2) DEPTNO NUMBER(2) SQL> DESC EMPVIEW2; Name Null? Type ----------------------------------------- -------- ----------------------------



EMPNO NOT NULL NUMBER(3) ENAME NOT NULL VARCHAR2(30) JOB VARCHAR2(20) SAL NUMBER(7,2) DEPTNO NUMBER(2) SQL> INSERT INTO EMPVIEW VALUES(666,'hari','programmer',4000,40); 1 row created. SQL> INSERT INTO EMPVIEW VALUES(667,'sarath','programmer',5000,40); 1 row created. SQL> INSERT INTO EMPVIEW2 VALUES(668,'hilary','analyst',5000,10); 1 row created. SQL> INSERT INTO EMPVIEW2 VALUES(669,'sardar g','operator',4000,40); 1 row created. SQL> select * from empview; EMPNO ENAME JOB SAL DEPTNO ---------- ---------- -------------------- ---------- ---------- 222 mahesh analyst 5000 10 333 john project_leader 7000 20 444 satheesh manager 10000 30 555 mathew managing director 15000 30 112 ajay operator 2000 111 suresh programmer 3000 10 666 hari programmer 4000 40 667 sarath programmer 5000 40 668 hilary analyst 5000 10 669 sardar g operator 4000 40 10 rows selected. SQL> select * from empview2; EMPNO ENAME JOB SAL DEPTNO ---------- ---------- -------------------- ---------- ---------- 333 john project_leader 7000 20 444 satheesh manager 10000 30 555 mathew managing director 15000 30

THE BASE TABLE AFTER INSERTING VALUES INTO THE VIEWS SQL> select empno,ename,job,mgr,doj,sal from emp; EMPNO ENAME JOB MGR DOJ SAL COMM DEPTNO ---------- ---------- -------------------- ---------- --------- ---------- ---------- ---------- 222 mahesh analyst 333 31-JAN-84 5000 500 10 333 john project_leader 444 31-JAN-83 7000 20 444 satheesh manager 555 31-JAN-82 10000 30



555 mathew managing director 26-JAN-82 15000 30 112 ajay operator 222 22-FEB-84 2000 111 suresh programmer 222 22-FEB-84 3000 300 10 666 hari programmer 4000 40 667 sarath programmer 5000 40 668 hilary analyst 5000 10 669 sardar g operator 4000 40 10 rows selected.



PL/SQL

SOLVED PROBLEMS



Exercise no 1

Aim:

To find the greatest among three numbers

.

Algorithm:

Step 1: Start.

Step 2: Declare variables a,b and c as integers and initialize them.

Step 3: Check whether:

a > b and a > c, then print a is greater

elsif check b>c then b is greater

else c is greater

Step 5: Stop.

Program:

Declare

a number(10):='&a';

b number(10):='&b';

c number(10):='&c';

Begin

if ((a>b) and (a>c)) then

dbms_output.put_line('a is greater');

elsif (b>c) then

dbms_output.put_line('b is greater');

else

dbms_output.put_line('c is greater');

end if;

End;

/

Output:

Enter value for a: 4

old 2: a number(10):='&a';

new 2: a number(10):='4';

Enter value for b: 3

old 3: b number(10):='&b';

new 3: b number(10):='3';

Enter value for c: 2

old 4: c number(10):='&c';

new 4: c number(10):='2';

a is greater




Exercise no 2

Aim:

To write a PL/SQL code for factorial of a number.

Algorithm:

Step 1: Start.

Step 2: Declare variables n(the number),i(initialized to 1) and f(initialized to 1) get

value to n.

Step 3: Check if i is lesser than or equal to n, if not then goto Step 6.

Step 4: Set f as product of current f and i.

Increment value to i by 1.

Step 5: Goto Step 3.

Step 6: Print factorial(f) of n.

Step 7: Stop.

Program:

Declare

n number(2):='&n';

i number(2):=1;

f number(30):=1;

Begin

while (i<=n)

Loop

f:=f*i;

i:=i+1;

End Loop;

dbms_output.put_line('Factorial of '|| n ||' :-'||f);

End;

/

Output:

Enter value for n: 3

old 2: n number(2):='&n';

new 2: n number(2):='3';

Factorial of 3 :-6




Exercise no 3

Aim:

To write a PL/SQL block to generate Fibonacci series.

Algorithm:

Step 1: Start.

