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* ordered by join? ranked aggregate? how to?
@ 2009-09-14 14:25 wstrzalka <wstrzalka@gmail.com>
0 siblings, 2 replies; 5+ messages in thread
From: wstrzalka @ 2009-09-14 14:25 UTC (permalink / raw)
To: pgsql-sql
What I need is to join 2 tables
CREATE TABLE master(
id INT4
);
CREATE TABLE slave (
master_id INT4,
rank INT4,
value TEXT);
What I need is to make the query:
SELECT m.id, array_agg(s.value) AS my_problematic_array
FROM master AS m LEFT JOIN slave AS s ON (m.id = s.master_id)
GROUP BY m.id;
return the 'my_problematic_array' in order specified by slave.rank
As you probably can guest I don't have any idea know how to do it :/
^ permalink raw reply [nested|flat] 5+ messages in thread
* Re: ordered by join? ranked aggregate? how to?
@ 2009-09-15 07:49 A. Kretschmer <andreas.kretschmer@schollglas.com>
parent: wstrzalka <wstrzalka@gmail.com>
1 sibling, 0 replies; 5+ messages in thread
From: A. Kretschmer @ 2009-09-15 07:49 UTC (permalink / raw)
To: pgsql-sql
In response to wstrzalka :
> What I need is to join 2 tables
>
> CREATE TABLE master(
> id INT4
> );
>
>
> CREATE TABLE slave (
> master_id INT4,
> rank INT4,
> value TEXT);
>
>
> What I need is to make the query:
>
> SELECT m.id, array_agg(s.value) AS my_problematic_array
> FROM master AS m LEFT JOIN slave AS s ON (m.id = s.master_id)
> GROUP BY m.id;
>
> return the 'my_problematic_array' in order specified by slave.rank
>
> As you probably can guest I don't have any idea know how to do it :/
test=*# select * from master;
id
----
1
2
(2 rows)
test=*# select * from slave;
master_id | rank | value
-----------+------+-------
1 | 5 | 5
1 | 3 | 3
1 | 7 | 7
(3 rows)
test=*#
select id, array_agg(unnest) from (
select id, unnest(my_problematic_array) from (
SELECT m.id, array_agg(s.value) AS my_problematic_array FROM master AS m LEFT JOIN slave AS s ON (m.id = s.master_id) GROUP BY m.id
) foo order by 1,2
) bar group by 1;
id | array_agg
----+-----------
1 | {3,5,7}
2 | {NULL}
(2 rows)
Andreas
--
Andreas Kretschmer
Kontakt: Heynitz: 035242/47150, D1: 0160/7141639 (mehr: -> Header)
^ permalink raw reply [nested|flat] 5+ messages in thread
* Re: ordered by join? ranked aggregate? how to?
@ 2009-09-15 07:56 A. Kretschmer <andreas.kretschmer@schollglas.com>
parent: wstrzalka <wstrzalka@gmail.com>
1 sibling, 1 reply; 5+ messages in thread
From: A. Kretschmer @ 2009-09-15 07:56 UTC (permalink / raw)
To: pgsql-sql
In response to wstrzalka :
> What I need is to join 2 tables
>
> CREATE TABLE master(
> id INT4
> );
>
>
> CREATE TABLE slave (
> master_id INT4,
> rank INT4,
> value TEXT);
>
>
> What I need is to make the query:
>
> SELECT m.id, array_agg(s.value) AS my_problematic_array
> FROM master AS m LEFT JOIN slave AS s ON (m.id = s.master_id)
> GROUP BY m.id;
Faster solution (compared to my other email):
test=# select id, array_agg(value) from (SELECT m.id, s.value FROM
master AS m LEFT JOIN slave AS s ON (m.id = s.master_id) order by 1,2)
foo group by 1;
id | array_agg
----+-----------
1 | {3,5,7}
2 | {NULL}
(2 rows)
Andreas
--
Andreas Kretschmer
Kontakt: Heynitz: 035242/47150, D1: 0160/7141639 (mehr: -> Header)
^ permalink raw reply [nested|flat] 5+ messages in thread
* Re: ordered by join? ranked aggregate? how to?
