Re: sub-select with aggregate - Mailing list pgsql-sql

From Stephan Szabo
Subject Re: sub-select with aggregate
Date
Msg-id 20021023081036.M5349-100000@megazone23.bigpanda.com
Whole thread Raw
In response to Re: sub-select with aggregate  (Tomasz Myrta <jasiek@klaster.net>)
Responses Re: sub-select with aggregate
Re: sub-select with aggregate
List pgsql-sql
On Wed, 23 Oct 2002, Tomasz Myrta wrote:

> Uz.ytkownik Stephan Szabo napisa?:
> > But you can't do that anyway, because you don't expose group_id
> > in the original query.  I assume user_id was a mistake then and was
> > meant to be group_id or that both were meant to be in the
> > select list.
> Yes, I meant group_id, but in orginal query I didn't have to add
> group_id to select list.

Without group_id in the select list you couldn't do a where
group_id = <something> if the select was a view.

> > In the first case changing the order means that the output
> > group_id column is X.group_id rather than users.group_id
> > (using removes one of them which is why group_id isn't
> > ambiguous.  In the second it uses on to get both group_ids
> > and exposes the one from X.
> The problem isn't ambigous columns, but speed.
> I think Postgres first performs sub-query with all records from table
> (it takes very long time). After this Postgres permforms joining table
> with sub-query. The question is: How to speed up query like this? How to
> give param group_id from first table (users) to subquery?

Did you see the other two queries I gave?  On 7.3, both of those queries
appear (according to explain output) to do the limiting of group_id
inside the subquery rather than doing the subquery with all rows.
The explanation above was why I believe it was different from your
original query.




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