Step 2: Declare variables.

n – for limit

a =0

b=1

c=1

Step 3: Check if c lesser than n, if false then goto Step 5.

Print value to c. Set c as sum of a and b.

Set a = b .

Set b = c.

Step 4: goto step 3

Step 5: stop

Program:

Declare

n number(8):='&n';

a number(8):=0;

b number(8):=1;

c number(8):=1;

Begin

dbms_output.put_line('Limit is '||n||'');

dbms_output.put_line('**************');

dbms_output.put_line('The series is');

dbms_output.put_line('**************');

dbms_output.put_line(a);

while (c<n)

Loop

dbms_output.put_line(c);

c:=a+b;

a:=b;

b:=c;

End Loop;

End;

/

Output:

Enter value for n: 4

old 2: n number(8):='&n';

new 2: n number(8):='4';

Limit is 4

**************



The series is

**************

0

1

1

2

3




Exercise no 4

Table Used

EMPNO EMPNAME SAL DEPTNO COMM

------- ---------- ------ ----------- --------

1111 AISWARYA 5000 5 300.5

1112 EISHA 6000 8 342.15

1113 NASHA 7000 4 242.15

1114 ABHIJEET 1000 10 150

1115 CLARK 10000 9 350

Aim:

Write a PL/SQL block to display empno, empname, deptno and salary of employee no

is 1113.

Algorithm:

Step1: Start

Step2: declare variables en(empno),

name(empname), sal(sal) ,dno(deptno),

com(comm) of %type.

Step3: select the values for en,name,sal,dno,com from EMP table with empno=1113

Step4: print the values of en,name,sal,dno

Step5: stop

Program:

declare

en emp.empno%type;

name emp.empname%type;

sal emp.sal%type;

dno emp.deptno %type;

com emp.comm%type;

begin

select empno, empname, sal, deptno into en, name,

sal, dno from emp where empno=1113;

dbms_output.put_line(‘employee no is’|| en);

dbms_output.put_line(‘name ’||name);

dbms_output.put_line(‘salary’|| to_char(sal));

dbms_output.put_line(‘dept-no’|| to-char(dno));

end;

/

Output:

EMPLOYEE NO IS 1113

NAME NASHA

SALARY 7000

DEPT-NO 4




Exercise no 5

Aim:

Write a PL/SQL block to display empno, empname and salary of employee who is

getting salary 1000 and who is working in department no 10.

Algorithm:

Step1: Start

Step2: declare variables en(empno),

name(empname), sal(sal) ,dno(deptno),

com(comm) of %type.

Step3: select the values for en,name,sal,dno,com from EMP table with empno=1000

and deptno = 10

Step4: print the values of en,name,sal,dno

Step5: stop

Program:

declare

en emp.empno%type;


sal emp.sal%type;

dno emp.deptno %type;

com emp.comm%type;

begin

select empno, empname, sal, deptno into en, name, sal, dno from

emp where sal=1000 and deptno=10;

dbms_output.put_line(‘employee no is’|| en);


dbms_output.put_line(‘salary’|| to_char(sal));

dbms_output.put_line(‘dept-no’|| to-char(dno));

end;

Output:

EMPLOYEE NO IS 1114

NAME ABHIJEET

SALARY 1000

DEPT-NO 10




Exercise no 6

Aim:

Write a PL/SQL block to display and calculate da, hra, ta for the employee CLARK. hra

is calculated based on 10% of basic, da is calculated based on 6% of basic and ta is

calculated based on 10% of commission. total salary is calculated based on the sum of

da, hra, ta and basic salary.

Algorithm:

Step1: Start

Step2: declare variables name,sl,cm,da,hra,ta,tsal

Step3: select the values for name,sl,cm from EMP table with empname=clerk

Step4: compute da= sl*0.06, hra=sl*.01,ta=cm*.01,tsal = sl+da+hra+ta

Step5: print the values of name,sl,da,hra,ta,tsal

Step5: stop

Program:

declare


sl emp.sal%type;

cm emp.comm%type;

da number(6,2);

hra number(6,2);

ta number(6,2);

tsal number(8,2);

begin

select empname, sal, comm into name, sl, cm from emp where

empname like ‘clark’;

da:=sl*0.06;

hra:=sl*0.1;

ta:=cm*0.1;

tsal:=sl+da+hra+ta;


dbms_output.put_line(‘salary’|| to_char(sl));

dbms_output.put_line(‘da’|| to_char(da));

dbms_output.put_line(‘hra’|| to_char(hra));

dbms_output.put_line(‘ta’|| to_char(ta));

dbms_output.put_line(‘totalsalary’|| to_char(tsal));

end;

Output:

NAME CLARK

SALARY 10000

DA 600

HRA 1000

TA 50

TOTAL SALARY 11650

PL/SOL procedure successfully completed.



