Re: grouping subsets

Поиск
Список
Период
Сортировка
От A. Kretschmer
Тема Re: grouping subsets
Дата
Msg-id 20100722100226.GE10348@a-kretschmer.de
обсуждение исходный текст
Ответ на grouping subsets  (Rainer Stengele <rainer.stengele@diplan.de>)
Ответы Re: grouping subsets  (Richard Huxton <dev@archonet.com>)
Список pgsql-sql
In response to Rainer Stengele :
> Hi,
> 
> having a table similar to
> 
> | 1 | B | [2010-07-15 Do] |
> | 1 | B | [2010-07-16 Fr] |
> |---+---+-----------------|
> | 2 | C | [2010-07-17 Sa] |
> | 2 | C | [2010-07-18 So] |
> |---+---+-----------------|
> | 1 | B | [2010-07-19 Mo] |
> | 1 | B | [2010-07-20 Di] |
> | 1 | B | [2010-07-21 Mi] |
> | 1 | B | [2010-07-22 Do] |
> |---+---+-----------------|
> | 3 | D | [2010-07-23 Fr] |
> 
> a simple group by gives me:
> 
> | 6 | B |
> | 4 | C |
> | 3 | D |
> 
> 
> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the
columchanges.
 
> Is there a way to get
> 
> | 2 | B |
> | 4 | C |
> | 4 | B |
> | 3 | D |
> 
> by SQL only?

I think, the problem is that there are 2 identical groups. I think, you
can write a pl/pgsql-proc, selecting all ordered by the date-field and
walking through the result to do the grouping, checking if the 2nd
column is different from the previous. 

With plain SQL it's maybe possible too, but i don't know how ...

Regards, Andreas
-- 
Andreas Kretschmer
Kontakt:  Heynitz: 035242/47150,   D1: 0160/7141639 (mehr: -> Header)
GnuPG: 0x31720C99, 1006 CCB4 A326 1D42 6431  2EB0 389D 1DC2 3172 0C99


В списке pgsql-sql по дате отправления:

Предыдущее
От: Rainer Stengele
Дата:
Сообщение: grouping subsets
Следующее
От: Richard Huxton
Дата:
Сообщение: Re: grouping subsets