> > -----Mensaje original-----
> > De: Scott Marlowe [mailto:scott.marlowe@gmail.com] Enviado el:
> > Miércoles, 18 de Junio de 2008 17:47
> > Para: Fernando Hevia
>
> > >
> > > 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?
> >
>
> Take a look at this example (tried on 8.2.7 & 8.1.11):
>
> create or replace function test(p1 integer, p2 integer)
> returns integer[] as $BODY$ declare
> retval integer[];
> begin
> raise info 'called test(%, %)', p1, p2;
> retval[0] = p1 + p2;
> retval[1] = p1 * p2;
> retval[2] = p1 - p2;
> return retval;
> end;
> $BODY$
> LANGUAGE 'plpgsql' IMMUTABLE;
>
> -- In this case function test is called three times:
> pg=# select (test(1, 2))[0] as sum, (test(1, 2))[1] as prod,
> (test(1, 2))[2] as dif;
> INFO: called test(1, 2)
> INFO: called test(1, 2)
> INFO: called test(1, 2)
> sum | prod | dif
> -----+------+-----
> 3 | 2 | -1
> (1 row)
>
>
> -- In this case function test is called only once:
> pg=# select res[0] as sum, res[1] as prod, res[2] as dif from
> pg-# (select (test(1, 2))::integer[] as res) t ;
> INFO: called test(1, 2)
> sum | prod | dif
> -----+------+-----
> 3 | 2 | -1
> (1 row)
>
> I assume the second form will perform better since test is
> being called only once.
> I might be missing something in this assumption but at first
> glance it seems pretty straightforward.
>
> Regards,
> Fernando.
>
--Follow up--
When I use one of the inner-query columns as a condition for the outer-query
the function is being called again:
pg=# select res[0] as sum, res[1] as prod, res[2] as dif from
pg-# (select (test(1, 2))::integer[] as res) t
pg-# where res[0] = 3;
INFO: called test(1, 2)
INFO: called test(1, 2)sum | prod | dif
-----+------+----- 3 | 2 | -1
(1 row)
Seems this blows away my theory, at least part of it.
