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Re: a JOIN on same table, but 'slided over' - Mailing list pgsql-general

From Gurjeet Singh
Subject Re: a JOIN on same table, but 'slided over'
Date
Msg-id 65937bea0706260644y794881c3q86a055ec564bf30e@mail.gmail.com
Whole thread Raw
In response to Re: a JOIN on same table, but 'slided over'  ("Gurjeet Singh" <singh.gurjeet@gmail.com>)
Responses Re: a JOIN on same table, but 'slided over'  ("news.gmane.org" <nis@superlativ.dk>)
Re: a JOIN on same table, but 'slided over'  (Rafal Pietrak <rafal@zorro.isa-geek.com>)
List pgsql-general
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I missed the ORDER BY clause... Here it goes:

select    t1.id as id, t2.id as "id+1",
        t1.thread as thread, t2.thread as "thread+1",
        t1.info as info, t2.info as "info+1"
from test as t1, test as t2
where t2.id = ( select min(id) from test as t3 where t3.id > t1.id )
order by t1.id asc;

Also note that this query is much cheaper that the 'distinct on' query by more than two orders on magnitude ( 217.86 vs. 98040.67):

postgres=# explain
postgres-# select
postgres-#     distinct on (t1.id)
postgres-#     t1.*, t2.*
postgres-# from
postgres-#     test t1
postgres-#     join test t2 on t2.id > t1.id
postgres-# order by t1.id asc, t2.id asc;
                                           QUERY PLAN
------------------------------------------------------------------------------------------------
 Unique  (cost=95798.00..98040.67 rows=1160 width=80)
   ->  Sort  (cost=95798.00..96919.33 rows=448533 width=80)
         Sort Key: t1.id, t2.id
         ->  Nested Loop  (cost=0.00..13827.29 rows=448533 width=80)
               ->  Seq Scan on test t1  (cost=0.00..21.60 rows=1160 width=40)
               ->  Index Scan using test_id_key on test t2  (cost=0.00..7.06 rows=387 width=40)
                     Index Cond: (t2.id > t1.id)
(7 rows)
Time: 5.003 ms
postgres=# explain
postgres-# select       t1.id as id, t2.id as "id+1",
postgres-#              t1.thread as thread, t2.thread as "thread+1",
postgres-#              t1.info as info, t2.info as "info+1"
postgres-# from test as t1, test as t2
postgres-# where t2.id = ( select min(id) from test as t3 where t3.id > t1.id )
postgres-# order by t1.id asc;
                                                QUERY PLAN
----------------------------------------------------------------------------------------------------------
 Sort  (cost=214.96..217.86 rows=1160 width=80)
   Sort Key: t1.id
   ->  Hash Join  (cost= 36.10..155.92 rows=1160 width=80)
         Hash Cond: ((subplan) = t2.id)
         ->  Seq Scan on test t1  (cost=0.00..21.60 rows=1160 width=40)
         ->  Hash  (cost=21.60..21.60 rows=1160 width=40)
               ->  Seq Scan on test t2  (cost=0.00..21.60 rows=1160 width=40)
         SubPlan
           ->  Result  (cost=0.13..0.14 rows=1 width=0)
                 InitPlan
                   ->  Limit  (cost= 0.00..0.13 rows=1 width=4)
                         ->  Index Scan using test_id_key on test t3  (cost=0.00..51.02 rows=387 width=4)
                               Index Cond: (id > $0)
                               Filter: (id IS NOT NULL)
(14 rows)
Time: 4.125 ms


Best regards,
--
gurjeet[.singh]@EnterpriseDB.com
singh.gurjeet@{ gmail | hotmail | indiatimes | yahoo }.com

17°29'34.37"N  78°30'59.76"E - Hyderabad *
18°32'57.25"N  73°56'25.42"E - Pune

Sent from my BlackLaptop device

On 6/26/07, Gurjeet Singh <singh.gurjeet@gmail.com > wrote:
Hi Rafal,

    Just a note that this is not standard SQL... 'distinct on' is an extension to SQL provided by postgres.

Following query utilizes the standard SQL to get the same results:

select    t1.id as id, t2.id as "id+1",
        t1.thread as thread, t2.thread as "thread+1",
        t1.info as info, t2.info as "info+1"
from test as t1, test as t2
where t2.id = ( select min(id) from test as t3 where t3.id > t1.id);

HTH
--
gurjeet[.singh]@EnterpriseDB.com
singh.gurjeet@{ gmail | hotmail | indiatimes | yahoo }.com

17°29'34.37"N  78°30'59.76"E - Hyderabad *
18°32'57.25"N  73°56' 25.42 "E - Pune

Sent from my BlackLaptop device


On 6/26/07, Rafal Pietrak < rafal@zorro.isa-geek.com> wrote:
Marvelous! Thenx!

-R

On Tue, 2007-06-26 at 10:06 +0200, hubert depesz lubaczewski wrote:
> On 6/26/07, Rafal Pietrak < rafal@zorro.isa-geek.com> wrote:
>         Is there an SQL construct to get it?
>
> select
>     distinct on (t1.id)
>     t1.*, t2.*
> from
>     test t1
>     join test t2 on t2.id > t1.id
> order by t1.id asc, t2.id asc
>
> should do the trick.
>
> depesz
>
> --
> http://www.depesz.com/ - nowy, lepszy depesz

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