DATE command in SQL

                                                                                             edited  by Sukamal

1:

 select sysdate

from dual

output:

SYSDATE

13-FEB-19

2:

select to_char(sysdate,'dd/mm/yy')

from dual

output:

TO_CHAR(SYSDATE,'DD/MM/YY')

13/02/19

3:

 select to_char(sysdate,'dd/mm/yyyy')

from dual

output:

TO_CHAR(SYSDATE,'DD/MM/YYYY')

13/02/2019

4:

select to_char(sysdate,'dy/mon/yyyy')

from dual

output:

TO_CHAR(SYSDATE,'DY/MON/YYYY')

wed/feb/2019

5:

select to_char(sysdate,'day/month/yyyy')

from dual

output:

TO_CHAR(SYSDATE,'DAY/MONTH/YYYY')

wednesday/february /2019

6:

select to_char(sysdate,'dd/mm/yy,hh:mi:ss')

from dual

output:

TO_CHAR(SYSDATE,'DD/MM/YY,HH:MI:SS')

13/02/19,02:48:55

7:

create table employee(id varchar2(20) primary key,name varchar2(20),dob date);

insert into employee values('1','vk',to_date('05/11/1988','dd/mm/yyyy'));

insert into employee values('2','msd',to_date('02/07/1982','dd/mm/yyyy'));

select *

from employee

output:

ID	NAME	DOB
2	msd	02-JUL-82
1	vk	05-NOV-88

8:Find the name of the employee who is younger than msd.

select name

from employee

where dob>(select dob from employee where name ='msd');

output:

NAME

vk

9:Find the name of employee who was born before 1985.

select name

from employee

where dob <to_date('1985','yyyy');

output:

NAME

msd

10:Find the name and month of dob of each employee.

 select name,to_char(dob,'mm') as month

from employee

output:

NAME	MONTH
msd	07
vk	11

11:Find the age of all employee.

select floor((sysdate-dob)/364.25)

from employee;

output:

FLOOR((SYSDATE-DOB)/364.25)

30

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