Exercise no 7

Aim:

Write a PL/SQL block to display total salary of all employees.

Algorithm:

Step1: Start

Step2: declare the variables n1(empno),sl(sal),na(empname),com(comm) of %type

Step3: Initialize the cursor c1 by selecting empno,empname,sal,comm from emp.

Step4: open the cursor c1

Step5: fetch the values of n1,na,sl,com into cursor c1.

Step6: calculate ts = sl + com

Step7: print the values of n1,na,ts

Step8: close the cursor c1

Step9: Stop

Program:

declare

n1 emp.empno%type;

sl emp.sal%type;

na emp.empname%type;

com emp.comm%type;

cursor c1 is select empno, empname, sal, comm from emp;

ts number(8,2);

begin

open c1;

dbms_output.put_line(‘……………………..’);

loop

fetch c1 into n1,na,sl,com;

exit when (c1%notfound);

ts:=sl+nvl(com,0);

dbms_output.put_line(‘empno’|| to_char(n1));

dbms_output.put_line(‘name ’||na);

dbms_output.put_line(‘total salary’|| to_char(ts));

dbms_output.put_line(‘……………………..’);

end loop;

close c1;

end;

Output:

EMPNO 1111

NAME AISHWARYA

TOTAL SALARY 5300.5

……………………………………………

EMPNO 1112

NAME EISHA

TOTAL SALARY 6772.5

……………………………………………

EMPNO 1113



NAME NASHA

TOTAL SALARY 15300.5

……………………………………………

EMPNO 1114

NAME ABHIJEETH

TOTAL SALARY 8307.5

……………………………………………



Exercise no 8

Aim:

Create a procedure to evaluate student grade by accepting student name.

Algorithm:

Step 1: Start procedure.

Step 2: Define procedure sgrade taking parameter name and program variable

mk(mark).

Step 3: Select mark of student into an integer(mk) whose name had been passed.

Step 4: If mk greater than 90 then

Print “Grade:=X”.

Else if mk greater than 80 then

Print “Grade:=A”.


Print “Grade:=B”.


Print “Grade:=C”.

Else

Print “Grade:=F”.

Step 5: Stop procedure.

Program:

create or replace procedure sgrade(n_name IN varchar2)

IS

mk number(3);

Begin

select mark into mk from student where name=n_name;

if mk>90 then

dbms_output.put_line('Grade:=X');

else if mk>80 then

dbms_output.put_line('Grade:=A');

else if mk>60 then

dbms_output.put_line('Grade:=B');

else if mk>50 then

dbms_output.put_line('Grade:=C');

else

dbms_output.put_line('Grade:=F');

end if;

end if;

end if;

end if;

End sgrade;

/

Procedure created.



Output:

SQL> execute sgrade('rajesh');

Grade:=B


SQL> execute sgrade('jeevan');

Grade:=A




Exercise no 9

Table Used

ENO ENAME DNAME DOJ DESIG

----- --------- --------- ----------------- ----------------

100 rajesh MCA 25-MAY-05 lecturer

102 joy ME 20-APR-06 lecturer

104 ranjith EC 12-JAN-06 lecturer

105 mable CSE 12-JAN-04 lecturer

Aim:

Write a procedure in PL/SQL to check if employee is eligible for promotion depending

on his period of service.

Algorithm:

Step1: Start.

Step2: Define procedure des_ch taking parameter employee number (eid)

Step3: Declare variables dt (experience), jd(joining date), sd(system date).

Step4: Select doj field value from relation emp_proc into program variable jd for an

employee (whose employee id is eid).

Step5: Select system date into program variable sd.

Step6: Compute the number of months between jd and sd dates.

Step7: Check if dt greater than or equal to 3,

if true then update design=ass.prof

Print “Your designation changed”.

Else

Print “you have to wait”.

Step8: Stop.

