Re: RE:Re: RE:Re: A complex SQL query

 This is much cleaner way
select * from (
select usr_id, 
       employer, 
       start_date,
       end_date,
       rank() OVER (PARTITION BY employer ORDER BY start_date DESC) AS rnk
from <tablename> ) t
where t.rnk=1;
Thanks
    On Sunday, September 8, 2019, 7:40:05 AM GMT+5:30, jj08 <[email protected]> wrote:  
 
 
Hello Vicor!

It works!!!

Thank you very much.





--From: [email protected]
--To: [email protected]
--CC: [email protected]
--Date: 9/7/2019 2:33:40 PM
--Subject: Re: RE:Re: A complex SQL query
сб, 7 сент. 2019 г. в 21:23, jj08 <[email protected]>:

To Janis or other members of the forum:

My sample table contains other columns as well.

One of the columns that I did not include in my first message is "position".
The last table has been modified as follows:

+--------+----------+------------+------------------------+
| usr_id | employer | position   + start_date | end_date  |
+--------+----------+------------+------------------------+
| A      | Goo      | CTO        |            | 201904    |         
| A      | Micro    | Mgr        | 201704     | 201903    |

| B      | Micro    | Engg_Mgr   | 201706     | -         |
| B      | Goo      | Researcher | 201012     | 201705    |
| B      | Micro    | Postdoc    | 201001     | 201011    |
+--------+----------+------------+------------------------+

SELECT DISTINCT ON (usr_id)
       *
  FROM employment
 WHERE employer='Micro'
 ORDER BY usr_id, start_date DESC; -- 
Victor Yegorov
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