dbms record

INDEX

Basic Queries

1.Create student table with rno, name, class, mid1, mid2 and final marks in GE. Apply necessary constraints

2. Create course table with course id, course name(GE), instructor. Apply necessary constraints.

3. Insert rows in both the tables

4. Add a column total, average and course id to student table

5.Make course id as foreign key in student table

6.Increase the width of name column to 15.

7. Change the marks of a particular student.

8. Add five marks grace in m1 subject to all students who scored less than 10.

9. Update total and average columns.

10. Delete the students of solar processing course and rollback.

11. List all students data

12. Display all the students belonging to NCE.

13. Display all the students of NCE who scored more than 150 marks.

14. Display all students information in ascending order

15. List all the students who do not belong to NCE.

16. List all students having average between 60 and 80

17. List all students belonging to Mr.Smith.

18. List the rollno and class of the students

19. List the students belonging to course id 631 and 633.

20. Count the number of students in each course.

21. List the course id having minimum of 2 students.

22. Find the names of the students that begin with D.

23. Find the names of the students that ends with K.

24. Find the names of the students who have S in their name.

25.Display all students name in uppercase.

26. Determine the names and length of each name.

27. Display all students having 4 characters.

28. Display student details in descending order.

29. List the rno, name, course name, course id.

30. To display all records of course table even if there is no matching course id in student.

31. Display the maximum and minimum total and average.

32. Find the highest marks obtained in PCMT.

33. To display the second highest and second lowest marks.

34. Display all students who have not opted for wed designing.

Queries on Emp and Dept tables

1.Simple Queries

2.Updating the Queries

3.Order By Clause

4.Data Functions

5.Group By Clause

6.Subqueries

7.Group Functions

8.Joins

9.Set Operators

10.Views

11.Co-related subquery and multicolumn subquery

12.Miscellaneous

12.DDL Queries

INTERACTIVE SQL:PART -1

1.SQL Statement for creating the tables:

a)Table Name: client_master

b)Table Name: product_master

c) Table Name: salesman_master

d) Table Name: sales_order

e) Table Name: sales_order_details

2.SQL Statement for inserting data into their respective tables:

a)Data for client_master table

b)Data for product_master table

c)Data for salesman_master table

d)Data for sales_order table

e)Data for sales_order_details table

3.SQL Statement for retrieving records from a table

4.SQL Statement for updating records in a table

5.SQL Statement for deleting records in a table

6.SQL Statement for altering the table structure

7.SQL Statement for renaming the table


1.Generate SQL Statements to perform the computations on table data.

2.SQL Statement for Date Manipulation


1.SQL Statement for using Having and Group By Clauses

2.Exercise on Joins and Correlation

3.SQL Statements for exercise on Sub-queries

PL/SQL QUERIES

Simple PL/SQL Queries

a)Pl/Sql to add two numbers

b)PL/Sql to find factorial of a number

c)PL/Sql for demo on for loop

d)PL/Sql for case structure

e)PL/Sql for simple loop.

f).PL/Sql to increase the vaue by 10.

g)PL/SQL for performing arithmetic operations

h)PL/SQL to find square, cube, double of a number

i)PL/SQL program to swap two numbers

j)PL/SQL program to find multiplication table

k)PL/SQL to determine whether a year is leap year or not

l)PL/SQL to insert item numbers from 1 to 5 using a loop.

m)PL/SQL to delete an item whose itemnum=4

n)PL/SQL program to display name, salary , commission and emp no of an employee based on the id.

o)PL/SQL program for inverting a number.

p)PL/SQL program to calculate the area of a circle for value of radius ranging from 3 to 7.

q)Write a program to print empno, ename, job and salary of an employee given empno.

r)Write a program to set the values for DA, HRA, GROSS fields of all the records of the given table.

s)Write a PL/SQL block that will accept a account number from the user , check if the users balance is less than the minimum balance then deduct Rs.100 from the balance as a fine.

t)Write a PL/SQL block that will accept a account number and incremented amount from the user, and increments the balance with the incremented amount.

u)Write a PL/SQL block that withdraws an amount of Rs.1000 then deposit’s an amount of Rs.10,000, updates the current balance ,then checks to see if the current balance of all the accounts does not exceed 50,000 if so then undo the deposit made.

PL/SQL Queries on Cursors

a)Write a PL/SQL block using the cursor to display details of all employee’s from emp2 table whose sum of sal and comm is > 2000.b)Write a PL/SQL block that display’s employee name and the salary of the first 5 employee’s having the highest salary.

c)Write a pl/sql program to find the sum of the salaries in a particular department.

d)PL/SQL to print all employees begin with S.

e)PL/SQL to print names of all managers

PL/SQL Queries on Procedures

a)PL/SQL to update salary of an employee whose empno and increment mentioned.

PL/SQL Queries on Exceptional Handling

a)Write a PL/SQL block which accepts the empno from the user and display the details of the employee. When the user enters an empno that is not in the emp table then the PL/SQL block must display an appropriate message to the user.

b)PL/SQL block that retrieves course table into cursor. Input course id to search. If course exists, print the information. If the course id does not exist, print user defined message by throwing user defined exception.

c)Write a program to handle ZERO_DIVIDE exception

PL/SQL Queries on Functions

a)Write a PL/SQL function which checks for an employee number in the table and returns 1 if present otherwise returns 0.

PL/SQL Queries on Tiggers

a)PL/SQL that creates a trigger that inserts or update values of ename, job as lower case strings.

BASIC QUERIES

1.Create student table with rno, name, class, mid1, mid2 and final marks in GE. Apply necessary constraints.

SQL> create table student(

2 rno number(3),

3 name varchar2(10),

4 class varchar2(5),

5 mid1 number(2) ,

6 mid2 number(2) ,

7 finalmarks number(3)

8 );

Table created.

2. Create course table with course id, course name(GE), instructor. Apply necessary constraints.

SQL> create table course(

2 courseid varchar(10) primary key,

3 coursename varchar2(10),

4 instructor varchar2(10)

5 );

3. Insert rows in both the tables

Student table:

SQL> insert into student(

2 rno,name,class,mid1,mid2,finalmarks)

3 values(101,'smith','ncs',35,41,78);

1 row created.

SQL>insert into student(


3 values(102,'john','nce',10,45,81) ;

1 row created.



3 values(103,'ashok','nct',47,46,91);

1 row created.

SQL>insert into student(


3 values(104,'Karthik','ncc',37,36,81);

1 row created.

SQL>1 insert into student(


3 values(105,'durga','nce',48,49,92)

4 ;

1 row created.



3 values(105,'durga','nce',48,49,92);

1 row created.

SQL> select * from student

2 ;

RNO NAME CLASS MID1 MID2 FINALMARKS

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

101 smith ncs 32 42 77

102 john nce 20 46 79

103 ashok nct 45 44 81

104 Karthik ncc 33 36 83

105 durga nce 41 43 90

Course table:

SQL> insert into course(

2 courseid,coursename,instructor)

3 values(631,'pcmt','Mrsmith');

1 row created.



3 values(632,'webdesign','john');

1 row created.



3 values(633,'solar','joe');

1 row created.



3 values(634,'acounting','latha');

1 row created.



3 values(635,'statics','rani');

1 row created.



3 values(636,'horti','vamshi');

1 row created

SQL> select * from course;

Output:

COURSEID COURSENAME INSTRUCTOR

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

631 pcmt Mrsmith

632 webdesign john

633 solar joe

634 acounting latha

635 statics rani

636 horti vamshi

4. Add a column total, average and course id to student table

SQL> alter table student

2 add(total number(5),

3 average number(3),

4 courseid number(10)

5 );

Table altered.

Output:

SQL> select * from student;

RNO NAME CLASS MID1 MID2 FINALMARKS TOTAL AVERAGE COURSEID

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

101 smith ncs 32 42 77 151 50

102 john nce 20 46 79 145 48

103 ashok nct 45 44 81 170 56

104 Karthik ncc 33 36 83 152 50

105 durga nce 41 43 90 174 58

5. SQL> alter table student

2 (add constraint fk foriegn key(courseid)

3 refernces courses(courseid));

1 row updated

6.Increase the width of name column to 15.

SQL> alter table student

2 modify

3 (name varchar2(15));

Table altered.

Output:

Name Null? Type

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

RNO NOT NULL NUMBER(3)

NAME VARCHAR2(15)

CLASS VARCHAR2(5)

MID1 NUMBER(2)

MID2 NUMBER(2)

FINALMARKS NUMBER(3)

TOTAL NUMBER(5)

AVERAGE NUMBER(3)

COURSEID NUMBER(10)

7. Change the marks of a particular student.

SQL> update student

2 set finalmarks=95

3 where name='john';

1 row updated.

Output:



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

101 smith ncs 32 42 77 152 50 634

102 john nce 20 46 79 145 48 631

103 ashok nct 45 44 81 170 56 636

104 Karthik ncc 33 36 83 152 50 632

105 durga nce 41 43 90 174 58 633

8. Add five marks grace in m1 subject to all students who scored less than 10.

SQL> select mid1+5

2 from student

3 where mid1<10;

Output :

MID1+5

--------

14

9. Update total and average columns.

SQL> update student

2 set total=mid1+mid2+finalmarks,

3 average=total/3;

5 rows updated.

Output:



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

101 smith ncs 32 42 77 151 50

102 john nce 20 46 79 145 48

103 ashok nct 45 44 81 17056

104 Karthik ncc 33 36 83 152 50

105 durga nce 41 43 90 174 58

10. Delete the students of solar processing course and rollback.

SQL> delete from course

2 where coursename='solar';

1 row deleted.

Output:

SQL> select * from course;


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

631 pcmt Mrsmith

632 webdesign john

634 accounting latha

635 statics rani

636 horti vamshi

SQL>rool back;

OUTPUT:


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

631 pcmt Mrsmith

632 webdesign john

633 solar joe

634 acounting latha

635 statics rani

636 horti vamshi

11. List all students data



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

101 smith ncs 32 42 77 151 50 634

102 john nce 20 46 79 145 48 631

103 ashok nct 45 44 81 170 56 636

104 Karthik ncc 33 36 83 152 50 632

105 durga nce 41 43 90 174 58 633

12. Display all the students belonging to NCE.

SQL> select *

2 from student

3 where class='nce';

Output:


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

102 john nce 10 45 81

105 durga nce 48 49 92

13. Display all the students of NCE who scored more than 150 marks.

SQL> select *

2 from student

3 where class='nce' and total>150;

Output:


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

102 john nce 10 45 81 163 54 632

105 durga nce 48 49 92 189 63 634

14. Display all students information in ascending order

1 select *

2 from student

3 order by 2;

Output:


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

103 ashok nct 47 46 91

102 john nce 10 45 81

105 durga nce 48 49 92

101 smith ncs 35 41 78

104 Karthik ncc 37 36 81

15. List all the students who do not belong to NCE.

SQL> select *

2 from student

3 where not(class='nce');

Output:


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

101 smith ncs 35 41 78 154 51 634

103 ashok nct 47 46 91 184 61 636

104 Karthik ncc 37 36 81 154 51 632

16. List all students having average between 60 and 80

SQL> select *

2 from student

3 where average>60 and average<80;

Output:


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

103 ashok nct 47 46 91 184 61 636

105 durga nce 48 49 92 189 63 633

17. List all students belonging to Mr.Smith.

SQL> select name

2 from student

3 where courseid=(select courseid from courses where instructor='Mrsmith');

Output:

NAME

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

John

18. List the rollno and class of the students

SQL> select rno,class

2 from student;

Output:

RNO CLASS

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

101 ncs

102 nce

103 nct

104 ncc

105 nce

19. List the students belonging to course id 631 and 633.

SQL>select courseid

2 from course

3 where courseid=631 or courseid=633;

no rows selected.

Output:

NAME

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

john

durga

20. Count the number of students in each course.

SQL> select courseid, count(*)

2 from student

3 group by courseid;

Output:

COURSEID COUNT(*)

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

631 2

632 1

633 1

634 1

21. List the course id having minimum of 2 students.

SQL> select courseid,count(*)

2 from student

3 group by courseid having count(*)>=2;

Output:

COURSEID COUNT(*)

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

631 2

22. Find the names of the students that begin with D.

SQL> select name

2 from student

3 where name like 'd%';

Output:

NAME

----------

Durga

23. Find the names of the students that ends with K.

SQL> select name

2 from student

3 where name like '%k';

Output:

NAME

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

Karthik

24. Find the names of the students who have S in their name.

SQL> select name

2 from student

3 where name like '%s%';

Output:

NAME

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

ashok

25.Display all students name in uppercase.

SQL> select upper(name)

2 from student;

Output:

UPPER(NAME)

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

SMITH

JOHN

ASHOK

KARTHIK

DURGA

26. Determine the names and length of each name.

SQL> select name, length(name)

2 from student;

Output:

NAME LENGTH(NAME)

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

smith 5

john 4

ashok 5

Karthik 7

durga 5

27. Display all students having 4 characters.

SQL> select name

2 from student

3 where name like'____';

Output:

NAME

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

john

Karthik

28. Display student details in descending order.

SQL> select *

2 from student

3 order by 2 desc;

Output:

RNO NAME CLASS MID1 MID2 FINALMARKS TOATL AVERAGE COURSEID

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

104 Karthik ncc 37 36 81 154 51 632

101 smith ncs 35 41 78 154 51 634

105 durga nce 48 49 92 189 63 633

102 john nce 14 45 81 163 54 631

103 ashok nct 47 46 91 184 61 636

29. List the rno, name, course name, course id.

SQL> select rno,name,student.courseid,coursename

2 from student,course

3 where student.courseid=course.courseid;

RNO NAME COURSEID COURSENAME

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

101 smith 634 acounting

102 john 631 PCMT

103 ashok 636 horti

104 Karthik 632 webdesign

105 durga 633 solar

30. To display all records of course table even if there is no matching course id in student.

SQL> select *

2 from course c,student s

3 where c.courseid=s.courseid;

Output:

COURSEID COURSENAME INSTRUCTOR RNO NAME CLASS MID1 MID2 FINALMARKS

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

TOATL AVERAGE COURSEID

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

634 acounting latha 101 smith ncs 35 41 78

154 51 634

631 pcmt Mrsmith 102 john nce 14 45 81

163 54 631

636 horti vamshi 103 ashok nct 47 46 91

184 61 636

632 webdesign john 104 Karthik ncc 37 36 81

154 51 632

633 solar joe 105 durga nce 48 49 92

189 63 633

31. Display the maximum and minimum total and average.

SQL> select max(total)"max-toatl",

2 min(total)"min-total",

3 max(average)"max-avg",

4 min(average)"miv-avg"

5 from student;

Output:

max-toatl min-total max-avg miv-avg

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

189 154 63 51

32. Find the highest marks obtained in PCMT.

SQL> select max(toatl)

2 from student

3 where courseid=(select courseid from courses where coursename='pcmt');

Output:

MAX(TOATL)

----------

163

33. To display the second highest and second lowest marks.

Sql>select max(total)

2 from student

3 where total not in(select max(total)from student)

Output:

MAX(TOTAL)

---------

184

34. Display all students who have not opted for wed designing.

SQL> select rno,name,class,coursename

2 from course c, student s

3 where c.courseid=s.courseid and coursename <> 'web designing';

RNO NAME CLASS COURSENAME

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

101 smith ncs acounting

102 john nce pcmt

103 ashok nct horti

105 durga nce solar

Queries on Emp and Dept tables:

1.Simple queries

1)List the name, job and salary from emp.

SQL> select ename,job,sal

2 from emp;

Output:-

ENAME JOB SAL

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

SMITH CLERK 800

ALLEN SALESMAN 1600

WARD SALESMAN 1250

JONES MANAGER 2975

MARTIN SALESMAN 1250

BLAKE MANAGER 2850

CLARK MANAGER 2450

SCOTT ANALYST 3000

KING PRESIDENT 5000

TURNER SALESMAN 1500

ADAMS CLERK 1100

JAMES CLERK 950

FORD ANALYST 3000

MILLER CLERK 1300

14 rows selected.

2.Select names of all managers.

SQL> select ename

2 from emp

3 where job='MANAGER';

Output:

ENAME

----------

JONES

BLAKE

CLARK

3.Display the details of the employee joined in Feb 1981.

SQL> select*

2 from emp

3 where to_char(hiredate,'mon-yyyy')='feb-1981';

Output:

EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO

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

7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30

7521 WARD SALESMAN 7698 22-FEB-81 1250 500 30

4.List the total salary of each employee.

SQL> select sum(sal+comm)

2 from emp;

Output:

SUM(SAL+COMM)

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

7800

5.List all employees who receive no commission.

SQL> select*

2 from emp

3 where comm null;

Output:


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

7369 SMITH CLERK 7902 17-DEC-80 800 20

7566 JONES MANAGER 7839 02-APR-81 2975 20

7698 BLAKE MANAGER 7839 01-MAY-81 2850 30

7782 CLARK MANAGER 7839 09-JUN-81 2450 10

7788 SCOTT ANALYST 7566 19-APR-87 3000 20

7839 KING PRESIDENT 17-NOV-81 5000 10

7876 ADAMS CLERK 7788 23-MAY-87 1100 20

7900 JAMES CLERK 7698 03-DEC-81 950 30

7902 FORD ANALYST 7566 03-DEC-81 3000 20

7934 MILLER CLERK 7782 23-JAN-82 1300 10

10 rows selected.

6.List all rows by converting null values in commission to 0

SQL> select empno,ename,nvl(comm,0)

2 from emp

3 where comm is null;

Output:

EMPNO ENAME NVL(COMM,0)

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

7369 SMITH 0

7566 JONES 0

7698 BLAKE 0

7782 CLARK 0

7788 SCOTT 0

7839 KING 0

7876 ADAMS 0

7900 JAMES 0

7902 FORD 0

7934 MILLER 0

10 rows selected.

7.To list names & salary greater than 2000.

SQL> select ename,sal

2 from emp

3 where sal>2000;

Output:

ENAME SAL

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

JONES 2975

BLAKE 2850

CLARK 2450

SCOTT 3000

KING 5000

FORD 3000

14 rows selected.

8.List names and salary less than 2000.


2 from emp

3 where sal<2000;

Output:

ENAME SAL

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

SMITH 800

ALLEN 1600

WARD 1250

MARTIN 1250

TURNER 1500

ADAMS 1100

JAMES 950

MILLER 1300

8 rows selected.

9.To list distinct jobs from employee.

SQL> select distinct(job)

2 from emp;

Output:

JOB

---------

ANALYST

CLERK

MANAGER

PRESIDENT

SALESMAN

10.List all employees who are neither clerks or managers.

SQL> select*

2 from emp

3 where job not in('clerk','manager');

Output:


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

7369 SMITH CLERK 7902 17-DEC-80 800 20




7654 MARTIN SALESMAN 7698 28-SEP-81 1250 1400 30





7844 TURNER SALESMAN 7698 08-SEP-81 1500 0 30

7876 ADAMS CLERK 7788 23-MAY-87 1100 20

7900 JAMES CLERK 7698 03-DEC-81 950 30

7902 FORD ANALYST 7566 03-DEC-81 3000 20

7934 MILLER CLERK 7782 23-JAN-82 1300 10

14 rows selected.

11)List all emps who earn between 1000 and 3000.

SQL>select*

2 from emp

3 where sal between 1000 and 3000;

Output:


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




7654 MARTIN SALESMAN 7698 28-SEP-81 1250 1400 30




7844 TURNER SALESMAN 7698 08-SEP-81 1500 0 30

7876 ADAMS CLERK 7788 23-MAY-87 1100 20

7902 FORD ANALYST 7566 03-DEC-81 3000 20

7934 MILLER CLERK 7782 23-JAN-82 1300 10

11 rows selected.

12)List all employees whose name start with ‘A’.

SQL> select ename

2 from emp

3 where ename like 'A%';

Output:

ENAME

----------

ALLEN

ADAMS

13)List all employees whose name end with ‘S’.

SQL> select ename

2 from emp

3 where ename like '%S';

Output:

ENAME

----------

JONES

ADAMS

JAMES

14)List all employees whose names contain ‘S’.

SQL> select ename

2 from emp

3 where ename like '%S%';

Output:

ENAME

----------

SMITH

JONES

SCOTT

ADAMS

JAMES

15)Select records whose names have only 4 characters.

SQL> select ename

2 from emp

3 where ename like '____';

Output:

ENAME

----------

WARD

KING

FORD

16)List the names and job who are clerk.

SQL> select ename,job

2 from emp

3 where job='CLERK';

Output:

ENAME JOB

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

SMITH CLERK

ADAMS CLERK

JAMES CLERK

MILLER CLERK

17)List empno, name, length of each name. Display the name in uppercase, lowercase, titlecase and use rpad and lpad.

SQL> select empno,upper(ename),lower(ename),initcap(ename),length(ename),lpad(ename,10,'*'),rpad(ename,10,'*')

2 from emp;

Output:

EMPNO UPPER(ENAM LOWER(ENAM INITCAP(EN LENGTH(ENAME) LPAD(ENAME RPAD(ENAME

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

7369 SMITH smith Smith 5 *****SMITH SMITH*****

7499 ALLEN allen Allen 5 *****ALLEN ALLEN*****

7521 WARD ward Ward 4 ******WARD WARD******

7566 JONES jones Jones 5 *****JONES JONES*****

7654 MARTIN martin Martin 6 ****MARTIN MARTIN****

7698 BLAKE blake Blake 5 *****BLAKE BLAKE*****

7782 CLARK clark Clark 5 *****CLARK CLARK*****

7788 SCOTT scott Scott 5 *****SCOTT SCOTT*****

7839 KING king King 4 ******KING KING******

7844 TURNER turner Turner 6 ****TURNER TURNER****

7876 ADAMS adams Adams 5 *****ADAMS ADAMS*****

EMPNO UPPER(ENAM LOWER(ENAM INITCAP(EN LENGTH(ENAME) LPAD(ENAME RPAD(ENAME

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

7900 JAMES james James 5 *****JAMES JAMES*****

7902 FORD ford Ford 4 ******FORD FORD******

7934 MILLER miller Miller 6 ****MILLER MILLER****

14 rows selected.