Program:

create or replace procedure des_ch(eid IN varchar2)

IS

dt number;

jd date;

sd date;

Begin

select doj into jd from emp_proc where eno=eid;

select sysdate into sd from dual;

dt:=months_between(sd,jd) / 12;

if dt>=3 then

update emp_proc set desig='ass.prof' where eno=eid;

dbms_output.put_line('Your designation changed');

else

dbms_output.put_line('you have to wait');

end if;

End des_ch;



Output:

SQL> execute des_ch(100);

you have to wait



you have to wait



you have to wait



Your designation changed


select * from emp_proc;

ENO ENAME DNAME DOJ DESIG

----- --------- --------- ----------------- ----------------

100 rajesh MCA 25-MAY-05 lecturer

102 joy ME 20-APR-06 lecturer

104 ranjith EC 12-JAN-06 lecturer

105 mable CSE 12-JAN-04 ass.prof



Exercise no 10

Aim:

Write a function in PL/SQL to find student grade by accepting student name as

argument.

Algorithm:

Step 1: Start

Step 2: Define function Sff taking parameter as name.

Step 3: Declare variables mk(mark) and g(grade).

Step 4: Select mark of student into an mk whose name had been passed.

Step 5: If mk greater than 90 then

g=x

Print “Grade:=X”.


g=A

Print “Grade:=A”.


g=B

Print “Grade:=B”.


g=C

Print “Grade:=C”.

Else g=C.

Print “Grade:=F”.

Step 6: Return value to grade g.

Step 7: Stop

Program:

create or replace function Sff(n_name IN varchar2) return varchar2

IS

g varchar2(1);

mk number(3);

Begin

select mark into mk from student where name=n_name;

if mk>90 then

g:='x';

dbms_output.put_line('Grade: ' || g );

else if mk>80 then

g:='A';


else if mk>60 then

g:='B';




else if mk>50 then

g:='C';


else

g:='F';


end if;

end if;

end if;

end if;

return g;

End Sff;

Output:

SQL> select Sff('ntini') from dual;

SFF('NTINI')

--------------------------------------

A

Grade: A

SQL> select Sff('rajesh') from dual;

SFF('RAJESH')

--------------------------------------

X

Grade: X



Exercise no 11

Tables Used

T4

NO CHARA

------ ---------

5 Rajesh

13 Rejeev

10 Reneesh

T5

NO CHARA

Aim:

Create a trigger that may insert a tuple into table5 when a tuple is inserted into table4,

the trigger has to check whether the new tuple has first component 10 or less, and if

so insert the tuple into table5.

Program:

create or replace trigger tr1

after insert on t4

referencing new as newrow

for each row when (newrow.no<11)

begin

insert into t5 values(:newrow.no,:newrow.chara);

end tr1;

Output:

SQL>run;

trigger created.

SQL>select * from T5;

NO CHARA

------ ---------

5 Rajeesh

10 Reneesh



Exercise no 12

Tables Used

manager

MANAGERID MNAME

--------------------- -------------

41 Sanjitha

101 Abro

1 Sameer

employee

EMPNO ENAME

Aim:

Create a trigger that may insert a tuple into EMPLOYEE(empno,ename) when a tuple is

updated into MANAGER(managerid,mname) the trigger has to check whether the new

tuple has a first component 50 or less, and if so insert the tuple into EMPLOYEE.

create or replace trigger tr2

after update on manager

referencing new as newrow

for each row when (newrow.manager<51)

begin

insert into employe values

(:newrow.managerid,:newrow.mname);

end tr2;

Output:

SQL>run;

trigger created.

SQL> update manager set managerid =50 where mname=’Sanjith’;

1 row updated.

SQL>select * from employe;

MANAGERID MNAME

------------- ----------

50 Sanjitha



Database connectivity

Solved Problems



Exercise no 1

Aim: Create a database application in visual basic to manipulate student’s information.

Option Explicit Public LoginSucceeded As Boolean Private Sub cmdCancel_Click() LoginSucceeded = False Me.Hide End Sub Private Sub cmdOK_Click() If txtPassword = "ide" Then LoginSucceeded = True Unload frmLogin ICEmain.Show Else MsgBox "Invalid Password, try again!", , "Login" txtPassword.Text = "" End If End Sub

Private Sub Form_Load() txtUserName = "IDE" End Sub



Private Sub emd_Click() editregfrm.Show End Sub Private Sub exit_Click() Unload ICEmain End Sub Private Sub memb_Click() viewfrm.Show End Sub Private Sub newreg_Click() regfrm.Show End Sub Private Sub rm_Click() deletefrm.Show End Sub

