Oracle Assessment

A WORK REPORT SUBMITTED

IN PARTIAL FULLFILLMENT OF THE REQUIREMENT FOR THE DEGREE

Bachelor of Computer Application

Dezyne E’cole College

106/10, CIVIL LINES

AJMER

RAJASTHAN - 305001 (INDIA)

(JUNE, 2015)

www.dezyneecole.com

SUBMITTED BY

POOJA JAIN

CLASS: BCA 3RD YEAR

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PROJECT ABSTRACT

I am POOJA JAIN Student of 3rd year doing my Bachelor Degree in Computer

Application.

In the following pages I gave compiled my work learnt during my 3rd year at college. The

subject is RDBMS. These assessments are based on Relational Database Management

System that is useful to work with front end (user interface) application. Take example of

an online form if the user filling up an online form (E.g. SBI Form, Gmail registration Form)

on to the internet whenever he/she clicks on the submit button the field value is transfer to

the backend database and stored in Oracle, MS-Access, My SQL, SQL Server.

The purpose of a database is to store and retrieve related information later on. A database

server is the key to solving the problems for information management.

In these assessment we are using Oracle 10g as the Relation Database Software.

The back-end database is a database that is accessed by user indirectly through an

external application rather than by application programming stored within the database

itself or by low level manipulation of the data (e.g. through SQL commands).

Here in the following assessment we are performing the following operations:

1. Creating tables to store the data into tabular format(schemas of the data base)

2. Fetching the data from the database(Using Select Query)

3. Joining the multiple data tables into one(To reduces the redundancy of the data)

4. Nested Queries(Queries within Queries)

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Content Select Statement…………………………………………………………………………....8-17

Grouping Having………………………………………………………………………..….18-22

Functions……………………..……………………..……………………..………………..23-30

Coverage Joins……………………..……………………..……………………..….……..31-35

Nested and Correlated Subqueries……………………..……………………..…………36-38

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1. Create an Employee Table (Emp) with Following Fields:

FIELDS DATA TYPE SIZE

EMPNO NUMBER 4

ENAME VARCHAR2 20

DEPTNO NUMBER 2

JOB VARCHAR2 20

SAL NUMBER 5

COMM NUMBER 4

MGR NUMBER 4

HIREDATE DATE -

Solution create table Emp1 (EMPNO NUMBER (4), ENAME VARCHAR2(20), DEPTNO NUMBER (2), JOB VARCHAR2(20), SAL NUMBER (5), COMM NUMBER (4), MGR NUMBER (4), HIREDATE DATE); How to display structure of Employee Table

Solution

Desc Emp1

Output

Insert Atleast 5 records.

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Solution insert into emp1 (EMPNO,ENAME,DEPTNO,JOB,SAL,COMM,MGR,HIREDATE) values(:Empno,:Ename,:Deptno,:Job,:Sal,:Comm,:Mgr,:Hiredate) Output

How to Fetch Records in Emp Table Solution Select * from emp1 Output

2. Create a Department Table (Dept) with Following Fields:

FIELDS DATA TYPE SIZE

DEPTNO NUMBER 2

DNAME VARCHAR2 20

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LOC (location) VARCHAR2 20

Solution create table dept1 (DEPTNO NUMBER (2), DNAME VARCHAR2 (20), LOC VARCHAR2(20)); How to display structure of Department Table

Solution Desc dept1 Output Desc dept1

Insert Atleast 5 records. Solution insert into dept1 (DEPTNO,DNAME,LOC) values(:Deptno,:Dname,:Loc) Output

How to Fetch Records in Department Table Solution Select * from dept1 Output

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3. Create a SalGrade Table with Following Fields:

FIELDS DATA TYPE SIZE

GRADE NUMBER 1

LOSAL NUMBER 5

HISAL NUMBER 5

Solution create table SalGrade1 (GRADE NUMBER (1), LOSAL NUMBER (5), HISAL NUMBER (5)); How to display structure of SalGrade Table

Solution:

Desc SalGrade1

Output:

Insert Atleast 5 records. Solution insert into SalGrade1 (GRADE,LOSAL,HISAL) values(:Grade,:Losal,:Hisal)

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Output

How to Fetch Records from SalGrade Table Solution Select * from Salgrade1 Output

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SELECT STATEMENT

1. List all the information about all Employees.

Solution select * from emp1 Output

2. Display the Name of all Employees along with their Job.

Solution select ename,job from emp1

Output

3. List all the Employee Names who is working with Department Number is 20.

Solution select * from dept1 where deptno=20

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Output

4. List the Name of all ‘ANALYST’ and ‘SALESMAN’.

Solution Select ENAME from emp1 where JOB='Analyst' or JOB='Salesman'

Output

5. Display the details of those Employees who have joined before the end of Sept.

1981.

Solution select * from emp1 where HIREDATE<='30-sep-1981' Output

6. List the Employee Name and Employee Number, who is ‘MANAGER’.

Solution select ENAME,EMPNO from emp1 where JOB='Manager'

Output

7. List the Name and Job of all Employees who are not ‘CLERK’.

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Solution select ENAME,JOB from emp1 where JOB<>'Clerk'

Output

8. List the Name of Employees, whose Employee Number is 7369,7521,7839,7934

or 7788.

Solution select ENAME from emp1 where EMPNO in(7369,7521,7839,7934,7788)

Output

9. List the Employee detail who does not belongs to Department Number 10 and

30. Solution select * from emp1 where DEPTNO<>10 and DEPTNO<>30 Output

10. List the Employee Name and Salary, whose Salary is vary from 1000 to 2000.

Solution select ENAME,SAL from emp1 where SAL between 1000 and 2000

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Output

11. List the Employee Names, who have joined before 30-Jun-1981 and after Dec-1981.

Solution select ENAME from emp1 where HIREDATE<='30-jun-1981' or HIREDATE>='1-jan-1982'

Output

12. List the Commission and Name of Employees, who are availing the Commission.

Solution select COMM,ENAME from emp1 where COMM=0

Output

13. List the Name and Designation (job) of the Employees who does not report to anybody.

Solution select ENAME, JOB from emp1 where MGR is null Output

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14. List the detail of the Employees, whose Salary is greater than 2000 and

Commission is NULL. Solution select * from emp1 where SAL>2000 and COMM=0 Output

15. List the Employee details whose Name start with ‘S’. Solution select * from emp1 where ENAME like'S%' Output

16. List the Employee Names and Date of Joining in Descending Order of Date of Joining. The column title should be “Date Of Joining”. Solution select ENAME,HIREDATE as "Date Of Joining" from emp1 order by "Date Of Joining" asc Output

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17. List the Employee Name, Salary, Job and Department Number and display it in Descending Order of Department Number, Ascending Order of Name and Descending Order of Salary.

Solution select ENAME,JOB,SAL,DEPTNO from emp1 order by DEPTNO desc,ENAME asc,SAL desc Output

18. List the Employee Name, Salary, PF, HRA, DA and Gross Salary; Order the result

in Ascending Order of Gross Salary. HRA is 50% of Salary, DA is 30% and PF is 10%.

Solution select ENAME,SAL+SAL*50/100+SAL*30/100-SAL*10/100 as "Gross Salary" from emp1 order by "Gross Salary" asc Output

19. List Salesman from dept No 10.

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Solution select JOB,DEPTNO from emp1 where DEPTNO=10 and JOB='Salesman' Output

20. List Clerks from 20 and salesman from 30. In the list the lowest earning employee

must at top. Solution select * from emp1 where JOB='Clerk' and DEPTNO=20 or JOB='Salesman' and DEPTNO=10 order by SAL asc Output

21. List different departments from Employee table.

Solution select distinct DEPTNO from emp1

Output

22. List employees having “S” at the end of their Name.

Solution select * from emp1 where ENAME like'%s' Output

23. List employee who are not managed by anyone

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Solution select * from emp1 where MGR is null Output

24. List employees who are having “TT” or “LL” in their names.

Solution select * from emp1 where ENAME like'%tt%' or ENAME like '%ll%'

Output

25. List employees earning salaries below 1500 and more than 3000.

Solution select * from emp1 where SAL<1500 or SAL>3000 Output

26. List employees who are drawing some commission. Display their total salary as

well.

Solution select EMPNO,ENAME,JOB,DEPTNO,SAL,COMM,MGR,HIREDATE, sum(SAL) from emp1 where COMM<>0 groupby EMPNO,ENAME,JOB,DEPTNO,COMM,MGR,HIREDATE,SAL Output

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27. List employees who are drawing more commission than their salary. Only those

records should be displayed where commission is given (also sort the output in the descending order of commission). Solution select * from emp1 where COMM>SAL and COMM>0 order by COMM desc

Output

28. List the employees who joined the company in the month of “FEB”.

Solution select * from emp1 where HIREDATE between'1-feb-1981' and '28-feb-1981' Output

29. List employees who are working as salesman and having names of four

characters.

Solution select * from emp1 where JOB='Salesman' and ENAME like'____' Output

30. List employee who are managed by 7839.

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Solution select * from emp1 where MGR=7839 Output

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GROUPING, HAVING ETC.

1. List the Department number and total number of employees in each department. Solution Select count(EMPNO),DEPTNO from emp1 group by DEPTNO Output

2. List the different Job names (Designation) available in the EMP table. Solution select distinct JOB from emp1 Output

3. List the Average Salary and number of Employees working in Department number 20.

Solution select avg(SAL),DEPTNO,count(EMPNO) from emp1 where DEPTNO=20 group by DEPTNO Output

4. List the Department Number and Total Salary payable at each Department.

Solution

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select sum(SAL+COMM),DEPTNO from emp1 group by DEPTNO Output

5. List the jobs and number of Employees in each Job. The result should be in Descending Order of the number of Employees. Solution select job,count(empno) from emp group by job order by count(empno) desc Output

6. List the Total salary, Maximum Salary, Minimum Salary and Average Salary of Employees job wise, for Department number 20 only. Solution select sum(SAL+COMM),max(SAL),min(SAL),avg(SAL),JOB,DEPTNO from emp1 where DEPTNO=20 group by JOB,DEPTNO Output

7. List the Average Salary of each Job, excluding ‘MANAGERS’. Solution select avg(SAL),JOB from emp1 where JOB<>'Manager' group by JOB

Output

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8. List the Average Monthly Salary for each Job within each department.

Solution Select avg(SAL),JOB,DEPTNO from emp1 group by JOB,DEPTNO Output

9. List the Average Salary of all departments, in which more than five people are working. Solution select avg(SAL),DEPTNO from emp1 group by DEPTNO having count(EMPNO)>=5 Output

10. List the jobs of all Employees where Maximum Salary is greater than or equal to 5000. Solution select JOB from emp1 group by JOB having max(SAL)>=5000 Output

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11. List the total salary and average salary of the Employees job wise, for department number 20 and display only those rows having average salary greater than 1000 Solution Select sum(SAL+COMM),avg(SAL),JOB,DEPTNO from emp1 where DEPTNO=20group by JOB,DEPTNO having avg(SAL)>1000 Output

12. List the total salaries of only those departments in which at least 2 employees are working. Solution select sum(SAL+COMM),DEPTNO,count(EMPNO) from emp1 group by DEPTNO having count(EMPNO)=2 Output

13. List the Number of Employees Managed by Each Manager Solution select count(EMPNO),MGR from emp1 group by MGR Output

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14. List Average Commission Drawn by all Salesman

Solution select avg(COMM),JOB from emp1 where JOB='Salesman' group by JOB Output

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FUNCTIONS

1. Calculate the remainder for two given numbers. (213,9) Solution select mod(213,9) from dual Output

2. Calculate the power of two numbers entered by the users at run time of the query.

Solution select power(:n1,:n2) from dual

Output

3. Enter a number and check whether it is negative or positive.

Solution select sign(:n1) from dual

Output

4. Calculate the square root for the given number. (i.e. 10000).

Solution select sqrt(10000) from dual Output

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5. Enter a float number and truncate it up to -2 places of decimal. Solution select trunc(:n1,-2) from dual

Output

6. Find the rounding value of 563.456, up to 2 places of decimal.

Solution select round(563.456,2) from dual Output

7. Accept two numbers and display its corresponding character with the

appropriate title. Solution select chr(:a1) "First", chr(:a2) "second" from dual

Output

8. Input two names and concatenate it separated by one space.

Solution

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select (:n||' '||:n2) from dual

Output

9. Display all the Employee names-with first character in upper case from EMP table. Solution select initcap(ename) from emp Output

10. Display all the department names in upper and lower cases.

Solution select upper(DNAME),lower(DNAME) from dept1 Output

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11. Extract the character S and A from the left and R and N from the right of the

Employee name from EMP table. Solution select ENAME,ltrim(ENAME,'S'),ltrim(ENAME,'A'),rtrim(ENAME,'r'), rtrim(ENAME,'n') from emp1

Output

12. Change all the occurrence of cl with p in job domain.

Solution select ENAME,JOB,replace(JOB,'Cl','p') from emp1 Output

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13. Display the information of those Employees whose name has the second character A. Solution select ENAME, instr(ENAME,'a') from emp1 where instr(ENAME,'a')=2 Output

14. Display the ASCII code of the character entered by the user.

Solution select ascii(:n1) from dual

Output

15. Display the Employee names along with the location of the character A in the

Employees name from EMP table where the job is CLERK. Solution select ENAME,JOB,instr(ENAME,'a') from emp1 where JOB='Clerk' Output

16. Find the Employee names with maximum number of characters in it.

Solution

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select max(length(ENAME)) from emp1 Output

17. Display the salary of those Employees who are getting salary in three figures.

Solution select ENAME,substr(SAL,1,3) from emp1 Output

18. Display only the first three characters of the Employees name and their H RA

(salary * .20), truncated to decimal places. Solution select substr(ENAME,1,3),trunc(SAL*20/100) from emp1 Output

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19. List all the Employee names, who have more than 20 years of experience in the

company. Solution select ENAME from emp1 where round(months_between (sysdate,HIREDATE)/12)>20 Output

20. Display the name and job for every Employee, while displaying jobs, 'CLERK'

should be displayed as 'LDC' and 'MANAGER' should be displayed as 'MNGR'. The other job title should be as they are. Solution select ENAME,JOB,decode(JOB,'Clerk','LDC','Manager','mngr') from emp1 Output

21. Display Date in the Following Format Tuesday 31 December 2002.

Solution select to_char(sysdate,'day yy-month-yyyy') from dual Output

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22. Display the Sunday coming After 3 Months from Today’s Date

Solution select sysdate,next_day(add_months(sysdate,3),1) from dual

Output

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Coverage Joins

1. List Employee Name, Job, Salary, Grade & the Location where they are working Solution select ENAME,JOB,SAL,GRADE,LOC from emp1 join dept1 on emp1.DEPTNO=dept1.DEPTNO join SalGrade1 on SAL between LOSAL and HISAL

Output

2. Count the Employees For Each Salary Grade for Each Location Solution Select count(EMPNO),GRADE,LOC From emp1 join dept1 On emp1.DEPTNO=dept1.DEPTNO join SalGrade1 On SAL between LOSAL and HISAL Group by GRADE,LOC Output

3. List the Average Salary for Those Employees who are drawing salaries of grade 1 Solution

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Select avg(SAL) From emp1 join SalGrade1 On SAL between LOSAL and HISAL Where GRADE=3 Output

4. List Employee Name, Job, Salary Of those employees who are working in Accounting or Research department but drawing salaries of grade 1 Solution Select e.ENAME,e.JOB,e.SAL,d.DNAME,s.GRADE From emp1 e join dept1 d On e.DEPTNO=d.DEPTNO Join SalGrade1 s On SAL between LOSAL and HISAL Where (d.DNAME='Sales' or d.DNAME='Management') and s.GRADE=1 Output