18)List the empno, ename, salary per day, salary per month, salary per year.

SQL>select empno,ename,sal/30 "per day",sal "per month",sal*12 "per year"

2 from emp;

Output:

EMPNO ENAME per day per month per year

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

7369 SMITH 26.6666667 800 9600

7499 ALLEN 53.3333333 1600 19200

7521 WARD 41.6666667 1250 15000

7566 JONES 99.1666667 2975 35700

7654 MARTIN 41.6666667 1250 15000

7698 BLAKE 95 2850 34200

7782 CLARK 81.6666667 2450 29400

7788 SCOTT 100 3000 36000

7839 KING 166.666667 5000 60000

7844 TURNER 50 1500 18000

7876 ADAMS 36.6666667 1100 13200

7900 JAMES 31.6666667 950 11400

7902 FORD 100 3000 36000

7934 MILLER 43.3333333 1300 15600

14 rows selected.

19)Ceil the salary of employees.

SQL> select ceil(sal)

2 from emp;

Output:

CEIL(SAL)

----------

800

1600

1250

2975

1250

2850

2450

3000

5000

1500

1100

CEIL(SAL)

----------

950

3000

1300

2)Update:

1)To set salary of smith to 1000.

SQL>update emp

2 set sal=1000

3* where ename='SMITH'

4 ;

Output:

1 row updated.

To view the update.

SQL> select *

2 from emp

3 where ename='SMITH';

Output:

EMPNO ENAME JOB MGR HIREDATE SAL COMM

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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 1000

20

2)To increase the salary by 10%.

SQL> update emp

2 set sal=sal+sal*0.1;

Output:

14 rows updated.

To view the update.

SQL> select*

2 from emp;

Output:


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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 1100

20

7499 ALLEN SALESMAN 7698 20-FEB-81 1760 300

30

7521 WARD SALESMAN 7698 22-FEB-81 1375 500

30

7566 JONES MANAGER 7839 02-APR-81 3272.5

20

7654 MARTIN SALESMAN 7698 28-SEP-81 1375 1400

30

7698 BLAKE MANAGER 7839 01-MAY-81 3135

30

7782 CLARK MANAGER 7839 09-JUN-81 2695

10

7788 SCOTT ANALYST 7566 19-APR-87 3300

20

7839 KING PRESIDENT 17-NOV-81 5500

10

7844 TURNER SALESMAN 7698 08-SEP-81 1650 0

30

7876 ADAMS CLERK 7788 23-MAY-87 1210

20

7900 JAMES CLERK 7698 03-DEC-81 1045

30

7902 FORD ANALYST 7566 03-DEC-81 3300

20

7934 MILLER CLERK 7782 23-JAN-82 1430

10

14 rows selected.

3)Update salary and commission for manager with 2000 and 800.

SQL>update emp

2 set sal=2000,comm=800


Output:

3 rows updated.

To view the update.

SQL> select*

2 from emp;

Output:


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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 1100

20


30


30


20


30


30


10


20


10


30

7876 ADAMS CLERK 7788 23-MAY-87 1210

20

7900 JAMES CLERK 7698 03-DEC-81 1045

30

7902 FORD ANALYST 7566 03-DEC-81 3300

20

7934 MILLER CLERK 7782 23-JAN-82 1430

10

14 rows selected.

4)Delete the records of clerk.

SQL> delete from

2 emp where job='CLERK';

Output:

4 rows deleted.

To view whether rows are delted or not.

SQL> select *

2 from emp;

Output:


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

DEPTNO

----------


30


30

7566 JONES MANAGER 7839 02-APR-81 2975

20


30


30


20


10


30

7902 FORD ANALYST 7566 03-DEC-81 3000

20

10 rows selected.

3.Order by caluse

1)List the employees in ascending order account to experience.

SQL> select ename,job,months_between(sysdate,hiredate)/12 as "no of months"

2 from emp order by 3;

Output:

ENAME JOB no of months

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

ADAMS CLERK 26.2891545

SCOTT ANALYST 26.3832405

MILLER CLERK 31.6224878

JAMES CLERK 31.7595846

FORD ANALYST 31.7595846

KING PRESIDENT 31.8052835

MARTIN SALESMAN 31.9423803

TURNER SALESMAN 31.9961437

CLARK MANAGER 32.2434555

BLAKE MANAGER 32.3482943

JONES MANAGER 32.4289394

WARD SALESMAN 32.5418426

ALLEN SALESMAN 32.547219

SMITH CLERK 32.7219502

14 rows selected.

2)List the emp details in descending order a/c to sal.

SQL> select *

2 from emp

3 order by sal desc;

Output:


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

DEPTNO

----------


10


20

7902 FORD ANALYST 7566 03-DEC-81 3000

20


20


30


10


30


30

7934 MILLER CLERK 7782 23-JAN-82 1300

10


30


30

7876 ADAMS CLERK 7788 23-MAY-87 1100

20

7900 JAMES CLERK 7698 03-DEC-81 950

30

7369 SMITH CLERK 7902 17-DEC-80 800

20

14 rows selected.

3)Emp details in ascending order working in dept no 10and 30.

SQL> select *

2 from emp

3 where deptno in(10,30) order by ename desc;

Output:


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

DEPTNO

----------


30


30

7934 MILLER CLERK 7782 23-JAN-82 1300

10


30


10

7900 JAMES CLERK 7698 03-DEC-81 950

30


10


30


30

9 rows selected.

4)List the names and empno of manager who earn more than 2600 and display in alphabetical order.

SQL> select empno,ename

2 from emp

3 where job='MANAGER' and sal>2600 order by ename;

Output:

EMPNO ENAME

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

7698 BLAKE

7566 JONES

4.Data Functions

1)List all employees who joined in a particular year.

SQL> select *

2 from emp

3 where to_char(hiredate,'yy')=81;

Output:


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

DEPTNO

----------


30


30


20


30


30


10


10


30

7900 JAMES CLERK 7698 03-DEC-81 950

30

7902 FORD ANALYST 7566 03-DEC-81 3000

20

10 rows selected.

2)Add 2 months to date of join of employee with id=7839.

SQL> update emp set hiredate=add_months(hiredate,2)

2 where empno=7839;

Output:

1 row updated.

To view the update.

SQL> select *

2 from emp;

Output:


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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 800

20


30


30


20


30


30


10


20

7839 KING PRESIDENT 17-JAN-82 5000

10


30

7876 ADAMS CLERK 7788 23-MAY-87 1100

20

7900 JAMES CLERK 7698 03-DEC-81 950

30

7902 FORD ANALYST 7566 03-DEC-81 300

20

7934 MILLER CLERK 7782 23-JAN-82 1300

10

14 rows selected.

SQL> select empno,hiredate

2 from emp

3 where empno=7839;

EMPNO HIREDATE

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

7839 17-JAN-82

3)To find the employee hired recently.

SQL> select*

2 from emp

3 where hiredate=(select max(hiredate) from emp);

Output:


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

DEPTNO

----------

7876 ADAMS CLERK 7788 23-MAY-87 1100

20

4)To list ename and hiredate whose hiredate matches with minimum hiredate.

SQL> select ename,hiredate

2 from emp

3 where hiredate=(select min(hiredate)from emp);

Output:

ENAME HIREDATE

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

SMITH 17-DEC-80

5)To list the employees experience in days in company.

SQL> select ename,(hiredate-sysdate)/24"experience"

2 from emp;

Output:

ENAME experience

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

SMITH -498.19434

ALLEN -495.48601

WARD -495.40268

JONES -493.77768

MARTIN -486.31934

BLAKE -492.56934

CLARK -490.94434

SCOTT -401.77768

KING -484.23601

TURNER -487.15268

ADAMS -400.36101

JAMES -483.56934

FORD -483.56934

MILLER -481.44434

14 rows selected.

6)To display emps who joined before 30-june-81 and 31-dec-81 from emp.

SQL> select*

2 from emp

3 where hiredate not between '30-JUNE-81' and '31-DEC-81';

Output:


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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 800

20


30


30


20


30


10


20

7876 ADAMS CLERK 7788 23-MAY-87 1100

20

7934 MILLER CLERK 7782 23-JAN-82 1300

10

9 rows selected.

5.Group by Clause.

1)To llist the deptno, no. of emps in each department.

SQL> select deptno,count(*)"total_employees"

2 from emp

3 group by deptno;

Output:

DEPTNO total_employees

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

10 3

20 5

30 6

2)List the total, maximum, minimum and average salary of emps job wise.

SQL> select sum(sal),max(sal),min(sal),avg(sal),job

2 from emp

3 group by job;

Output:

SUM(SAL) MAX(SAL) MIN(SAL) AVG(SAL) JOB

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

6000 3000 3000 3000 ANALYST

4150 1300 800 1037.5 CLERK

8275 2975 2450 2758.33333 MANAGER

5000 5000 5000 5000 PRESIDENT

5600 1600 1250 1400 SALESMAN

3)To list the job and no. of emp’s in each job.

SQL> select job,count(*)

2 from emp

3 group by job;

Output:

JOB COUNT(*)

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

ANALYST 2

CLERK 4

MANAGER 3

PRESIDENT 1

SALESMAN 4

4)To list the department and no. of clerks in each department.

SQL> select deptno,count(*)

2 from emp

3 where job='CLERK'

4 group by deptno;

Output:

DEPTNO COUNT(*)

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

10 1

20 2

30 1

5)To list the job which are done by minimum of 2 persons.

SQL> select job

2 from emp

3 group by job having count(*)>=2;

Output:

JOB

---------

ANALYST

CLERK

MANAGER

SALESMAN

6)List the no. of employees under each manager.

SQL> select mgr,count(*)

2 from emp

3 group by mgr;

Output:

MGR COUNT(*)

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

7566 2

7698 5

7782 1

7788 1

7839 3

7902 1

6 rows selected.

7)List the no. of employees along with their total salary in deptno 10 & 20.

SQL> select deptno,count(*),sum(sal)

2 from emp

3 group by deptno having deptno in(10,20);

Output:

DEPTNO COUNT(*) SUM(SAL)

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

10 3 8750

20 5 10875

8)Find the highest and lowest salary for each job.

SQL> select job,max(sal),min(sal)

2 from emp

3 group by job;

Output:

JOB MAX(SAL) MIN(SAL)

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

ANALYST 3000 3000

CLERK 1300 800

MANAGER 2975 2450

PRESIDENT 5000 5000

SALESMAN 1600 1250

9)Find the employees having similar jobs.

SQL> select job,count(*)

2 from emp

3 group by job having count(*)>1;

Output:

JOB COUNT(*)

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

ANALYST 2

CLERK 4

MANAGER 3

SALESMAN 4

10)To perform group on dept & sort on sum of salaries.