Dim db As ADODB.Connection Dim gender As String Private Sub cancelcmd_Click() Unload regfrm End Sub

Private Sub femaleopt_Click() If femaleopt.Value = True Then gender = femaleopt.Caption End If End Sub Private Sub Form_Load() Set db = New ADODB.Connection db.Open ("provider=MSDASQL.1;data source=rvs;user id=system;



password=sa") coursecmb.AddItem ("MCA") coursecmb.AddItem ("MBA") coursecmb.AddItem ("BTech CSE") coursecmb.AddItem ("BTech IT") coursecmb.AddItem ("BTech EEE") coursecmb.AddItem ("BTech EC") datetxt.Text = Date End Sub Private Sub maleopt_Click() If maleopt.Value = True Then gender = maleopt.Caption End If End Sub Private Sub savecmd_Click() db.Execute ("insert into lab(regno,name,course,gender) values('" & regtxt.Text & "','" & nametxt.Text & "','" & coursecmb.Text & "','" & gender & "')") MsgBox ("Thank You , Record saved") regtxt.Text = Clear nametxt.Text = Clear regtxt.Text = Clear End Sub

Dim s As String Private Sub cancelcmd_Click() Unload editregfrm End Sub Private Sub editcmd_Click() Editfrm.Show End Sub Private Sub Form_Load() Set db = New ADODB.Connection Set rs = New ADODB.Recordset db.Open ("provider=MSDASQL.1;data source=rvs;user id=system; password=sa") Set rs = db.Execute("select regno from lab") While rs.EOF <> True



editregcmb.AddItem (rs![regno]) rs.MoveNext Wend s = editregcmb.Text End Sub

Private Sub Form_Load() ecoursecmb.AddItem ("MCA") ecoursecmb.AddItem ("MBA") ecoursecmb.AddItem ("BTech CSE") ecoursecmb.AddItem ("BTech IT") ecoursecmb.AddItem ("BTech EEE") ecoursecmb.AddItem ("BTech EC") eregtxt.Text = editregfrm.editregcmb.Text eregtxt.Enabled = False Set db = New ADODB.Connection Set rs = New ADODB.Recordset db.Open ("provider=MSDASQL.1;data source=rvs;user id=system; password=sa") Set rs = db.Execute("select* from lab where regno='" & eregtxt.Text & "'") enametxt.Text = rs![Name] ecoursecmb.Text = rs![course] If rs![gender] = "Male" Then eoptionmale.Value = True Else eoptionfemale.Value = True End If End Sub Dim ecmb As String Dim egender As String Dim db As ADODB.Connection Dim rs As ADODB.Recordset Private Sub ecancelcmd_Click()



Unload Editfrm End Sub Private Sub eoptionfemale_Click() If eoptionmale.Value = True Then egender = eoptionmale.Caption End If End Sub Private Sub eoptionmale_Click() If eoptionfemale.Value = True Then egender = eoptionfemale.Caption End If End Sub Private Sub eupdatecmd_Click() If eoptionmale.Value = True Then egender = eoptionmale.Caption End If If eoptionfemale.Value = True Then egender = eoptionfemale.Caption End If ecmb = ecoursecmb.Text Set rs = db.Execute("update lab set name = '" & enametxt.Text & "', course ='" & ecmb & "',gender ='" & egender & "' where regno='" & eregtxt.Text & "' ") MsgBox ("your record is edited") Unload Editfrm End Sub

Private Sub Form_Load() Set db = New ADODB.Connection Set rs = New ADODB.Recordset db.Open ("provider=MSDASQL.1;data source=rvs;user id=system; password=sa") Set rs = db.Execute("select regno from lab") While rs.EOF <> True deleteregcmb.AddItem (rs![regno]) rs.MoveNext Wend End Sub



Dim db As ADODB.Connection Dim rs As ADODB.Recordset Private Sub cancelcmb_Click() Unload deletefrm End Sub Private Sub deletecmb_Click() db.Execute ("delete from lab where regno='" & deleteregcmb.Text & "' ") MsgBox ("your record with Regno '" & deleteregcmb.Text & "' is deleted") End Sub

Private Sub Form_Load() With Adodc1 .ConnectionString = "dsn=rvs" .UserName = "system" .Password = "sa" .RecordSource = "select * from lab" End With Set DataGrid1.DataSource = Adodc1 End Sub

dbms theory

Documents