Hi Arup, Two ways come to mind for me. They're pretty much the same as Szymon's, just minus the sample table creation. I would suggest creating a view instead, so you can just select from it whenever you please.
create view vw_employee as select * from employees where ((age(joining_date::date) like '5 years%') or (age(joining_date::date) like '10 years%') ) or create view vw_employee as select * from employees where ((to_char(joining_date, 'YYYY-MM') = to_char((now() - interval '5 years'), 'YYYY-MM') ) or (to_char(joining_date, 'YYYY-MM') = to_char((now() - interval '10 years'), 'YYYY-MM'))) And then to check the employees who have completed 5 or 10 years, you'll just do: select * from vw_employee This is done off the top of my head so there will likely be syntax errors, but I hope this can give you a general idea. - Rebecca On Mon, Jun 30, 2014 at 12:00 PM, Szymon Guz <mabew...@gmail.com> wrote: > > On 30 June 2014 12:38, Arup Rakshit <arupraks...@rocketmail.com> wrote: > >> I have employee table. Where I have a column joining_date. Now I am >> looking for a way to get all employee, who completed 5 years, 10 years >> current month. How to do so ? I am not able to figure this out. >> >> Regards, >> Arup Rakshit >> > > Hi, > take a look at this example: > > I've created a sample table: > > create table users(id serial, joining_date date); > > and filled it with sample data: > > insert into users(joining_date) select now() - (j::text || 'days' > )::interval from generate_series(1,10000) j; > > Then the query showing up all users who complete 5 and 10 years this month > can look like: > > with u as ( > select id, date_trunc('month', age(now()::date, joining_date)) age > from users > ) > select * > from u > where u.age in ('5 years', '10 years'); > > > - Szymon >