SQL> select sum(sal),deptno

2 from emp

3 group by deptno order by sum(sal);

Output:

SUM(SAL) DEPTNO

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

8750 10

9400 30

10875 20

11)Find all departments who have more than three employees


2 from emp

3 group by deptno having count(*)>3;

Output:

DEPTNO COUNT(*)

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

20 5

30 6

6.Subqueries:

1.List ename, deptno who works in ‘sales’.

SQL>select ename,deptno

2 from emp

3 where deptno=(select deptno from dept where dname='SALES');

Output:

ENAME DEPTNO

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

ALLEN 30

WARD 30

MARTIN 30

BLAKE 30

TURNER 30

JAMES 30

6 rows selected.

2)Display the information about employees working in new York.

SQL> select*

2 from emp

3 where deptno=(select deptno from dept where loc='NEW YORK');

Output:


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

DEPTNO

----------


10


10

7934 MILLER CLERK 7782 23-JAN-82 1300

10

3)List all the employees in dallas having salary greater than scott.

SQL>select empno,ename,emp.deptno,loc,sal

2 from emp,dept

3 where loc='DALLAS' and sal>(select sal from emp where ename='SCOTT');

Output:

EMPNO ENAME DEPTNO LOC SAL

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

7839 KING 10 DALLAS 5000

4)To list ename, job and empno who have same job as clerk and salary greater than manager.

SQL> select empno,job,ename,sal

2 from emp

3 where job='CLERK';

Output:

EMPNO JOB ENAME SAL

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

7369 CLERK SMITH 800

7876 CLERK ADAMS 1100

7900 CLERK JAMES 950

7934 CLERK MILLER 1300

5)To list the deptno in which maximum employees work.


2 from emp

3 group by deptno having count(*)=(select max(count(*)) from emp group by deptno);

Output:

DEPTNO COUNT(*)

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

30 6

6) To print names who draw maximum salary.


2 from emp

3 where sal=(select max(sal) from emp);

Output:

ENAME SAL

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

KING 5000

7)To find the second maximum salary.

SQL> select max(sal)smax

2 from emp

3 where sal<(select max(sal) from emp);

Output:

SMAX

----------

3000

8)To list the jobs, which are done by maximum no. of persons.

SQL> select job

2 from emp

3 group by job having count(*)=(select max(count(*)) from emp group by job);

Output:

JOB

---------

CLERK

SALESMAN

9)Find employees who earn more than the lowest salary in deptno 30.

SQL> select empno,ename

2 from emp

3 where sal>any(select min(sal) from emp where deptno=30);

Output:

EMPNO ENAME

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

7499 ALLEN

7521 WARD

7566 JONES

7654 MARTIN

7698 BLAKE

7782 CLARK

7788 SCOTT

7839 KING

7844 TURNER

7876 ADAMS

7902 FORD

7934 MILLER

12 rows selected.

10)Find the employees who earn lowest salary in each department.

SQL>select*

2 from emp

3 where sal in(select min(sal) from emp group by deptno);

Output:


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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 800

20

7900 JAMES CLERK 7698 03-DEC-81 950

30

7934 MILLER CLERK 7782 23-JAN-82 1300

10

11)Find employees who earn less than every employee in deptno 30.

SQL> select*

2 from emp

3 where sal<all(select min(sal) from emp where deptno=30);

Output:


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

7369 SMITH CLERK 7902 17-DEC-80 800 20

12)Find the job with highest avg salaries.

SQL> select job,sal

2 from emp

3 where sal in(select max(avg(sal)) from emp group by job);

Output:

JOB SAL

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

PRESIDENT 5000

13)List all employees in deptno 10 having the same job as any one in deptno 30.

SQL> select*

2 from emp

3 where deptno=10 and job in(select job from emp where deptno=30);

Output:


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

7934 MILLER CLERK 7782 23-JAN-82 1300 10


14)Display the list of employees who have same job & salary as turner.

SQL> select ename,job,sal

2 from emp

3 where(job,sal) in (select job,sal from emp where ename='TURNER');

Output:

ENAME JOB SAL

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

TURNER SALESMAN 1500

15 )Display the information about employees working in the department in the which miller work.

SQL> select *

2 from emp

3 where deptno=(select deptno from emp where ename='MILLER');

Output:

EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTN

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



7934 MILLER CLERK 7782 23-JAN-82 1300 10

16)Display the emp details whose salary is greater than any of their managers.

SQL> select*

2 from emp

3 where sal>any(select sal from emp where job='ANALYST');

Output:


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


7.Group Functions:

1)Count the managers of the company.

SQL>select count(*)

2 from emp


Output:

COUNT(*)

----------

3

2)Find the highest, lowest and difference of salary.

SQL> select max(sal)"max",min(sal)"min",max(sal)-min(sal)"diff"

2 from emp;

Output:

max min diff

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

5000 800 4200

3)Find the average salary of salesman.

SQL> select avg(sal)

2 from emp

3 where job='SALESMAN';

Output:

AVG(SAL)

----------

1400

4)Find the average annual salary of all employees.

SQL> select avg(sal*12)

2 from emp;

Output:

AVG(SAL*12)

-----------

24878.5714

5)Find the no. of employees in deptno 30 who receive commission.

SQL> select count(*)"no_of_emp"

2 from emp

3 where deptno=30 and comm is not null;

Output:

no_of_emp

----------

4

8.Joins:

1) List empno, name, designation, salary, deptname and location.

SQL> select empno,ename,job,sal,dname,loc

2 from emp,dept

3 where emp.deptno=dept.deptno;

Output:

EMPNO ENAME JOB SAL DNAME LOC

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

7369 SMITH CLERK 800 RESEARCH DALLAS

7499 ALLEN SALESMAN 1600 SALES CHICAGO

7521 WARD SALESMAN 1250 SALES CHICAGO

7566 JONES MANAGER 2975 RESEARCH DALLAS

7654 MARTIN SALESMAN 1250 SALES CHICAGO

7698 BLAKE MANAGER 2850 SALES CHICAGO

7782 CLARK MANAGER 2450 ACCOUNTING NEW YORK

7788 SCOTT ANALYST 3000 RESEARCH DALLAS

7839 KING PRESIDENT 5000 ACCOUNTING NEW YORK

7844 TURNER SALESMAN 1500 SALES CHICAGO

7876 ADAMS CLERK 1100 RESEARCH DALLAS

7900 JAMES CLERK 950 SALES CHICAGO

7902 FORD ANALYST 3000 RESEARCH DALLAS

7934 MILLER CLERK 1300 ACCOUNTING NEW YORK

14 rows selected.

2)List the employee names and their managers.

SQL> select e1.ename"WORKER",e2.ename"MANAGER"

2 from emp e1,emp e2

3 where e1.empno=e2.mgr;

Output:

WORKER MANAGER

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

FORD SMITH

BLAKE ALLEN

BLAKE WARD

KING JONES

BLAKE MARTIN

KING BLAKE

KING CLARK

JONES SCOTT

BLAKE TURNER

SCOTT ADAMS

BLAKE JAMES

JONES FORD

CLARK MILLER

13 rows selected.

3)List the empno, ename, salary, deptname where salary < 5000.

SQL> select empno,ename,sal,dname

2 from emp,dept

3 where emp.deptno=dept.deptno and sal<5000;

Output:

EMPNO ENAME SAL DNAME

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

7369 SMITH 800 RESEARCH

7499 ALLEN 1600 SALES

7521 WARD 1250 SALES

7566 JONES 2975 RESEARCH

7654 MARTIN 1250 SALES

7698 BLAKE 2850 SALES

7782 CLARK 2450 ACCOUNTING

7788 SCOTT 3000 RESEARCH

7844 TURNER 1500 SALES

7876 ADAMS 1100 RESEARCH

7900 JAMES 950 SALES

7902 FORD 3000 RESEARCH

7934 MILLER 1300 ACCOUNTING

13 rows selected.

9.Set Operators:

1)List jobs of deptno 10 7 20.

SQL> select job

2 from emp

3 where deptno=10

4 union

5 select job from emp where deptno=20;

Output:

JOB

---------

ANALYST

CLERK

MANAGER

PRESIDENT

2)Common jobs in deptno 10 & 20.

SQL> select job

2 from emp

3 where deptno=10

4 intersect


Output:

JOB

---------

CLERK

MANAGER

3)List unique jobs to deptno 10 but not in 20.

SQL>select job from emp where deptno=10

2 minus


Output:

JOB

---------

PRESIDENT

4)All jobs in deptno 10 & 20.

SQL> select job from emp where deptno=10

2 union all

3 select job

4 from emp

5 where deptno=20;

Output:

JOB

---------

MANAGER

PRESIDENT

CLERK

CLERK

MANAGER

ANALYST

CLERK

ANALYST

8 rows selected.

10.Views:

1)Create a view of salaries and ename of emp table.

SQL> create view vv_sal_name as

2 select ename,sal

3 from emp;

Output:

View created.

2)Update the salary of view.

SQL> update vv_sal_name set sal=800

2 where ename='SMITH';

Output:

1 row updated.

3)Drop the view.

SQL> drop view vv_sal_name;

Output:

View dropped.

11.Co-related subquery and multicolumn subquery:

1)List the name, salary, deptno of employees earning salary more than the average salary of their department.

SQL>select ename,sal,deptno

2 from emp e

3 where sal>(select avg(sal)

4 from emp m

5 where e.deptno=m.deptno);

Output:

ENAME SAL DEPTNO

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

ALLEN 1600 30

JONES 2975 20

BLAKE 2850 30

SCOTT 3000 20

KING 5000 10

FORD 3000 20

6 rows selected.

2)To print the employees who draw firstthree maximum salaries.

SQL> select rownum,ename,sal

2 from(select ename,sal from emp order by sal desc)

3 where rownum<=3;

Output:

ROWNUM ENAME SAL

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

1 KING 5000

2 SCOTT 3000

3 FORD 3000

3)To print all the employees who have atleast one person reporting to them.

SQL>select ename,job

2 from emp e

3 where exists(select mgr from emp where mgr=e.empno);

Output:

ENAME JOB

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

JONES MANAGER

BLAKE MANAGER

CLARK MANAGER

SCOTT ANALYST

KING PRESIDENT

FORD ANALYST

6 rows selected.

12.Miscellaneous:

1)Create a table to hold records whose salary < 2000.

SQL> create table emp3 as

2 select*

3 from emp

4 where sal<2000;

Output:

Table created.

2)Create table to hold ename, empno, job and salary columns.

SQL> create table emp4 as

2 select empno,ename,job,sal

3 from emp;

Output:

Table created.

3)Insert into the above table where ename starts with ‘s’.

SQL> insert into emp4

2 select empno,ename,job,sal

3 from emp

4 where ename like's%';

Output:

0 rows created.

13.DDL Queries:

1)Create employee table.

SQL>create table employee(

2 empno number(4),

3 ename varchar2(20),

4 job varchar2(15),

5 sal number(5),

6 comm number(3),

7 hiredate date,

8 deptno number(2));

Output:

Table created.

2)Create depts table.

SQL>create table depts(

2 deptno number(2),

3 dname varchar2(20),

4 loc varchar2(10));

Output:

Table created.

