This document is a work report by Simran Kaur, a third-year BCA student, detailing her assessments on the relational database management system (RDBMS) using Oracle 10g. It includes the creation of tables, data insertion, and various SQL queries to manipulate and retrieve data. The report provides practical examples of operations such as creating employee and department tables, fetching records, and performing complex queries.

1. 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
SIMRAN KAUR
CLASS: BCA 3RD
YEAR

2. 1
PROJECT ABSTRACT
I am SIMRAN KAUR 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)

3. 2
Contents
Select Statements................................................................................................................................8
Grouping,Having etc. ........................................................................................................................18
Functions. ............................................................................................................................................23
Coverage Joins...................................................................................................................................31
Nested and Co-related sub queries...............................................................................................36

4. 3
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 EMPLOYEE
(EMPNO NUMBER(4),
ENAME VARCHAR2(20),
JOB VARCHAR(20),
SAL NUMBER(5),
COMM NUMBER(4),
DEPTNO NUMBER(2),
MGR NUMBER(4),
HIREDATE DATE)
Output:
To check the table structure:
Solution:
Desc employee
Insert Atleast 5 records.
Solution:
Insert into employee(empno,ename,job,deptno,sal,comm,mgr,hiredate)

5. 4
values(:empno,:ename,:job,:deptno,:sal,:comm,:mgr,:hiredate)
Output:
How to fetch the record from the table:
Solution:
Select * from employee
Output:
2. Create a Department Table (Dept) with Following Fields:
FIELDS DATA TYPE SIZE
DEPTNO NUMBER 2
DNAME VARCHAR2 20
LOC (location) VARCHAR2 20
Solution:
Create table dept
(deptno number(2),
dname varchar2(20),

6. 5
loc varchar2(20))
Output:
To check the table structure:
Solution:
Desc dept
Insert Atleast 5 records.
Solution:
Insert into dept(deptno,dname,loc)
values(:deptno,:dname,:loc)
Output:
How to fetch the record from the table:
Select * from dept
Output:

7. 6
3. Create a SalGrade Table with Following Fields:
FIELDS DATA TYPE SIZE
GRADE NUMBER 1
LOSAL NUMBER 5
HISAL NUMBER 5
Solution:
Create table salgrade
(grade number(1),
lowsal number(5),
highsal number(5))
Output:
To check the table structure:
Solution:
Desc salgrade
Insert Atleast 5 records.
Solution:
Insert into salgrade(grade,lowsal,highsal)
values(:grade,:lowsal,:highsal)
Output:
How to fetch the record from the table:
Solution:
Select * from salgrade

8. 7
Output:

9. 8
SELECT STATEMENT
1. List all the information about all Employees.
Solution:
Select * from employee
Output:
2. Display the Name of all Employees along with their Salary.
Solution:
Select ename,sal from emp
Output:
3. List all the Employee Names who is working with Department Number is 20.
Solution:
Select ename from employee
where deptno=20
Output:
4. List the Name of all ‘ANALYST’ and ‘SALESMAN’.

10. 9
Solution:
Select ename from employee
Where job in('analyst','salesman')
Output:
5. Display the details of those Employees who have joined before the end of Sept.
1981.
Solution:
Select * from employee
where hiredate<='30-sep-1981'
Output:
6. List the Employee Name and Employee Number, who is ‘MANAGER’.
Solution:
Select ename,empno from employee
where job='manager'
Output:
7. List the Name and Job of all Employees who are not ‘CLERK’.
Solution:
Select ename,job from employee
where job!='clerk'

11. 10
Output:
8. List the Name of Employees, whose Employee Number is 7369,7521,7839,7934
or 7788.
Solution:
Select ename from employee
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 employee where deptno not in(10,30)
Output:
10.List the Employee Name and Salary, whose Salary is vary from 1000 to 2000.
Solution:
Select ename,sal from employee
where sal between 1000 and 2000

12. 11
Output:
11.List the Employee Names, who have joined before 30-Jun-1981 and after Dec-
1981.
Solution:
Select ename from employee
where hiredate<'30-jun-1981' or hiredate>'31-dec-1981'
Output:
12.List the Commission and Name of Employees, who are availing the
Commission.
Solution:
Select comm,ename from employee where comm is not null
Output:
13.List the Name and Designation (job) of the Employees who does not report to
anybody.
Solution:
Select ename,job from employee
where mgr is null

13. 12
Output:
14.List the detail of the Employees, whose Salary is greater than 2000 and
Commission is NULL.
Solution:
Select * from employee
where sal>2000 and comm is null
Output:
15.List the Employee details whose Name start with ‘S’.
Solution:
Select * from employee
where ename like 's%'
Output:
16.List the Employee Names and Date of Joining in Descending Order of Date of
Joining. Th0e column title should be “Date Of Joining”.
Solution:
Select ename,hiredate as "Date of Joining" from employee
order by hiredate
Output:

14. 13
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,sal,job,deptno from employee
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+sal*50/100+sal*30/100-sal*10/100+comm as "gross salary"
from employee
order by "gross salary" asc
Output:
19.List Salesman from dept No 50.
Solution:
Select job,deptno from employee

15. 14
where job='salesman' and deptno=50
Output:
20.List Clerks from 20 and salesman from 50. In the list the lowest earning
employee must at top.
Solution:
Select job,deptno from employee
where job in('clerk','salesman')and deptno in(20,50)
Output:
21.List different departments from Employee table.
Solution:
Select distinct(deptno) from employee
Output:
22.List employees having “N” at the end of their Name.
Solution:
Select ename from employee
where ename like '%n'
Output:

16. 15
23.List employee who are not managed by anyone.
Solution:
Select * from employee
where mgr is null
Output:
24.List employees who are having “AR” or “HN” in their names.
Solution:
Select ename from employee
where ename like '%ar%' or ename like '%hn%'
Output:
25.List employees earning salaries below 1500 and more than 3000.
Solution:
Select ename,sal from employee
where sal<1500 or sal>3000
Output:

17. 16
26.List employees who are drawing some commission. Display their total salary as
well.
Solution:
Select ename,sal+comm from employee
where comm is not null
Output:
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 comm,sal from employee
where comm>sal
order by comm desc
Output:
28.List the employees who joined the company in the month of “FEB”.
Solution:
Select ename,hiredate from employee
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.

18. 17
Solution:
Select ename,job from employee
where job='salesman' and ename like '____'
Output:
30.List employee who are managed by 7839.
Solution:
Select ename,mgr from employee
where mgr=7839
Output:

19. 18
GROUPING, HAVING ETC.
1. List the Department number and total number of employees in each department.
Solution:
Select deptno,count(Empno)
from employee
group by deptno
Output:
2. List the different Job names (Designation) available in the EMP table.
Solution:
Select distinct job from employee
group by job
Output:
3. List the Average Salary and number of Employees working in Department
number 20.
Solution:
Select avg(sal),count(empno),deptno from employee
where deptno=20
group by deptno
Output:

20. 19
4. List the Department Number and Total Salary payable at each Department.
Solution:
Select deptno,sum(Sal) from employee
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 employee
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 10 only.
Solution:
Select sum(sal),max(Sal),min(sal),avg(sal) from employee
where deptno=10
group by job

21. 20
Output:
7. List the Average Salary of each Job, excluding ‘MANAGERS’.
Solution:
Select avg(sal),job
from employee
where job!='manager'
group by job
Output:
8. List the Average Monthly Salary for each Job within each department.
Solution:
Select avg(sal),job,deptno
from employee
group by job,deptno
Output:
9. List the Average Salary of all departments, in which more than two people are
working.
Solution:

22. 21
Select avg(sal),deptno,count(empno)
from employee
group by deptno
having count(empno)>2
Output:
10. List the jobs of all Employees where Maximum Salary is greater than or equal
to 5000.
Solution:
Select job,max(sal)
from employee
group by job
having max(sal)>=5000
Output:
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
from employee
where deptno=20
group by job
having avg(sal)>1000
Output:

23. 22
12.List the Number of Employees Managed by Each Manager
Solution:
Select count(empno),mgr
from employee
group by mgr
Output:
13.List Average Commission Drawn by all Salesman
Solution:
Select avg(comm),job
from employee
where job='salesman'
group by job
Output:

24. 23
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(:first,:second) from dual
Output:
3. Enter a number and check whether it is negative or positive.
Solution:
Select sign(:n) from dual
Output:
4. Calculate the square root for the given number. (i.e. 10000).

25. 24
Solution:
Select sqrt(100000) from dual
Output:
5. Enter a float number and truncate it up to 1 and -2 places of decimal.
Solution:
Select trunc(:n,-2) as "upto2",trunc(:n,1) as "upto1" from dual
Output:
6. Find the rounding value of 563.456, up to 2, 0 and -2 places of decimal.
Solution:
Select round(563.456,-2) as "upto-2",round(563.456,2) as "upto2",round(563.456) as
"upto0" from dual
Output:
7. Accept two numbers and display its corresponding character with the
appropriate title.
Solution:
Select chr(:n) as "first",chr(:n1) as "second" from dual

26. 25
Output:
8. Input two names and concatenate it separated by two spaces.
Solution:
Select (:first)||' '||(:second) from dual
Output:
9. Display all the Employee names-with first character in upper case from EMP
table.
Solution:
Select initcap(ename) from employee
Output:
10. Display all the department names in upper and lower cases.
Solution:
Select upper(dname),lower(dname) from dept

27. 26
Output:
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,rtrim(ltrim(ename,'sa'),'rn') from employee
Output:
12.Change all the occurrences of CL with P in job domain.
Solution:
Select job,replace(job,'clerk','perk') from employee
Output:

28. 27
13.Display the information of those Employees whose name has the second
character A.
Solution:
Select ename,instr(ename,'a',2) from employee
Output:
14.Display the ASCII code of the character entered by the user.
Solution:
Select chr(:n) 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 employee
where job='clerk'
Output:

29. 28
16.Find the Employee names with maximum number of characters in it.
Solution:
Select max(length(ename)) from employee
Output:
17. Display the salary of those Employees who are getting salary in four figures.
Solution:
Select sal from employee
where length(sal)=4
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) as "total salary"
from employee
Output:

30. 29
19.List all the Employee names, who have more than 20 years of experience in the
company.
Solution:
Select ename from employee
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',job) job_details
from employee
Output:

31. 30
21.Display Date in the Following Format Tuesday 31 December 2002.
Solution:
Select to_char(sysdate,'day yy month yyyy')
from dual
Output:
22.Display the Sunday coming After 3 Months from Today’s Date.
Solution:
Select sysdate,next_day(sysdate,'sunday')
from dual
Output:

32. 31
Coverage Joins
1. List Employee Name, Job, Salary, Grade & the Location where they are working
Solution:
Select ename,job,sal,grade,loc from employee
join salgrade on sal between lowsal and highsal
join dept using(deptno)
Output:
2. Count the Employees For Each Salary Grade for Each Location
Solution:
Select loc,count(empno),grade from employee
join salgrade on sal between lowsal and highsal
join dept using(deptno)
group by grade,loc
Output:

33. 32
3. List the Average Salary for Those Employees who are drawing salaries of grade
3
Solution:
Select avg(sal) from employee
join salgrade on sal between lowsal and highsal
where grade=3
group by grade
Output:
4. List Employee Name, Job, Salary Of those employees who are working in
Accounting or Research department but drawing salaries of grade 3
Solution:
Select empno,job,sal from employee
join dept on employee.deptno=dept.deptno
join salgrade on sal between lowsal and highsal
where dname in('accounting','research') and grade=3
Output:
5. List employee Names, Manager Names & also Display the Employees who are
not managed by anyone
Solution:
Select m.ename as "manager",e.ename as "employee" from employee e
left outer join emp m
on m.empno=e.mgr
Output:

34. 33
6. List Employees who are drawing salary more than the salary of MARTIN.
Solution:
Select e.*
from employee e join employee f
on e.sal>f.sal
where f.ename='martin'
Output:
7. List Employees who have joined the company before their managers
Solution:
Select e.ename,e.hiredate,m.ename,m.hiredate from employee e
join employee 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

35. 34
Solution:
Select ename,job,sal,deptno,dname,loc from employee
join dept using(deptno)
where loc='newyork'
Output:
9. List Employee Name, Job, Salary, Hire Date and Location Of all employees
reporting in Manager or Sales Department
Solution:
Select ename,job,sal,hiredate,loc from employee
join dept using(deptno)
where dname='sales' and job='manager'
Output:
10. List Employee Name, Job, Salary, Department Name, Location for Employees
drawing salary more than 2000 and working at New York or America
Solution:
Select ename,job,sal,dname,loc from employee join dept using(deptno)
where sal>2000 and loc in('newyork','america')
Output:
11. List Employee Name, Job, Salary, Department Name, Location Of all
employees, also list the Department Details in which no employee is working
Solution:
Select ename,job,sal,dname,loc from employee

36. 35
right outer join dept using(deptno)
Output:
12. List all Employee details and also calculate the Average Salary and Total Salary
Given to All Employees
Solution:
Select sum(sal+comm),avg(sal),empno,ename,deptno,mgr
from employee
group by empno,ename,deptno,mgr
Output:

37. 36
Nested and Correlated sub queries
1. List Employees who are working in the Sales Department (Use Nested).
Solution:
Select * from employee
where deptno=(select deptno from dept where dname='sales')
Output:
2. List Departments in which at least one employee is working (Use Nested).
Solution:
Select dname from dept
where deptno in(select deptno from employee)
Output:
3. Find the Names of employees who do not work in the same department of Scott.
Solution:
Select ename from employee
where deptno <>(select deptno from employee where ename='scott')
Output:

38. 37
4. List departments (dept details) in which no employee is working (use nested).
Solution:
Select * from dept
where deptno not in(select deptno from employee)
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 employee
where sal>(select avg(sal) from employee where deptno=20)and deptno<>20
Output:
6. List Employee names drawing Second Maximum Salary
Solution:
Select * from employee
where sal=(select max(sal) from employee where sal<(select max(sal) from
employee))
Output:
7. List the Employee Names, Job & Salary for those employees who are drawing
minmum salaries for their department (Use Correlated)
Solution:
Select ename,job,sal from employee o
where sal=(select min(sal) from employee i
where o.deptno=i.deptno)

39. 38
Output:
8. List the highest paid employee for each department using correlated sub query.
Solution:
Select * from employee e
where sal=(select max(sal) from employee i
where i.deptno=e.deptno)
Output:
9. List Employees working for the same job type as of JOHN and drawing more than
him (use Self Join)
Solution:
Select e.*
from employee e join employee f
on e.sal>f.sal and e.job=f.job
where f.ename='john'
Output: