The document is a detailed assessment report by Vishwajeet Sikhwal, a third-year Bachelor of Computer Applications student. It focuses on database management using Oracle 10g, including creating tables, fetching data, and various SQL queries and operations. The report covers essential SQL functionalities and operations on employee and department records, illustrating examples and solutions throughout.

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:
VISHWAJEET SIKHWAL
CLASS: BCA 3RD YEAR

2. 1
PROJECT ABSTRACT
I am VISHWAJEET SIKHWAL 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 Clause.............................................................................................................. 7
Group Having ........................................................................................................... 16
Functions.................................................................................................................. 21
Coverage Joins ........................................................................................................ 29
Nested And Corelated Sub Queries ....................................................................... 33

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 EMP (
EMPNO NUMBER (4),
ENAME VARCHAR2 (20),
DEPTNO NUMBER (2),
JOB VARCHAR2 (20),
SAL NUMBER (5),
COMM NUMBER (4),
MGR NUMBER (4),
HIREDATE DATE);
To Check the Table Structure
Solution:
Desc emp
Output:
Insert Atleast 5 records.
Solution:

5. 4
Insert into emp values(:empno,:ename,:deptno,:job,:sal,:comm,:mgr,:hiredate)
Output:
2. Create a Department Table (Dept) with Following Fields:
Solution:
CREATE TABLE DEPT
(DEPTNO NUMBER (2),
DNAME VARCHAR2 (20),
LOC VARCHAR2 (20));
To check the table structure
Solution:
Desc dept
Output:
Insert At least 5 records.
FIELDS DATA TYPE SIZE
DEPTNO NUMBER 2
DNAME VARCHAR2 20
LOC (location) VARCHAR2 20

6. 5
Solution:
Insert into dept values(:deptno,:dname,:loc)
Output:
How to fetch records
Solution:
Select * from dept
Output:
3. Create a SalGrade Table with Following Fields:
FIELDS DATA TYPE SIZE
GRADE NUMBER 1
LOSAL NUMBER 5
HISAL NUMBER 5
Insert At least 5 records.
Solution:
CREATE TABLE SALGRADE
(GRADE NUMBER (1),
LOSAL NUMBER (5),
HISAL NUMBER (5));
To check the table structure
Solution:
Desc salgrade
Output

7. 6
Insert At least 5 records.
Solution:
Insert into salgrade(:grade,:losal,:hisal)
Output:
How to fetch records
Solution:
Select * from salgrade
Output:

8. 7
SELECT CLAUSE
1. List out employee records from the table.
Solution:
Select * from emp
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 * from emp where deptno=20
Output:
4. List the Name of all ‘ANALYST’ and ‘SALESMAN’.

9. 8
Solution:
Select * from emp 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 emp where hiredate <='30-sep-1981'
Output:
6. List the Employee Name and Employee Number, who is ‘MANAGER’.
Solution:
Select empno,ename from emp where job='manager'
Output:
7. List the Name and Job of all Employees who are not ‘CLERK’.
Solution:
Select ename,job from emp where job!='clerk'
Output:

10. 9
8. List the Name of Employees, whose Employee Number is 7369,7521,7839,7934
or 7788.
Solution:
Select ename from emp 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 emp 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 emp where sal between 1000 and 2000
Output:

11. 10
11.List the Employee Names, who have joined before 30-Jun-1981 and after Dec-
1981.
Solution:
Select ename from emp where hiredate <'30-jun-1981' or hiredate>'31-dec-1981'
Output:
12.List the Commission and Name of Employees, who are availing the
Commission.
Solutions:
Select ename,comm from emp where comm is not null
Output:
13.List the Name and Designation (job) of the Employees who does not report to
anybody.
Solutions:
Select ename,job from emp where mgr is null
Output:

12. 11
14.List the detail of the Employees, whose Salary is greater than 2000 and
Commission is NULL.
Solutions:
Select * from emp where sal>2000 and comm is null
Outputs:
15.List the Employee Names and Date of Joining in Descending Order of Date of
Joining. The column title should be “Date of Joining”.
Solutions:
Select ename, hiredate as "date of joining" from emp order by "date of joining" desc
Output:
16.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 emp order by deptno desc,ename asc,sal desc
Output:

13. 12
17.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 emp order by "gross salary" asc
Output:
18.List Clerk from dept No 30.
Solution:
Select * from emp where job='clerk' and deptno=30
Output:
19.List Clerks from 30 and salesman from 20. In the list the lowest earning
employee must at top.
Solution:
Select * from emp where job='clerk' and deptno=30 or job='salesman' and deptno=20
order by sal asc
Output:

14. 13
20.List different departments from Employee table.
Solution:
Select distinct deptno from emp
Output:
21.List employees having “S” at the end of their Name.
Solution:
Select * from emp where ename like '%s'
Output:
22.List employee who are not managed by anyone.
Solution:
Select * from emp where mgr is null
Output:
23.List employees who are having “TT” or “LL” in their names.
Solution:
Select * from emp where ename like '%tt%' or ename like '%ll%'
Output:

15. 14
24.List employees earning salaries below 1500 and more than 3000.
Solution:
Select * from emp where sal not between 1500 and 3000
Output:
25.List employees who are drawing some commission. Display their total salary as
well.
Solution:
Select ename,sal,sal+comm from emp where comm is not null
Output:
26.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 emp where sal<comm and comm is not null order by comm desc

16. 15
Output:
27.List the employees who joined the company in the month of “FEB”.
Solution:
Select * from emp where hiredate between '1-feb-1981' and '28-feb-1981'
Output:
28.List employees who are working as salesman and having names of four
characters.
Solution:
Select * from emp where job='salesman' and ename like '____'
Output:
29.List employee who are managed by 7839.
Solution:
Select * from emp where mgr=7839
Output:

17. 16
GROUPING, HAVING ETC.
1. List the Department number and total number of employees in each department.
Solution:
Select deptno,count(empno) from emp group by deptno
Output:
2. List the different Job names (Designation) available in the EMP table.
Solution:
Select distinct job from emp
Output:
3. List the Average Salary and number of Employees working in Department
number 20.
Solution:
Select avg(sal),count(empno) from emp where deptno=20
Output:
4. List the Department Number and Total Salary payable at each Department.

18. 17
Solution:
Select deptno,sum(sal+comm)from emp 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 10 only.
Solution:
Select sum(sal+comm),max(sal),min(sal),avg(sal) from emp where deptno=10 group
by job
Output:
7. List the Average Salary of each Job, excluding ‘MANAGERS’.
Solution:
Select job,avg(sal) from emp where job!=’manager’ group by job

19. 18
Output:
8. List the Average Monthly Salary for each Job within each department
Solution:
Select avg(sal) from emp group by deptno,job
Output:
9. List the Average Salary of all departments, in which more than two people are
working.
Solution:
Select avg(sal),count(empno)from emp 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 max(sal),job from emp group by job having max(sal)>=5000
Output:

20. 19
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 job,sum(sal+comm),avg(sal) from emp where deptno=20 group by job having
avg(sal)>1000
Output:
12.List the total salaries of only those departments in which at least 3 employees
are working.
Solution:
Select sum(sal+comm),count(empno),deptno from emp group by deptno having
count(empno)>=3
Output:
13.List the Number of Employees Managed by Each Manager
Solution:
Select count(empno),mgr from emp group by mgr
Output:

21. 20
14.List Average Commission Drawn by all Salesman
Solution:
Select avg(comm) from emp where job=’salesman’ group by job
Output:

22. 21
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(:n,:n2) 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).

23. 22
Solution:
Select sqrt(:n) from dual
Output:
5. Enter a float number and truncate it up to 1 and -2 places of decimal.
Solution:
Select trunc(:n,1) as “upto 1”,trunc(:n,-2) as “upto -2” 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(563.456,0) as “upto 0”,round(563.456,-2)
as “upto -2” from dual
Output:
7. Accept two numbers and display its corresponding character with the
appropriate title.
Solution:
Select chr(:n),chr(:n1) from dual

24. 23
Output:
8. Input two names and concatenate it separated by two spaces.
Solution:
Select concat(concat(:n,’ ‘),:n1)
Output:
9. Display all the Employee names-with first character in upper case from EMP
table.
Solution:
Select initcap(ename) from emp
Output:

25. 24
10.Display all the department names in upper and lower cases.
Solution:
Select upper(job),lower(job) from emp
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 ltrim(ename,’sa’) as left,rtrim(ename,’rn’)as right from emp
Output:
12.Change all the occurrences of CL with P in job domain.
Solution:
Select ename,job,replace(job,’cl’,’p’) from emp
Output:

26. 25
13.Display the information of those Employees whose name has the second
character A.
Solution:
Select ename from emp where instr(ename,’a’)=2
Output:
14.Display the ASCII code of the character entered by the user.
Solution:
Select ascii(: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 emp where job=’clerk’

27. 26
Output:
16.Find the Employee names with maximum number of characters in it.
Solution:
Select max(length(ename)) from emp
Output:
17.Display the salary of those Employees who are getting salary in three figures.
Solution:
Select sal from emp where length(sal)=3
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 totalsal from emp
Output:

28. 27
19.List all the Employee names, who have more than 20 years of experience in the
company.
Solution:
Select ename from emp 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) from emp
Output:
21.Display Date in the Following Format Tuesday 31 December 2002.
Solution:
Select to_char(sysdate,’day’),to_char(sysdate,’dd-mm-yyyy’) from dual
Output:

29. 28
22.Display the Sunday coming After 3 Months from Today’s Date.
Solution:
Select next_day(add_months(sysdate,3),1) from dual
Output:

30. 29
Coverage Joins
1. List Employee Name, Job, Salary, Grade & the Location where they are working
Solution:
Select e.ename,e.job,e.sal,s.grade,d.loc from emp e join dept d on e.deptno=d.deptno
join salgrade s on e.sal between losal and hisal
Output:
2. Count the Employees For Each Salary Grade for Each Location
Solution:
Select count(e.empno),s.grade,d.loc from emp e join salgrade s on e.sal between
losal and hisal join dept d on e.deptno=d.deptno group by s.grade,d.loc
Output:
3. List the Average Salary for Those Employees who are drawing salaries of grade
3

31. 30
Solution:
Select avg(e.sal),s.grade from emp e join salgrade s on e.sal between losal and hisal
where s.grade=3 group by s.grade
Output:
4. List Employee Name, Job, Salary Of those employees who are working in
Accounting or Research department but drawing salaries of grade 2
Solution:
Select e.ename,e.job,e.sal,d.dname,s.grade from emp e join dept d on
e.deptno=d.deptno join salgrade s on e.sal between losal and hisal where
(d.dname=’accounting’ or d.dname=’research’)and s.grade=2
Output:
5. List employee Names, Manager Names & also Display the Employees who are
not managed by anyone
Solution:
Select e.ename,m.ename from emp e left outer join emp m on e.empno=m.mgr
Output:
6. List Employees who are drawing salary more than the salary of SCOTT

32. 31
Solution:
Select e.ename from emp e join emp 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.* from emp e join emp m on e.empno=m.mgr 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,deptno,dname,loc from emp join dept using(deptno) where
loc=’newyork’
Output:
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 from emp join dept using(deptno) where
dname=’accounting’ or dname=’sales’
Output:

33. 32
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 emp join dept using(deptno) where sal>2000
and loc=’newyork’ or loc=’dallas’
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 e.ename,e.sal,d.dname,d.loc from emp e right outer join dept d on
e.deptno=d.deptno
Output:

34. 33
Nested and Corelated sub queries
1. List Employees who are working in the Sales Department (Use Nested)
Solution:
Select * from emp 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 emp)
Output:
3. Find the Names of employees who do not work in the same department of Scott.
Solution:
Select ename from emp where deptno<>(Select deptno from emp where
ename=’scott’)
Output:

35. 34
4. List departments (dept details) in which no employee is working (use nested)
Solution:
Select dname from dept where deptno<>all(Select deptno from emp)
Output:
5. List Employees who are drawing the Salary more than the Average salary of
DEPTNO 30. Also ensure that the result should not contain the records of
DEPTNO 30
Solution:
Select * from emp where sal>(Select avg(sal)from emp where deptno=30)and
deptno!=30
Output:
6. List Employee names drawing Second Maximum Salary
Solution:
Select ename from emp where sal=(Select (max(sal)) from emp where sal<(Select
(max(sal)) from emp))
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 emp e where e.sal=(Select min(sal)from where
e.sal=(Select min(sal) from emp where i.deptno=e.deptno)
Output:

36. 35
8. List the highest paid employee for each department using correlated sub query.
Solution:
Select * from emp e where e.sal=(Select max(sal) from emp i where
i.deptno=e.deptno)
Output:
9. List Employees working for the same job type as of BLAKE and drawing more
than him (use Self Join)
Solution:
Select e.* from emp e join emp f on e.job=f.job where e.sal>f.sal and e.ename=’blake’
Output: