Re: Subquery with multiple rows

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От David G. Johnston
Тема Re: Subquery with multiple rows
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Msg-id CAKFQuwa-pW2=qMxXMt-7yxc94+4UbRH0kdo=rrJoRE_L2+_RyA@mail.gmail.com
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Ответ на Subquery with multiple rows  (Weiss, Jörg <J.Weiss@dvz-mv.de>)
Ответы Re: Subquery with multiple rows  (Weiss, Jörg <J.Weiss@dvz-mv.de>)
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On Tue, Jun 14, 2016 at 2:22 AM, Weiss, Jörg <J.Weiss@dvz-mv.de> wrote:

Hi all!

 

How can I outsource a subquery?

 

An Example:

SELECT DISTINCT a.*,


​Lose the DISTINCT.  DISTINCT is a code smell.  In this case it is also pointless since a.* had better already be unique and its the only table in the query..​

​And, please don't top-post.

(       SELECT SUM(std)

        FROM all_std

        WHERE (a.status <=5 AND status = 5)

        AND foerd_id IN (SELECT foerd_id FROM foerds WHERE mass_id = '55896') 

) AS done_std

(       SELECT SUM(anerk_std)

        FROM all_std

        WHERE (a.status >5 AND status < 5)

        AND foerd_id IN (SELECT foerd_id FROM foerds WHERE mass_id = '55896')

) AS accepted_std

FROM table_a a

WHERE a.mass_id = '55896';

 

The sub-subquery is “SELECT foerd_id FROM foerds WHERE mass_id = '55896')” an delivers more than one row.

 

Now I want to run the subquery only one time.

 

I tried this:

 

SELECT DISTINCT a.*, 

(       SELECT SUM(std)

        FROM all_std

        WHERE (a.status <=5 AND status = 5)

        AND foerd_id IN (f.foerd_id) 

) AS done_std,

(       SELECT SUM(anerk_std)

        FROM all_std

        WHERE (a.status >5 AND status < 5)

        AND foerd_id IN (f.foerd_id)

) AS accepted_std

FROM table_a a,

(SELECT foerd_id FROM foerds WHERE mass_id = '55896')  f

WHERE a.mass_id = '55896';

 

But the result is not correct because I got one row for every row of the of “f.foerd_id”.

Allowed is only one row.

 

How must the SQL looks like to get the right result?



​The following should work on recent versions though you will need to play with the syntax.​

​SELECT a.*, my_sums.*
FROM table_a a
LATERAL JOIN (SELECT SUM(anerk_std) FILTER (...) AS accepted_std, SUM(std) FILTER (...) AS done_std FROM all_std WHERE all_std.mass_id = a.mass_id)​
 
​AS my_sums​
WHERE a.mass_id = $1

David J.

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