5. List employee Names, Manager Names & also Display the Employees who are not managed by anyone Solution Select e.ENAME,m.ENAME From emp1 e left outer join emp1 m On e.EMPNO=m.MGR where e.MGR is null Output

6. List Employees who are drawing salary more than the salary of SCOTT

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Solution Select e.* from emp1 e join emp1 f On e.SAL>f.SAL where f.ENAME='Scott' Output

7. List Employees who have joined the company before their managers Solution Select e.ENAME,e.HIREDATE,m.ENAME,m.HIREDATE From emp1 e join emp1 m On e.MGR=m.EMPNO where e.HIREDATE<m.HIREDATE Output

8. List Employee Name, Job, Salary, Department No, Department name and Location Of all employees Working at NEW YORK Solution Select ENAME,JOB,SAL,DNAME,LOC From emp1 join dept1 On emp1.DEPTNO=dept1.DEPTNO Where LOC='NewYork' Output

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9. List Employee Name, Job, Salary, Hire Date and Location Of all employees reporting in Accounting or Sales Department Solution Select ENAME,JOB,SAL,HIREDATE,LOC,DNAME From emp1 join dept1 On emp1.DEPTNO=dept1.DEPTNO Where DNAME='Account' or DNAME='Sales' Output

10. List Employee Name, Job, Salary, Department Name, Location for Employees drawing salary more than 2000 and working at New York or Dallas Solution Select ENAME,JOB,SAL,DNAME,LOC From emp1 join dept1 On emp1.DEPTNO=dept1.DEPTNO Where LOC in('Dallas','NewYork')and SAL>2000 Output

11. List Employee Name, Job, Salary, Department Name, Location Of all employees, also list the Department Details in which no employee is working

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Solution Select ENAME,JOB,SAL,DNAME,LOC From emp1 right outer join dept1 On emp1.DEPTNO=dept1.DEPTNO Output

12. List all Employee details and also calculate the Average Salary and Total Salary Given to All Employees Solution Select EMPNO,ENAME,DEPTNO,JOB,SAL,COMM,MGR, HIREDATE,sum(SAL+COMM),avg(SAL) From emp1 Group by EMPNO,ENAME,DEPTNO,JOB,SAL,COMM,MGR,HIREDATE Output

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Nested and Correlated subqueries

1. List Employees who are working in the Sales Department (Use Nested) Solution Select * from emp1 where DEPTNO=(select DEPTNO from dept1 where DNAME='Sales') Output

2. List Departments in which at least one employee is working (Use Nested) Solution Select DNAME from dept1 where DEPTNO in(select DEPTNO from emp1) Output

3. Find the Names of employees who do not work in the same department of Scott. Solution Select ENAME from emp1 where DEPTNO<>(select DEPTNO from emp1 where ENAME='Scott') Output

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4. List departments (dept details) in which no employee is working (use nested)

Solution Select DNAME from dept1 where DEPTNO <>all(select DEPTNO from emp1) Output

5. List Employees who are drawing the Salary more than the Average salary of DEPTNO 20. Also ensure that the result should not contain the records of DEPTNO 20 Solution Select * from emp1 where SAL>(select avg(SAL) from emp1 where DEPTNO=20)and DEPTNO!=20 Output

6. List Employee names drawing Second Maximum Salary Solution Select ENAME from emp1 where SAL=(select(max(SAL))from emp1 where SAL<(select max(SAL) from emp1)) Output

7. List the Employee Names, Job & Salary for those employees who are drawing minmum salaries for their department (Use Correlated)

Solution Select e.ENAME,e.JOB,e.SAL from emp1 e where SAL=(select min(SAL) from emp1 i where i.DEPTNO=e.DEPTNO)

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Output

8. List the highest paid employee for each department using correlated sub query. Solution Select * from emp1 e where SAL=(select max(SAL) from emp1 i where i.DEPTNO=e.DEPTNO) Output

9. List Employees working for the same job type as of SCOTT and drawing more than him (use Self Join) Solution Select e.* from emp1 e join emp1 f on e.JOB=f.JOB where e.SAL>f.SAL and f.ENAME='Scott' Output