RE: conditional 'order by'
- From: "Kenneth Naim" <kennaim@xxxxxxxxx>
- To: <eugene.pipko@xxxxxxxxxxxx>, <oracle-l@xxxxxxxxxxxxx>
- Date: Fri, 19 Dec 2008 12:46:20 -0500
Just use a case statement.
Order by case when substr(code,2,1)='F' then decode(code,'F',0,'H',1,2)
When substr(code,2,1)='H' then
decode(code,'H',0,'F',1,2)
Else .
End
_____
From: oracle-l-bounce@xxxxxxxxxxxxx [mailto:oracle-l-bounce@xxxxxxxxxxxxx]
On Behalf Of Eugene Pipko
Sent: Friday, December 19, 2008 11:57 AM
To: 'oracle-l@xxxxxxxxxxxxx'
Subject: conditional 'order by'
Hi all,
I have a query that needs to be ordered by based on the second letter of the
code passed in it.
For instance if the code is '9F9Q' then I'd order by 'F' , then by 'H', then
by any other.
If the code is '7H7Q' then I'd order by 'H', then by 'F', then by any other.
I know that I can do it in 2 steps, but is it possible to do in one sql
statement?
Select .
From .
Where ..
If substr(code,2,1)='F' then
Order by decode(code,'F',0,'H',1,2);
Elsif substr(code,2,1)='H' then
Order by decode(code,'H',0,'F',1,2);
End if;
Thanks,
Eugene
P Please consider the environment before printing this e-mail.
Other related posts:
