Re: using calculated column in where-clause

On Wed, Jun 18, 2008 at 1:35 PM, Fernando Hevia <fhevia(at)ip-tel(dot)com(dot)ar> wrote:
>
>> -----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
> improvement.

I was under the impresion from previous discussions that the query
planner flattened these out to be the same query. Do you get
different query plans when you re-arrange this way?