RE: conditional 'order by'

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

Else .

End    

From: oracle-l-bounce_at_freelists.org [mailto:oracle-l-bounce_at_freelists.org] On Behalf Of Eugene Pipko
Sent: Friday, December 19, 2008 11:57 AM To: 'oracle-l_at_freelists.org'
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.  

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