3)Modify the job column to hold 10 characters.

SQL> alter table employee

2 modify job varchar2(10);

Output:

Table altered.

4)Add primary key constraint to column.


2 add constraint pk_key primary key(empno);

Output:

Table altered.

5)Add foreign key constraint to dept setting primary key to depts..

SQL> alter table depts

2 add constraint pk_key1 primary key(deptno);

Output:

Table altered.

SQL>alter table employee

2 add(constraint fk_key foreign key(deptno) references depts);

Output:

Table altered.

6)Drop constraint:


2 drop constraint fk_key;

Output:

Table altered.

1.SQL Statement for creating the table.

a)Table Name:client_master

SQL> create table client_master(

2 clientno varchar2(6) primary key,

3 name varchar2(20) not null,

4 city varchar2(15),

5 pincode number(8),

6 state varchar2(15),

7 baldue number(10,2));

Output:

Table created.

b)Table Name:product_master

SQL> create table product_master(

2 productno varchar2(6) primary key,

3 description varchar2(15) not null,

4 profitpercent number(4,2) not null,

5 unitmeasure varchar2(10) not null,

6 qtyonhand number(8) not null,

7 reorderlvl number(8) not null,

8 sellprice number(8,2) not null,

9 costprice number(8,2) not null,

10 constraint ck_product check(productno like 'p%'),

11 constraint ck_sell check(sellprice <> 0),

12 constraint ck_cost check(costprice <> 0));

Output:

Table created.

c) Table Name:salesman_master

SQL>create table salesman_master(

2 salesmanno varchar2(6) primary key,

3 salesmanname varchar2(20) not null,

4 address1 varchar2(30) not null,

5 address2 varchar2(30),

6 city varchar2(20),

7 pincode number(8),

8 state varchar2(20),

9 salamt number(8,2) not null,

10 tgttoget number(6,2) not null,

11 ytdsales number(6,2) not null,

12 remarks varchar2(60),

13 constraint ck_salesman check(salesmanno like 's%'),

14 constraint ck_sal check(salamt <> 0),

15 constraint ck_target checktgttoget <> 0));

Output:

Table created.

d)Table Name:sales_order

SQL> create table sales_order(

2 orderno varchar2(6) primary key,

3 clientno varchar2(6) references client_master,

4 orderdate date,

5 delyaddr varchar2(25),

6 salesmanno varchar2(6) references salesman_master,

7 delytype char(1) default 'f',

8 billedyn char(1),

9 delydate date,

10 orderstatus varchar2(10),

11 constraint ck_order check(orderno like 'o%'),

12 constraint ck_dely_type check(delytype in ('p','f')),

13 constraint ck_ord_status

14 check(orderstatus in ('in process', 'fulfilled','backorder','cancelled')));

Output:

Table created.

e)Table Name: sales_order_details

SQL> create table sales_order_details(

2 orderno varchar2(6) references sales_order,

3 productno varchar2(6) references product_master,

4 qtyordered number(8),

5 qtydisp number(8),

6 productrate number(10,2),

7 primary key(orderno,productno));

Output:

Table created.

2. SQL Statement for inserting data into their respective tables.

a)Data for client_master table:

SQL> insert into client_master(clientno,name,city,pincodeno,state,baldue)

2 values('c00001','ivanbayross','mumbai',400054,'maharashtra',15000);

1 row created.


2 values('c00002','mamtamuzumdar','madras',780001,'tamil nadu',0);

1 row created.


2 values('c00003','chhayabankar','mumbai',400057,'maharashtra',5000);

1 row created.


2 values('c00004','ashwinijoshi','bangalore',560001,'karnataka',0);

1 row created.


2 values('c00005','hanselcolaco','mumbai',400060,'maharashtra',2000);

Output:

1 row created.


2 values('c00006','deepaksharma','mangalore',560050,'karnataka',0);

Output:

1 row created.

To view the client_master table.

SQL> select *

2 from client_master;

Output:

CLIENT NAME CITY PINCODE STATE BALDUE

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

c00001 ivan bayross mumbai 400054 maharashtra 15000

c00002 mamta muzumdar madras 780001 tamil nadu 0

c00003 chhaya bankar mumbai 400057 maharashtra 5000

c00004 ashwini joshi bangalore 560001 karnataka 0

c00005 hansel colaco mumbai 400060 maharashtra 2000

c00006 deepak sharma mangalore 560050 karnataka 0

b) Data for product_master table:

SQL> insert intoproduct_master

2 values('p00001','t-shirts',5,'piece',200,50,350,250);

Output:

1 row created.

SQL> insert into product_master

2 values('p03453','shirts',6,'piece',150,50,500,350);

Output:

1 row created.


2 values('p06734','cotton jeans',5,'piece',100,20,600,450);

Output:

1 row created.


2 values('p07865','jeans',5,'piece',100,20,750,500);

Output:

1 row created.


2 values('p07868','trousers',2,'piece',150,50,850,550);

Output:

1 row created.


2 values('p07885','pull overs',2.5,'piece',80,30,700,450);

Output:

1 row created.


2 values('p07965','denim shirts',4,'piece',100,40,350,250);

Output:

1 row created.


2 values('p07975','lycra tops',5,'piece',70,30,300,175);

Output:

1 row created.


2 values('p08865','skirts',5,'piece',75,30,450,300);

Output:

1 row created.

To view the product_master table.

SQL> select *

2 from product_master;

Output:

PRODUC DESCRIPTION PROFITPERCENT UNITMEASUR QTYONHAND REORDERLVL SELLPRICE

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

COSTPRICE

----------

p00001 t-shirts 5 piece 200 50 350

250

p03453 shirts 6 piece 150 50 500

350

p06734 cotton jeans 5 piece 100 20 600

450

P07865 jeans 5 piece 100 20 750

500

p07868 trousers 2 piece 150 50 850

550

p07885 pull overs 2.5 piece 80 30 700

450

p07965 denim shirts 4 piece 100 40 350

250

p08865 skirts 5 piece 75 30 450

300

p07975 lycra tops 5 piece 70 30 300

175

9 rows selected.

c) Data for sales_master table:

SQL> insert into salesman_master

2 values( 's00001','aman','a/14','worli','mumbai',400002,'maharashtra',3000,100,50,'good');

Output:

1 row created.


2 values('s00002','omkar','65','nariman','mumbai',400001,'maharashtra',3000,200,100,'good');

Output:

1 row created.


2 values('s00003','raj','p-7','bandra','mumbai',400032,'maharashtra',3000,200,100,'good');

Output:

1 row created.


2 values( 's00004','ashish','a/5','juhu','bombay',400044,'maharashtra',3500,200,150,'good');

Output:

1 row created.

To view the salesman_master table.

SQL> select *

2 from salesman_master;

Output:

SALESM SALESMANNAME ADDRESS1 ADDRESS2 CITY PINCODE

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

STATE SALAMT TGTTOGET YTDSALES REMARKS

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

s00001 aman a/14 worli mumbai 400002

maharashtra 3000 100 50 good

s00002 omkar 65 nariman mumbai 400001


s00003 raj p-7 bandra mumbai 400032


s00004 ashish a/5 juhu bombay 400044


4 rows selected.

d) Data for sales_order table:

SQL> insert into sales_order(orderno,orderdate,clientno,delytype,billedyn,salesmanno,delydate,orderstatus)

2 values('o19001','12-june-02','c00001','f','n','s00001','20-july-02','in process');

Output:

1 row created.

SQL> insert into sales_order(orderno,orderdate,clientno,delytype,billedyn,salesmanno,delydate,orderstatus)

2 values('o19002','25-june-02','c00002','p','n','s00002','27-july-01','cancelled');

Output:

1 row created.

SQL> insert into sales_order(orderno,orderdate,clientno,delytype,billedyn,salesmanno,deltdate,orderstatus)

2 values('o19003','18-feb-02','c00003','f','y','s00003','20-feb-02','fulfilled');

Output:

1 row created.

SQL> insert into sales_order(orderno,orderdate,clientno,delytype,billedyn,salesmanno,deltdate,orderstatus)

2 values('o19004','03-apr-02','c00001','f','y','s00001','07-apr-02','fulfilled');

Output:

1 row created.

SQL> insert into

sales_order(orderno,orderdate,clientno,delytype,billedyn,salesmanno,deltdate,orderstatus)

2 values('o46866','20-may-02','c00004','p','n','s00002','22-may-02','cancelled');

Output:

1 row created.

SQL> insert into

sales_order(orderno,orderdate,clientno,delytype,billedyn,salesmanno,deltdate,orderstatus)

2 values('o19008','24-may-02','c00005','f','n','s00004','26-july-96','in process');

Output:

To view the sales_order table.

SQL> select *

2 from sales_order;

Output:

ORDERN CLIENT ORDERDATE DELYADDR SALESM D B DELTDATE

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

ORDERSTATUS

----------

o19001 c00001 12-JUN-02 s00001 f n 20-JUL-02

in process

o19002 c00002 25-JUN-02 s00002 p n 27-JUL-01

cancelled

o19003 c00003 18-FEB-02 s00003 f y 20-FEB-02

fulfilled

o19004 c00001 03-APR-02 s00001 f y 07-APR-02

fulfilled

o46866 c00004 20-MAY-02 s00002 p n 22-MAY-02

cancelled

o19008 c00005 24-MaY-02 s00004 f n 26-JULY-96

in process

6 rows selected.

e) Data for sales_order_details table:

SQL> insert into sales_order_details(orderno,productno,qtyordered,qtydisp,productrate)

2 values('o19001','p00001',4,4,525);

Output:

1 row created.


2 values('o19001','p07965',2,1,8400);

Output:

1 row created.


2 values('o19001','p07885',2,1,5250);

Output:

1 row created.


2 values('o46865','p07868',3,3,150);

Output:

1 row created.

SQL> insert into sales_order_details(orderno,productno,qtyorder,qtydisp,productrate)

2 values('o19003','p03453',2,2,1050);

Output:

1 row created.


2 values('o19003','p06734',1,1,12000);

Output:

1 row created.


2 values('o46866','p07965',1,0,8400);

Output:

1 row created.

SQL>insert into sales_order_details(orderno,productno,qtyordered,qtydisp,productrate)

2 values('o46866','p07975',1,0,1050);

Output:

1 row created.

ORDERNO PRODUCTNO qtyordered qtydisp productrate

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

o19001 p00001 4 4 525

o19001 p07965 2 1 8400

o19001 p07885 2 1 5250

o46865 p07868 3 3 150

o19003 p03453 2 2 1050

o19003 p06734 1 1 12000

o46866 p07965 1 0 8400

o46866 p07975 1 0 1050

3.SQL Statement for retrieving records from a table.

a)Find out the names of all the clients.

SQL> select name


Output:

NAME

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

ivan bayross

mamta muzumdar

chhaya bankar

ashwini joshi

hansel colaco

deepak sharma

6 rows selected.

b)Retrieve the entire contents of the client_master table.

