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Re: using calculated column in where-clause

From: "Fernando Hevia" <fhevia(at)ip-tel(dot)com(dot)ar>
To: "'Patrick Scharrenberg'" <pittipatti(at)web(dot)de>,<pgsql-sql(at)postgresql(dot)org>
Subject: Re: using calculated column in where-clause
Date: 2008-06-18 19:35:48
Message-ID: 022001c8d17a$827b67c0$ (view raw, whole thread or download thread mbox)
Lists: pgsql-sql
> -----Mensaje original-----
> De: pgsql-sql-owner(at)postgresql(dot)org 
> [mailto:pgsql-sql-owner(at)postgresql(dot)org] En nombre de Patrick 
> Scharrenberg
> Enviado el: Martes, 17 de Junio de 2008 17:46
> Para: pgsql-sql(at)postgresql(dot)org
> Asunto: [SQL] using calculated column in where-clause
> Hi!
> I'd like to do some calculation with values from the table, 
> show them a new column and use the values in a where-clause.
> Something like this
> select a, b , a*b as c from ta where c=2;
> But postgresql complains, that column "c" does not exist.
> Do I have to repeat the calculation (which might be even more complex
> :-) ) in the "where"-clause, or is there a better way?

For complex calculations I have obtained better performance using nested
queries. For example:

select a, b, c select 
   ( select a, b, a*b as c from ta) subquery1
where c = 2;

This nesting is probably overhead in such a simple case as this, but in more
complex ones and specially with volatile functions it will provide an


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