SQL Lecture-6

8/10/2019 SQL Lecture-6

http://slidepdf.com/reader/full/sql-lecture-6 1/20

RDBMS Using Oracle

E-mail: [email protected](2) [email protected]

LECTURE WEEK 6

This Week Lecture Includes

Concept of Sub-query

Sub-query Examples

Sub-query Practice

Use of GROUP BY clause

GROUP BY Examples

Date Arithmetic

Date Functions



To count the number of rows in emp table enter:

SQL> select count(*) from emp;

COUNT(*)----------14

The COUNT Function

The count function is used to count the rowsselected by the query.

To count the

employees who are

getting salary.

SQL> select count(sal) from emp;

COUNT(SAL)----------14

To count the number

of different jobs held

by employees in

department 10.

SQL> select count (DISTINCT JOB) fromemp where deptno = 10;

COUNT (DISTINCT JOB)------------------

2



Group functions Vs Individual

Functions

A command

Select ename, avg(sal) from emp

is invalid. Because select ename will return multiplerecords and avg(sal) will return one average result of allsalaries.

IF we will enter this kind of command SQL*plus will

display a message saying that ENAME is “not a single-group group function”.

SQL> select ename, avg(sal) from emp;select ename, max(sal) from emp

*ERROR at line 1:ORA-00937: not a single-group group function

SQL> select ename, max(sal) from emp;select ename, max(sal) from emp

*ERROR at line 1:ORA-00937: not a single-group group function



Group function in a sub-query

Suppose we want to find out who makes thehighest salary in emp table. Since we cannot

write this SELECT ename, max(sal) from emp

In order to get this result we can put MAX(SAL)

function in sub query

Group function in a sub-query

To list the employee and job with the highestsalary enter:

SQL> select ENAME, JOB, SAL from emp

where sal = (select MAX(sal) from emp);

ENAME JOB SAL

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

Alan CLERK 900



Sub Query

Sub Query can be used with any of the following Operators includes = , != , <> , <= , >= , NOT IN and

IN, <operator> ANY , <operator> ALL etc

= != <> < > <= >= all are used with single

value output of a sub query, where as IN, NOT IN,

<operator> ANY , <operator> ALL keywords can be

used for single as well as for multi value output of a sub

query.

SUB QUERY PRACTICE:

Note:- You have understood the concept of SUBQUERY if you can

solve following queries by your own.

List the employee name and job with the lowestsalary?

List the employee name who has salary morethen the average salary of all Managers fromEMP table?

How many of employees have salary more thenthe SUM of Salaries of All CLERKS?



Select ename from emp wheresal>(select min(sal) from emp)

Select ename from emp where sal > (select avg(sal)from emp where job = ‘MANAGER’;

Select count (empno) from emp where sal> (selectsum(sal) from emp where job = ‘CLERK’);



Use of GROUP BY

SQL> select ename, max(sal) from emp*

ERROR at line 1:ORA-00937: not a single-group group function

Use of GROUP BY

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

DEPTNO SUM(SAL)--------- ----------10 270020 800

Suppose you want to find sum of salaries for eachdepartment.



SQL> select job, avg(sal) from emp group by job;

JOB AVG(SAL)-------------------- ----------CLERK 625MANAGER 800Manager 200

Use of GROUP BY

HAVING clause

As we can select specific rows with a WHEREclause, similarly we can select specfic groups

with a HEAVING clause.

HAVING clause will come after the GROUPBY clause.



HAVING clause

To list the average salary for all job groups with morethen two employees

SQL> select job , count (*) , avg (sal) from emp

group by JOB

HAVING count(*) > 2;

JOB COUNT(*) AVG(SAL)

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

CLERK 4 625

As an example, suppose we want to list the averagesalary for all job groups with more then two employees.

Can you solve this (A difficult One)

LIST all departments from emp table withat least two clerks?

SQL> select DEPTNO from EMP

where JOB = 'CLERK'GROUP BY DEPTNOHAVING count(*) >= 2;

DEPTNO----------10



Labeling GROUP Columns

SQL> select job, sum(sal) as "TOTAL SALARY"from emp group by job;

JOB TOTAL SALARY -------------------- ------------CLERK 2500MANAGER 800

Manager 200

DATE ARITHMETIC

We can perform arithmetic operations on datefields, the operations you may use are

Date + Number

Date – Number

Date – Date



We can perform arithmetic operations on date fields, theoperations you may use are

Date + Number(add a number of days in a date, producing a date)

Date – Number

(Subtracts a number of days in a date, producing a date)

Date – Date

(Subtract one date from another, producing a number)

SQL> Select sal, hiredate, hiredate + 365

as "New Hiredate" from emp

SAL HIREDATE New Hiredate---------- --------- ---------

800 17-DEC-80 17-DEC-811600 20-FEB-81 20-FEB-821250 22-FEB-81 22-FEB-82

446.25 02-APR-81 02-APR-821250 28-SEP-81 28-SEP-82

Date + Number

Add a number of days in a date,producing a date.

Try fore.g.(hiredate – 31)

Also try forDate - Numbere.g.(hiredate – 365) Ande.g.(hiredate – 180)



Date – Date

Subtract one date from another, producing anumber.

Date – Date return’s result in number of days.

SQL> select sysdate, hiredate, (sysdate - hiredate) from emp;

SYSDATE HIREDATE (SYSDATE-HIREDATE)--------- --------- ------------------23-NOV-02 17-DEC-80 8011.5064723-NOV-02 20-FEB-81 7946.50647

23-NOV-02 22-FEB-81 7944.5064723-NOV-02 02-APR-81 7905.50647

SQL> select sysdate, hiredate, (sysdate - hiredate) / 365 from emp;

SYSDATE HIREDATE (SYSDATE-HIREDATE)/365--------- --------- --------- -------------23-NOV-02 17-DEC-80 21.949328523-NOV-02 20-FEB-81 21.771246323-NOV-02 22-FEB-81 21.765766923-NOV-02 02-APR-81 21.6589176

Here you can also useNumber Functions

like ROUND, CEIL,FLOOR etc…



ADD_MONTHS Function

ADD_MONTHS function is used to add anynumber of months in a date.

ADD_MONTHS takes two arguments

a date

and a integer representing a number of month.It returns a date after adding that number in months.

SQL>select hiredate, ADD_MONTHS (hiredate, 1)from emp;

HIREDATE ADD_MONTH--------- ---------17-DEC-80 17-JAN-8120-FEB-81 20-MAR-81

22-FEB-81 22-MAR-8102-APR-81 02-MAY-8128-SEP-81 28-OCT-8101-MAY-81 01-JUN-81

Also try

ADD_MONTHS(hiredate, 4) ADD_MONTHS(hiredate, 12)

….….



GREATEST & LEAST

GREATEST Function is used to Find the Laterdate from two given dates

LEAST Function is used to Find the Earlier datefrom two given dates

For Example

SQL> select greatest('12-JAN-2002' , sysdate) from dual

GREATEST-----------14-NOV-03

SQL> select greatest('12-JAN-2002' , ’11-DEC-2003’) from dual

GREATEST-----------11-DEC-2003

SQL> select LEAST('12-JAN-2002' , ‘23-NOV-01’) from dual

LEAST-----------23-NOV-01

Similarly you can use LEAST Function

We can also use GREATEST & LEAST functions For

two date columns e.g.

select GREATEST (‘Hiredate' ,’any_date_column’)

from mytable;



COUNT the employees (FromEMP table) which were hired in

each year.

YEAR Employees

1980 11981 101987 2

Time5 Mins

Select

TO_CHAR(HIREDATE, 'YYYY'), COUNT(*)

from emp

group by

TO_CHAR(HIREDATE, 'YYYY');



DATE Functions Contd…

MONTHS_BETWEEN

This function is used to find the difference innumber of months between two date columnsor two given dates.

For Example

contd..

MONTHS_BETWEEN contd..

SQL> select months_between('12-JAN-

2002','12-JAN-2001') from dual;

MONTHS_BETWEEN('12-JAN-2002','12-JAN-2001')

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

12



MONTHS_BETWEEN

SQL> select sysdate, hiredate,

months_between(sysdate, hiredate) as Diff From emp;

SYSDATE HIREDATE DIFF

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

01-Nov-03 17-DEC-80 276.50282

01-Nov-03 20-FEB-81 264.406045

01-Nov-03 22-FEB-81 264.341529

01-Nov-03 02-APR-81 262.986691

SQL> select hiredate, LAST_DAY(HIREDATE) as "LAST

DATE" from emp

HIREDATE LAST DATE

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

17-DEC-80 31-DEC-80

20-FEB-81 28-FEB-81

22-FEB-81 28-FEB-81

02-APR-81 30-APR-81

LAST_DAYThis function is used to find out the last date of the month

for any given date. LAST_DAY (HIREDATE)



LAST_DAY

SQL> Select LAST_DAY('20-DEC-2002') as Result

from dual;

RESULT

---------

31-DEC-02

NEXT_DAY

This function is used to find the NEXT day

(given) of any date (given).

NEXT_DAY (HIREDATE, ‘FRIDAY’)

contd..



NEXT_DAY contd..

SQL> Select hiredate,NEXT_DAY(HIREDATE, 'FRIDAY')

as "NEXT DAY" from emp;

HIREDATE NEXT DAY

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

17-DEC-80 19-DEC-80

20-FEB-81 27-FEB-81

22-FEB-81 27-FEB-81

02-APR-81 03-APR-81 NEXT DAY column

is showing the FirstFRIDAY coming afterGiven (hiredate) date.

NEXT_DAY contd..

SQL> select hiredate,

to_char(

NEXT_DAY(HIREDATE,'FRIDAY'),

'DY DD MM YY') as "NEXT DAY" from emp

HIREDATE NEXT DAY

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

17-DEC-80 FRI 19 12 80

20-FEB-81 FRI 27 02 81

22-FEB-81 FRI 27 02 81

02-APR-81 FRI 03 04 81

Finding NEXT_DAY

and also displaying

DAY by using TO_CHAR

function.



NEXT_DAY

SQL> selectNEXT_DAY(‘15-Nov-2004', ‘THUSDAY')

as “EID DAY” from dual;

EID DAY

---------

16-NOV-04

Thanks

SQL Lecture-6

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