SQL> select *


Output:


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





c00005 hansel colaco mumbai 400060 maharastra 2000


6 rows selected.

c)Retrieve the list of names, city and the state of all the clients.

SQL> select name,city,state


Output:

NAME CITY STATE

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

ivan bayross mumbai maharashtra

mamta muzumdar madras tamil nadu

chhaya bankar mumbai maharashtra

ashwini joshi bangalore karnataka

hansel colaco mumbai maharastra

deepak sharma mangalore karnataka

6 rows selected.

d)List the various products available from the product_master table.

SQL> select description


Output:

DESCRIPTION

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

t-shirts

shirts

cotton jeans

jeans

trousers

pull overs

denim shirts

skirts

lycra tops

9 rows selected.

e)List all the clients who are located in Mumbai.

SQL> select *

2 from client_master

3 where city='mumbai';

Output:


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



c00005 hansel colaco mumbai 400060 maharashtra 2000

f)Find the names of salesman who have a salary equal to Rs.3000.

SQL>select salesman_name

2 from salesman_master

3 where salamt=3000;

Output:

SALESMANNAME

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

aman

omkar

raj

4.SQL Statement for update records in a table.

a)Change the city of clientno ‘c00005’ to ‘bangalore’.

SQL> update client_master

2 set city='bangalore'

3 where clientno='c00005';

Output:

1 row updated.

To check whether table is updated or not.

SQL> select *


Output:


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





c00005 hansel colaco bangalore 400060 maharashtra 2000


6 rows selected.

b)Change the baldue of clientno’c00001’ to Rs.1000.


2 set baldue=1000

3 where clientno='c00001';

Output:

1 row updated.


SQL> select *


Output:


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





c00005 hansel colaco bangalore 400060 maharashtra 2000


6 rows selected.

c)Change the cost price of ‘Trousers’ to Rs.950.00.

SQL>update product_master

2 set costprice=950.00

3 where description='trousers';

Output:

1 row updated.


SQL> select *


Output:

PRODUC DESCRIPTION PROFITPERCENT UNITMEASURE QTYONHAND REORDERLVL SELLPRICE

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

COSTPRICE

----------

p00001 t-shirts 5 piece 200 50 350

250

p03453 shirts 6 piece 150 50 500

350

p06734 cotton jeans 5 piece 100 20 600

450

p07865 jeans 5 piece 100 20 750

500


950


450

p07965 denim shirts 4 piece 100 40 350

250

p08865 skirts 5 piece 75 30 450

300


175

9 rows selected.

d)Change the city of the salesman to pune.


2 set city='pune';

Output:

6 rows updated.


SQL> select *

from client_master;

Output:


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

c00001 ivan bayross pune 400054 maharashtra 1000

c00002 mamta muzumdar pune 780001 tamil nadu 0

c00003 chhaya bankar pune 400057 maharashtra 5000

c00004 ashwini joshi pune 560001 karnataka 0

c00005 hansel colaco pune 400060 maharashtra 2000

c00006 deepak sharma pune 560050 karnataka 0

6 rows selected.

5.SQL Statement for deleting records in a table.

a)Delete all salesmen from the salesman_master whose salaries are equal to Rs.3500.

SQL> delete from salesman_master

2 where salamt=3500;

Output:

1 row deleted.

To check whether the row is deleted or not .

SQL> select *

2 from salesman_master;

Output:

SALESMANNO SALESMANNAME ADDRESS1 ADDRESS2 CITY PINCODE

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


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







b)Delete all products from product_master where the quantity on hand is equal to 100.

SQL> delete from product_master

2 where qtyonhand=100;

Output:

3 rows deleted.


SQL> select *

from product_master;

Output:


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

COSTPRICE

----------

p00001 t-shirts 5 piece 200 50 350

250

p03453 shirts 6 piece 150 50 500

350


950


450

p08865 skirts 5 piece 75 30 450

300


175

6 rows selected.

c)Delete from client_master where the column state holds the value ‘tamil nadu’.

SQL> delete from client_master

2 where state='tamil nadu';

Output:

1 row deleted.


SQL> select *


Output:


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




c00005 hansel colaco pune 400060 maharastra 2000


6.SQL Statement for altering the table structure.

a)Add a column called ‘telephone’ of data type ‘number’ and size=’10’ to the client_master table.

SQL> alter table client_master

2 add(telephone number(10));

Output:

Table altered.

To check whether the table is altered or not.

SQL> select *


Output:

CLIENT NAME CITY PINCODE STATE BALDUE TELEPHONE

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




c00005 hansel colaco pune 400060 maharastra 2000


b)Change the size of sellprice column in product_master to 10,2.

SQL> alter table product_master

2 modify(sellprice number(10,2));

Output:

Table altered.

To check whether the table is altered or not.

SQL> select *


Output:


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

COSTPRICE

----------

p00001 t-shirts 5 piece 200 50 350

250

p03453 shirts 6 piece 150 50 500

350


950


450

p08865 skirts 5 piece 75 30 450

300


175

6 rows selected.

7.SQL Statement for renaming the table.

a)Change the name of the salesman_master to sman_mast.

SQL> rename salesman_master to sman_mast;

Output:

Table renamed.

To view whether the table is renamed or not.

SQL>select *

2 from sman_mast;

Output:

SALESM SALESMANNAME ADDRESS1 ADDRESS2 CITY PINCODE

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


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







s00004 ashish a/5 juhu bombay 400044


4 rows selected.

9.INTERACTIVE SQL

1.Generate SQL Statements to perform the following computations on table data:

a) Listing of the names of all clients having ‘a’ as the second letter in their names.

SQL> select name


3 where name like'_a%';

Output:

NAME

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

mamta muzumdar

hansel colaco

b)Listing of clients who stay in a city whose first letter is’m’.

SQL>select clientno,name


3 where city like'm%'

4 ;

Output:

CLIENT NAME

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

c00001 ivan bayros

c00002 mamta muzumdar

c00003 chhaya bankar

c00005 hansel colaco

c00006 deepak Sharma

c)List all clients who stay in ’bangalore’ or ‘mangalore’.

SQL> select clientno,name


3 where city in('bangalore','mangalore');

Output:

CLIENT NAME

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

c00004 ashwini joshi

c00006 deepak sharma

d)List all clients whose baldue is greater than value 10000.



3 where baldue>10000;

Output:

CLIENT NAME

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

c00001 ivan bayros

e)Print the information from sales_order table for orders placed in the month of june.

SQL> select *

2 from sales_order

3 where to_char(orderdate,'mon')='jun';

Output:

ORDERN CLIENT ORDERDATE DELYADDR SALESM D B DELTDATE ORDERSTATU

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

o19001 c00001 12-JUN-02 s00001 f n 20-JUL-02

in process

f)Displayingthe order information of clintno ‘c00001’ and ‘c00002’.

SQL> select *

2 from sales_order

3 where clientno in('c00001','c00002');

Output:

ORDERN CLIENT ORDERDATE DELYADDR SALESM D B DELTDATE ORDERSTATU

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

o19001 c00001 12-JUN-02 s00001 f n 20-JUL-02

in process

g)List products whose selling price is greater than 500 and less than or equal to 750.

SQL> select productno,description

2 from product_master

3 where sellprice>500 and sellprice<750;

Output:

PRODUC DESCRIPTION

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

p06734 cotton jeans

p07885 pull overs

h)Listing of products whose selling price is more than 500 with the new selling price calculated as original selling price plus 15%.

SQL> select productno,description,sellprice,sellprice*15 new_price


3 where sellprice>500;

Output:

PRODUC DESCRIPTION SELLPRICE NEW_PRICE

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

p06734 cotton jeans 600 9000

p07865 jeans 750 11250

p07868 trousers 850 12750

p07885 pull overs 700 10500

i)Listing of names,city and state of clients who are not in the state of ‘maharashtra’.

SQL> select name,city,state


3 where state not in('maharashtra');

Output:

NAME CITY STATE

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

mamta muzumdar madras tamil nadu

ashwini joshi bangalore karnataka

deepak sharma mangalore karnataka

j)Count the total number of orders.

SQL> select count(orderno)"no. of order"

2 from sales_order;

Output:

no. of order

----------

4

k)Calculating the average price of all the products.

SQL> select avg(sellprice)


Output:

AVG(SELLPRICE)

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

538.888889

l)Determining the maximum and minimum price for the product prices.

SQL> select max(sellprice)max_price,min(sellprice)


Output:

MAX_PRICE MIN(SELLPRICE)

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

850 300

m)Count the number of products having price greater than or equal to 500.

SQL> select count(productno)


3 where sellprice<=1500;

Output:

COUNT(PRODUCTNO)

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

9

n)Find all the products whose qtyonhand is less than rereorder level.



3 where qtyonhand<reorderlvl;

Output:

no rows selected

2.SQL Statements for Date Manipulation:

a)Display the order number and day on which clients placed their order.

SQL> select orderno,to_char(orderdate,'day')

2 from sales_order;

Output:

ORDERN TO_CHAR(O

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

o19001 wednesday

o19002 tuesday

o19003 monday

o46866 Monday

b)Display the month (in alphabets) and date when the order must be delivered.

SQLselect to_char(deltdate,'month'),deltdate

2 from sales_order

3 order by to_char(deltdate,'month');

Output:

TO_CHAR(D DELTDATE

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

february 20-FEB-02

july 20-JUL-02

july 27-JUL-01

may 22-MAY-02

c)List the orderdate in the format ‘dd-month-yy’.

SQL> select to_char(orderdate,'dd-mm-yy')

2 from sales_order;

Output:

TO_CHAR(

--------

12-06-02

25-06-02

18-02-02

20-05-02

d)Find the date, 15 days after today’s date.

SQL> select sysdate+15

2 from dual;

Output:

SYSDATE+1

---------

08-OCT-13

10.INTERACTIVE SQL

1.SQL Statements for using having and group by clauses:

a)Printing the description and total quantity sold for each product.

SQL> select description,sum(qtydisp)

2 from product_master,sales_order_details

3 where product_master.productno=sales_order_details.productno

4 group by description;

Output:

DESCRIPTION SUM(QTYDISP)

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

cotton jeans 1

denim shirts 1

lycra tops 0

pull overs 1

shirts 2

t-shirts 4

6 rows selected.

b)Finding the value of each product sold.

SQL> select sales_order_details.productno,product_master.description,

2 sum(sales_order_details.qtydisp*sales_order_details.productrate)"sales per product"

3 from sales_order_details,product_master

4 where product_master.productno=sales_order_details.productno

5 group by sales_order_details.productno,product_master.description;

Output:

PRODUC DESCRIPTION sales per product

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

p00001 t-shirts 2100

p03453 shirts 2100

p06734 cotton jeans 12000

p07885 pull overs 5250

p07965 denim shirts 8400

p07975 lycra tops 0

6 rows selected.

c)Calculating the average quantity sold for each client that has a maximum order value of 15000.00.

SQLselect cm.clientno,cm.name,avg(sod.qtydisp)"avg.sales"

2 from sales_order_details sod,sales_order so,client_master cm

3 where cm.clientno=so.clientno and so.orderno=sod.orderno

4 group by cm.clientno,name having max(sod.qtyordered*sod.productrate)>15000;

Output:

CLIENT NAME avg.sales

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

c00001 ivan bayros 2

e)Listing the products and orders from customers who have ordered less than 5 units of’pull overs’.

SQL> select sales_order_details.productno,sales_order_details.orderno

2 from sales_order_details,sales_order,product_master

3 where sales_order.orderno=sales_order_details.orderno

4 and product_master.productno=sales_order_details.productno

5 and sales_order_details.qtyordered<5 and product_master.description='pull overs';

Output:

PRODUC ORDERN

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

p07885 o19001

g)Finding the products and their quantities for the orders placed by clientno ‘c00001’ and ‘c00002’.

SQL> select so.clientno,sod.productno,pm.description,sum(qtyordered)"units ordered"

2 from sales_order so,sales_order_details sod,product_master pm,client_master cm

3 where so.orderno=sod.orderno and sod.productno=pm.productno

4 and so.clientno=cm.clientno

5 group by so.clientno,sod.productno,pm.description

6 having so.clientno='c00001' or so.clientno='c00002';

Output:

CLIENT PRODUC DESCRIPTION units ordered

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

c00001 p00001 t-shirts 4

c00001 p07885 pull overs 2

c00001 p07965 denim shirts 2

c00002 p00001 t-shirts 10

3.SQL Statements for exercises on Sub-queries:

a)Finding the non-moving products i.e. products not being sold.



3 where productno not in(select productno from sales_order_details);

Output:

PRODUC DESCRIPTION

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

p07865 jeans

p07868 trousers

p08865 skirts

b)Finding the names and complete address for the customer who has placed order number ‘o19001’.

SQLselect name,city,state,pincode


3 where clientno in (select clientno from sales_order where orderno='o19001');

Output:

NAME CITY STATE PINCODE

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

ivan bayros mumbai maharashtra 400054

c)Find the clients who have placed orders before the month of may’02.



3 where clientno in(select clientno from sales_order where to_char(orderdate,'mm,yy')<'may,02');

Output:

CLIENT NAME

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

c00001 ivan bayros

c00002 mamta muzumdar

c00003 chhaya bankar

c00004 ashwini joshi

d)Find out if the product ‘lycra tops’ has been ordered by any client and print the clientno, name to whom it was sold.



3 where clientno in(select clientno

4 from sales_order where orderno in(select orderno from sales_order_details where productno in(se

lect productno from product_master where description='lycratops')));

Output:

no rows selected

e)Find the names of clients who have placed orders worth Rs.10000 or more.

SQL> select name from client_master

2 where clientno in(select clientno from sales_order

3 where orderno in(select orderno from sales_order_details

4 where (qtyordered*productrate)>=10000));

Output:

NAME

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

ivan bayros

chhaya banker

Simple pl/sql queries

a)Pl/Sql to add two numbers

set serveroutput on;declarenum1 number(2):=&first;num2 number(2):=&second;num3 number(2);beginnum3:=num1+num2;dbms_output.put_line(num1||'+'||num2||'='||num3);end;OUTPUT:Enter value for first: 10old 2: num1 number(2):=&first;new 2: num1 number(2):=10;Enter value for second: 20old 3: num2 number(2):=&second;new 3: num2 number(2):=20;10+20=30

PL/SQL procedure successfully completed.

b)PL/Sql to find factorial of a numberset serveroutput on;declarenum1 number(2):=&number;fact number(10):=1;i number(2):=1;beginwhile i<=num1loopfact:=fact*i;i:=i+1;end loop;dbms_output.put_line('factorial of'||num1||'is'||fact);end;OUTPUT:Enter value for number: 10old 2: num1 number(2):=&number;new 2: num1 number(2):=10;factorial of10is3628800


c)PL/Sql for demo on for loopset serveroutput on; declare s number(2):=&number; e number(2):=&number; i number(2); begin for i in s..e loop dbms_output.put_line('i= '||i); end loop; end;OUTPUT:Enter value for number: 2old 2: s number(2):=&number;new 2: s number(2):=2;Enter value for number: 3old 3: e number(2):=&number;new 3: e number(2):=3;i= 2i= 3


d)PL/Sql for case structureset serveroutput on;declareoper char(1):='&operation';num1 number(2):=&first;num2 number(2):=&second;num3 number(2);begincasewhen oper='+' thennum3:=num1+num2;when oper='-' thennum3:=num1-num2;when oper='*' thennum3:=num1*num2;when oper='/' thennum3:=num1/num2;when oper='%' thennum3:=mod(num1,num2);

elsedbms_output.put_line('invalid operation');end case;dbms_output.put_line(num1||oper||num2||'='||num3);end;OUTPUT:Enter value for operation: +old 2: oper char(1):='&operation';new 2: oper char(1):='+';Enter value for first: 2old 3: num1 number(2):=&first;new 3: num1 number(2):=2;Enter value for second: 3old 4: num2 number(2):=&second;new 4: num2 number(2):=3;2+3=5

e)PL/Sql for simple loop.set serveroutput on;declarei number:=0;beginloopdbms_output.put_line('i='|| i);i:=i+2;exit when i>10;end loop;end;OUTPUT:i=0i=2i=4i=6i=8i=10


f)PL/Sql to increase the vaue by 10.declaren number(2);beginn:=&number;

n:=n+10;dbms_output.put_line('value of n ='||n);end;OUTPUT:Enter value for number: 10old 4: n:=&number;new 4: n:=10;value of n =20


g)PL/SQL for performing arithmetic operations

declare

v_num1 number(2):=&first_num;

v_num2 number(2):=&second_num;

v_res number(3);

v_choice char(1):='&choice';

begin

case v_choice

when '+' then v_res:=v_num1+v_num2;

when '-' then v_res:=v_num1-v_num2;

when '*' then v_res:=v_num1*v_num2;

when '/' then v_res:=v_num1/v_num2;

when '%' then v_res:=mod(v_num1,v_num2);

end case;

dbms_output.put_line(v_num1||' '||v_choice||' '||v_num2||' '||' = '||v_res);

end;

OUTPUT:

Enter value for first_num: 8

old 2: v_num1 number(2):=&first_num;

new 2: v_num1 number(2):=8;

Enter value for second_num: 9

old 3: v_num2 number(2):=&second_num;

new 3: v_num2 number(2):=9;

Enter value for choice: +

old 5: v_choice char(1):='&choice';

new 5: v_choice char(1):='+';

8 + 9 = 17


h) PL/SQL to find square, cube, double of a number

declare

v_num number(3);

v_sq v_num%type;

v_cube v_sq%type;

v_double v_cube%type;

declare

v_num:=&p_num;

v_sq:=v_num*v_num;

v_cube:=v_sq*v_num;

v_double:=v_num*2;

dbms_output.put_line('number is '||v_num);

dbms_output.put_line('square is '||v_sq);

dbms_output.put_line('cube is'||v_cube);

dbms_output.put_line('double is'||v_double);

end;

OUTPUT:

Enter value for p_num: 5

old 7: v_num:=&p_num;

new 7: v_num:=5;

number is 5

square is 25

cube is125

double is10


i) PL/SQL program to swap two numbers

declare

a number(2);

b number(2);

begin

a:=&num1;

b:=&num2;

dbms_output.put_line('before swap a='||a||'b='||b);

a:=a+b;

b:=a-b;

a:=a-b;

dbms_output.put_line('after swap a='||a||'b='||b);

end;

OUTPUT:

Enter value for num1: 15

old 5: a:=&num1;

new 5: a:=15;

Enter value for num2: 9

old 6: b:=&num2;

new 6: b:=9;

before swap a=15b=9

after swap a=9b=15


j)PL/SQL program to find multiplication table

declare

v_multiplicand number(2):=&multiplicand;

v_res number(3);

v_count number(2);

begin

for v_count in 1..10 loop

v_res:=v_multiplicand*v_count;

dbms_output.put_line(v_multiplicand||'*'||v_count||'='||v_res);

end loop;

end;

OUTPUT:

Enter value for multiplicand: 2

old 2: v_multiplicand number(3):=&multiplicand;

new 2: v_multiplicand number(3):=2;

2*1=2

2*2=4

2*3=6

2*4=8

2*5=10

2*6=12

2*7=14

2*8=16

2*9=18

2*10=20


k)PL/SQL to determine whether a year is leap year or not

declare

v_year number(4):=&year;

begin

if(mod(v_year,4)=0 and mod(v_year,100)<>0 or mod(v_year,400)=0) then

dbms_output.put_line(v_year || ' is leap year');

else

dbms_output.put_line(v_year || ' is not leap year');

end if;

end;

OUTPUT:

Enter value for year: 1992

old 2: v_year number(4):=&year;

new 2: v_year number(4):=1992;

1992 is leap year


l)PL/SQL to insert item numbers from 1 to 5 using a loop.

create table item(itemnum number(2));

declare

v_num item.itemnum%type;

begin

for ctr in 1..5 loop

v_num:=ctr;

insert into item values(v_num);

end loop;

end;

OUTPUT:


SQL> select * from item;

ITEMNUM

----------

1

2

3

4

5

m)PL/SQL to delete an item whose itemnum=4

declare

v_num item.itemnum%type;

begin

select itemnum into v_num from item where itemnum=&p_num;

delete from item where itemnum=v_num;

end;

OUTPUT:

Enter value for p_num: 1

old 4: select itemnum into v_num from item where itemnum=&p_num;

new 4: select itemnum into v_num from item where itemnum=1;


SQL> select * from item;

ITEMNUM

----------

2

3

4

5

n)PL/SQL program to display name, salary , commission and emp no of an employee based on the id.

declare

eno emp.empno%type;

empname emp.ename%type;

esal emp.sal%type;

ecom emp.comm%type;

begin

select empno,ename,sal,comm into eno,empname,esal,ecom from emp where empno=&no;

dbms_output.put_line('EMPNO'|| eno);

dbms_output.put_line('ENAME'|| empname);

dbms_output.put_line('SALARY'|| esal);

dbms_output.put_line('COMM'|| ecom);

end;

OUTPUT:

Enter value for no: 7902

old 7: select empno,ename,sal,comm into eno,empname,esal,ecom from emp where empno=&no;

new 7: select empno,ename,sal,comm into eno,empname,esal,ecom from emp where empno=7902;

EMPNO7902

ENAMEFORD

SALARY3000

COMM


o)PL/SQL program for inverting a number.

declare

given_num varchar(5):='&num';

str_len number(2);

inv_num varchar(5);

begin

str_len:=length(given_num);

for cntr in reverse 1..str_len

loop

inv_num:=inv_num||substr(given_num,cntr,1);

end loop;

dbms_output.put_line('the given number is '||given_num);

dbms_output.put_line('the inverted number is '||inv_num);

end;

OUTPUT:

Enter value for num: 12

old 2: given_num varchar(5):='&num';

new 2: given_num varchar(5):='12';

the given number is 12

the inverted number is 21


p)PL/SQL program to calculate the area of a circle for value of radius ranging from 3 to 7.

SQL>create table areas(radius number(5),area number(14,2));

SQL>declare

pi constant number(4,2):=3.14;

radius number(5);

area number(14,2);

begin

radius:=&radius;

while radius<=7

loop

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

insert into areas values(radius,area);

radius:=radius+1;

end loop;

end;

OUTPUT:

Enter value for radius: 5

old 6: radius:=&radius;

new 6: radius:=5;


SQL> select * from areas;

RADIUS AREA

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

5 78.5

6 113.04

7 153.86

q)Write a program to print empno, ename, job and salary of an employee given empno.

SQL> declare

2 type emprec is record(eno emp.empno%type, n emp.ename%type, j emp.job%type,s emp.sal%type);

3 r emprec;

4 eno emp.empno%type:=&eno;

5 begin

6 select empno,ename,job,sal into r from emp where empno=eno;

7 dbms_output.put_line(r.eno||' '||r.n||' '||r.j||' '||r.s);

8 end;

9 /

Output:

Enter value for eno: 7369

old 4: eno emp.empno%type:=&eno;

new 4: eno emp.empno%type:=7369;

7369 SMITH CLERK 800


r)Write a program to set the values for DA, HRA, GROSS fields of all the records of the given table.

Condition DA HRA GROSSBasic < 2000 10% of Basic 5% of Basic Basic+DA+HRABasic >=2000 and <=4000

15% of Basic 10% of Basic Basic+DA+HRA

Basic>4000 20% of Basic 12% of Basic Basic+DA+HRA

SQL> create table employee(

2 empno number(3) primary key,

3 ename varchar2(10),

4 basic number(6,2),

5 da number(6,2),

6 hra number(6,2),

7 gross number(6,2));

Table created.

SQL> insert into employee(empno,ename,basic)

values(121,'Ravi',5000);

1 row created.


values(122,'Kiran',1500);

1 row created.


values(123,'Amar',6000);

1 row created.

Sql>begin

2 update employee set da=basic*0.1,hra=basic*0.05 where basic<2000;

3 update employee set da=basic*0.15,hra=basic*0.1 where basic>=2000 and basic<=4000;

4 update employee set da=basic*0.2,hra=basic*0.12 where basic>4000;

5 update employee set gross=basic+hra+da;

6 commit;

7 end;

SQL> /


OUTPUT:

SQL> select * from employee;

EMPNO ENAME BASIC DA HRA GROSS

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

121 Ravi 5000 1000 600 6600

122 Kiran 1500 150 75 1725

123 Amar 6000 1200 720 7920

s)Write a PL/SQL block that will accept a account number from the user , check if the users balance is less than the minimum balance then deduct Rs.100 from the balance as a fine.

declaremacno number;mcurbal number;minbal constant number:=520.00;fine constant number(4):=100;begin

macno:=&account_number;select bal into mcurbal from Account where acno=macno;ifmcurbal<minbal thenupdate account set bal=bal-fine where acno=macno;end if;end;OUTPUT:Enter value for account_number: 2005old 7: macno:=&account_number;new 7: macno:=2005;


t)Write a PL/SQL block that will accept a account number and incremented amount from the user, and increments the balance with the incremented amount.

declaremacno number;mcurbal number;minc number(5);beginmacno:=&account_number;minc:=&inc_number;select bal into mcurbal from Account where acno=macno;if (minc>0) thenupdate account set bal=bal+minc where acno=macno;end if;end;/OUTPUT:Enter value for account_number: 2005old 6: macno:=&account_number;new 6: macno:=2005;Enter value for inc_number: 1000old 7: minc:=&inc_number;new 7: minc:=1000;


u)Write a PL/SQL block that withdraws an amount of Rs.1000 then deposit’s an amount of Rs.10,000, updates the current balance ,then checks to see if the current balance of all the accounts does not exceed 50,000 if so then undo the deposit made.

declarembal number(7,2);macno number(4);beginmacno:=&account_no;update Account set bal=bal-1000 where acno=macno;update Account set bal=bal+10000whereacno=macno;savepoint s1;select sum(bal) into mbal from Account;if mbal> 50000 then rollback to s1;end if;commit;end;/OUTPUT:Enter value for account_no: 2005old 5: macno:=&account_no;new 5: macno:=2005;

PL/SQL procedure successfully completed.SQL> SELECT * FROM ACCOUNT;

ACNO BAL---------- ---------- 2005 10410

CURSORSa)Write a PL/SQL block using the cursor to display details of all employee’s from emp2 table whose sum of sal and comm is > 2000.

declaremempno emp2.empno%type;mename emp2.ename%type;msal emp2.sal%type;cursor c1 isselect empno,ename,sal+nvl(comm,0) from emp2 where sal+nvl(comm,0)>2000;beginopen c1;dbms_output.put_line('employeeno'||'empname'||'netsalary');loopfetch c1 into mempno,mename,msal;dbms_output.put_line(mempno||' '||mename||' '||msal);exit when c1%notfound;end loop;close c1;end;OUTPUT:employeenoempnamenetsalary7566 JONES 29757654 MARTIN 26507698 BLAKE 28507782 CLARK 24507788 SCOTT 30007839 KING 50007902 FORD 30007902 FORD 3000

PL/SQL procedure successfully completed.b)Write a PL/SQL block that display’s employee name and the salary of the first 5 employee’s having the highest salary.

declaremempno emp2.empno%type;mename emp2.ename%type;msal emp2.sal%type;cursor c1 isselect empno,ename,sal from emp2 order by sal desc;

beginopen c1;dbms_output.put_line('empname'||' netsalary');loopfetch c1 into mempno,mename,msal;dbms_output.put_line(mename||' '||msal);exit when (c1%rowcount -1)=5 or c1%notfound;end loop;close c1;end;/OUTPUT:empname netsalaryKING 5000SCOTT 3000FORD 3000JONES 2975BLAKE 2850CLARK 2450


c)Write a pl/sql program to find the sum of the salaries in a particular department.

Sql>declare

2 s number:=0;

3 pay emp.sal%type;

4 dno emp.deptno%type;

5 cursor c2 is select sal from emp where deptno=&dno;

6 begin

7 open c2;

8 loop

9 fetch c2 into pay;

10 exit when c2%notfound;

11 s:=s+pay;

12 end loop;

13 close c2;

14 dbms_output.put_line('sum of the salaries in deptno 10 is '||s);

15 end;

16 /

Output:

Enter value for dno: 10

old 5: cursor c2 is select sal from emp where deptno=&dno;

new 5: cursor c2 is select sal from emp where deptno=10;

sum of the salaries in deptno 10 is 8750


d)PL/SQL to print all employees begin with S.

declare

n emp.ename%type;

cursor c1 is select ename from emp where ename like 'S%';

begin

open c1;

fetch c1 into n;

while c1%found

loop

dbms_output.put_line(n);

fetch c1 into n;

end loop;

close c1;

end;

OUTPUT:

SMITH

SCOTT


e)PL/SQL to print names of all managers

declare

n emp.ename%type;

cursor c1 is select ename from emp where job='MANAGER' order by ename;

begin

open c1;

loop

fetch c1 into n;

exit when c1%notfound;

dbms_output.put_line(c1%rowcount||':'||n);

end loop;

close c1;

end;

OUTPUT:

1:BLAKE

2:CLARK

3:JONES


PROCEDURES

a)PL/SQL to update salary of an employee whose empno and increment mentioned.

create or replace procedure empupdate(enoemp.empno%type,incr number) as

begin

update emp set sal=sal+incr where empno=eno;

commit;

dbms_output.put_line('updated successfully');

end;

//Usage at sql prompt

Exec empupdate(7369,1000);

OUTPUT:

Procedure created.

SQL> Exec empupdate(7369,1000);

updated successfully


SQL> SELECT * FROM EMP

2 WHERE EMPNO=7369;


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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 1800

20

EXCEPTIONAL HANDLING

a)Write a PL/SQL block which accepts the empno from the user and display the details of the employee. When the user enters an empno that is not in the emp table then the PL/SQL block must display an appropriate message to the user.

declaremempno emp2.empno%type;beginselect empno into mempno from emp2 where empno=&empno;dbms_output.put_line(mempno || 'found!');ExceptionWhen no_data_found thendbms_output.put_line('Employee number not found!');end;/OUTPUT:Enter value for empno: 7902old 4: select empno into mempno from emp where empno=&empno;new 4: select empno into mempno from emp where empno=7902;

7902found!


b)PL/SQL block that retrieves course table into cursor. Input course id to search. If course exists, print the information. If the course id does not exist, print user defined message by throwing user defined exception.

declare

c_name course.course_name%type;

cursor c1 is select course_name from course where courseid='&id';

course_not_found exception;

begin

open c1;

fetch c1 into c_name;

if c1%found then

dbms_output.put_line('course name is'||c_name);

else

raise course_not_found;

end if;

exception when course_not_found then

dbms_output.put_line('course id not found');

end;

OUTPUT:

Enter value for id: 1600

old 3: cursor c1 is select course_name from course where courseid='&id';

new 3: cursor c1 is select course_name from course where courseid='1600';

course name ispcmt


c)Write a program to handle ZERO_DIVIDE exception

SQL> declare

2 x number:=&x;

3 y number:=&y;

4 begin

5 dbms_output.put_line('Division='||x/y);

6 exception when zero_divide then

7 dbms_output.put_line('check the denominator');

8 end;

9 /

Output:

Enter value for x: 5

old 2: x number:=&x;

new 2: x number:=5;

Enter value for y: 0

old 3: y number:=&y;

new 3: y number:=0;

check the denominator


FUNCTIONS

a)Write a PL/SQL function which checks for an employee number in the table and returns 1 if present otherwise returns 0.

create or replace function checkemp(eno number)return numberismempno emp2.empno%type;beginselect empno into mempno from emp2 where empno=eno;return 1;exceptionwhen no_data_found thenreturn 0;end;/SQL>exec dbms_output.put_line(checkemp(7499));

OUTPUT:Function created.

SQL> exec dbms_output.put_line(checkemp(7499));1


SQL> SELECT * FROM EMP 2 ;

EMPNO ENAME JOB MGR HIREDATE SAL COMM---------- ---------- --------- ---------- --------- ---------- ---------- DEPTNO---------- 7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30

TRIGGERS

a)PL/SQL that creates a trigger that inserts or update values of ename, job as lower case strings.

create or replace trigger lowername

before insert or update on emp for each row

begin

:new.ename:=lower(:new.ename);

:new.job:=lower(:new.job);

end;

update emp set job='MANAGER' where ename='SMITH';

OUTPUT:

Trigger created.

SQL> update emp set job='MANAGER' where ename='SMITH';

1 row updated.

SQL> SELECT * FROM EMP;


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

DEPTNO

----------

7369 smith manager 7902 17-DEC-80 1800

20

CREATE AN ACCOUNT TABLE

Query:

Creating a table account and inserting values in it.

create table account

(

acno number(5),

bal number(5)

);

Enter values into table ‘account’.

insert into account values(2005,510);

Displaying items in table ‘account’.select * from account;

ACNO BAL

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

2005 510