Re: Place of subselect

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От A. Kretschmer
Тема Re: Place of subselect
Дата
Msg-id 20081125075624.GB10443@a-kretschmer.de
обсуждение исходный текст
Ответ на Place of subselect  ("Guillaume Bog" <guibog@gmail.com>)
Ответы Re: Place of subselect  ("Guillaume Bog" <guibog@gmail.com>)
Список pgsql-general
am  Tue, dem 25.11.2008, um 15:34:57 +0800 mailte Guillaume Bog folgendes:
> Hi dear Postgres users.
>
> I have performance issues if I do the following pseudo-query:
>
> SELECT a, b, (SELECT count(*) FROM t2 WHERE something) AS c
> FROM t1 ORDER BY a LIMIT 10;
>
> After some tests, it seems to me that the subquery on t2 is computed for all
> rows of t1. As I don't "ORDER BY c", there is no need to compute c for every
> row. I know I can (or should ?) work with joins or with a subquery in the from
> clause, but I'd like to make sure there is no other way before changing my
> sqls.

Please check your presumption with explain analyse <your query>.

For example:

test=*# explain analyse select t1.*, (select count(1) from t2) from t1 order by 1 limit 5;
                                                  QUERY PLAN
--------------------------------------------------------------------------------------------------------------
 Limit  (cost=186.54..186.55 rows=5 width=4) (actual time=0.087..0.104 rows=3 loops=1)
   InitPlan
     ->  Aggregate  (cost=36.75..36.76 rows=1 width=0) (actual time=0.022..0.024 rows=1 loops=1)
           ->  Seq Scan on t2  (cost=0.00..31.40 rows=2140 width=0) (actual time=0.004..0.008 rows=1 loops=1)
   ->  Sort  (cost=149.78..155.13 rows=2140 width=4) (actual time=0.082..0.088 rows=3 loops=1)
         Sort Key: i
         ->  Seq Scan on t1  (cost=0.00..31.40 rows=2140 width=4) (actual time=0.046..0.056 rows=3 loops=1)
 Total runtime: 0.197 ms
(8 rows)


Both tables executes only one scan.


Andreas
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Andreas Kretschmer
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