@ 2009-09-15 08:10 wstrzalka <wstrzalka@gmail.com>
parent: A. Kretschmer <andreas.kretschmer@schollglas.com>
0 siblings, 1 reply; 5+ messages in thread
From: wstrzalka @ 2009-09-15 08:10 UTC (permalink / raw)
To: pgsql-sql
On 15 Wrz, 09:56, andreas.kretsch...@schollglas.com ("A. Kretschmer")
wrote:
> In response to wstrzalka :
>
>
>
> > What I need is to join 2 tables
>
> > CREATE TABLE master(
> > id INT4
> > );
>
> > CREATE TABLE slave (
> > master_id INT4,
> > rank INT4,
> > value TEXT);
>
> > What I need is to make the query:
>
> > SELECT m.id, array_agg(s.value) AS my_problematic_array
> > FROM master AS m LEFT JOIN slave AS s ON (m.id = s.master_id)
> > GROUP BY m.id;
>
> Faster solution (compared to my other email):
>
> test=# select id, array_agg(value) from (SELECT m.id, s.value FROM
> master AS m LEFT JOIN slave AS s ON (m.id = s.master_id) order by 1,2)
> foo group by 1;
> id | array_agg
> ----+-----------
> 1 | {3,5,7}
> 2 | {NULL}
> (2 rows)
>
> Andreas
> --
> Andreas Kretschmer
> Kontakt: Heynitz: 035242/47150, D1: 0160/7141639 (mehr: -> Header)
>
> --
> Sent via pgsql-sql mailing list (pgsql-...@postgresql.org)
> To make changes to your subscription:http://www.postgresql.org/mailpref/pgsql-sql
Yes. And No :)
The problem is that in original query the aggregate can be used more
then once :/
So in fact it's like:
SELECT m.id, array_agg(s1.value), array_agg(s2.value)
FROM master AS m LEFT JOIN slave AS s1 ON (m.id = s1.master_id AND
SOME_CONDITION_ON_S1)
LEFT JOIN slave AS s2 ON (m.id =
s2.master_id AND SOME_OTHER_CONDITION_ON_S1)
GROUP BY m.id;
^ permalink raw reply [nested|flat] 5+ messages in thread
* Re: ordered by join? ranked aggregate? how to?
@ 2009-09-15 08:12 wstrzalka <wstrzalka@gmail.com>
parent: wstrzalka <wstrzalka@gmail.com>
0 siblings, 0 replies; 5+ messages in thread
From: wstrzalka @ 2009-09-15 08:12 UTC (permalink / raw)
To: pgsql-sql
On 15 Wrz, 10:10, wstrzalka <wstrza...@gmail.com> wrote:
> On 15 Wrz, 09:56, andreas.kretsch...@schollglas.com ("A. Kretschmer")
> wrote:
>
>
>
> > In response to wstrzalka :
>
> > > What I need is to join 2 tables
>
> > > CREATE TABLE master(
> > > id INT4
> > > );
>
> > > CREATE TABLE slave (
> > > master_id INT4,
> > > rank INT4,
> > > value TEXT);
>
> > > What I need is to make the query:
>
> > > SELECT m.id, array_agg(s.value) AS my_problematic_array
> > > FROM master AS m LEFT JOIN slave AS s ON (m.id = s.master_id)
> > > GROUP BY m.id;
>
> > Faster solution (compared to my other email):
>
> > test=# select id, array_agg(value) from (SELECT m.id, s.value FROM
> > master AS m LEFT JOIN slave AS s ON (m.id = s.master_id) order by 1,2)
> > foo group by 1;
> > id | array_agg
> > ----+-----------
> > 1 | {3,5,7}
> > 2 | {NULL}
> > (2 rows)
>
> > Andreas
> > --
> > Andreas Kretschmer
> > Kontakt: Heynitz: 035242/47150, D1: 0160/7141639 (mehr: -> Header)
>
> > --
> > Sent via pgsql-sql mailing list (pgsql-...@postgresql.org)
> > To make changes to your subscription:http://www.postgresql.org/mailpref/pgsql-sql
>
> Yes. And No :)
>
> The problem is that in original query the aggregate can be used more
> then once :/
>
> So in fact it's like:
>
> SELECT m.id, array_agg(s1.value), array_agg(s2.value)
> FROM master AS m LEFT JOIN slave AS s1 ON (m.id = s1.master_id AND
> SOME_CONDITION_ON_S1)
> LEFT JOIN slave AS s2 ON (m.id =
> s2.master_id AND SOME_OTHER_CONDITION_ON_S1)
> GROUP BY m.id;
However it still may work as the order by is the same :) Will try ...
^ permalink raw reply [nested|flat] 5+ messages in thread
end of thread, other threads:[~2009-09-15 08:12 UTC | newest]
Thread overview: 5+ messages (download: mbox mbox.gz follow: Atom feed)
-- links below jump to the message on this page --
2009-09-14 14:25 ordered by join? ranked aggregate? how to? wstrzalka <wstrzalka@gmail.com>
2009-09-15 07:49 ` A. Kretschmer <andreas.kretschmer@schollglas.com>
2009-09-15 07:56 ` A. Kretschmer <andreas.kretschmer@schollglas.com>
2009-09-15 08:10 ` wstrzalka <wstrzalka@gmail.com>
2009-09-15 08:12 ` wstrzalka <wstrzalka@gmail.com>
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