Re: Variable column name
От | Bob Pawley |
---|---|
Тема | Re: Variable column name |
Дата | |
Msg-id | F5352B6A066D447A98E3E7E3DE0AC8EF@BobPC обсуждение исходный текст |
Ответ на | Re: Variable column name (Raymond O'Donnell <rod@iol.ie>) |
Список | pgsql-general |
-----Original Message----- From: Raymond O'Donnell Sent: Friday, September 02, 2011 10:38 AM To: Bob Pawley Cc: Bill Moran ; Postgresql Subject: Re: [GENERAL] Variable column name On 02/09/2011 18:33, Bob Pawley wrote: > > > -----Original Message----- >> From: Bill Moran > Sent: Thursday, September 01, 2011 8:19 AM > To: Bob Pawley > Cc: Postgresql > Subject: Re: [GENERAL] Variable column name > > http://www.postgresql.org/docs/9.0/static/plpgsql-statements.html > Section 39.5.4 > > If you're not familiar with plpgsql at all, you might want to start with > this: > http://www.postgresql.org/docs/9.0/static/plpgsql-structure.html > > > Thanks for the suggestion. > > Following is my interpretation of what I have read. > > I am getting an error -- "column "1" does not exist" > > Could someone point to what I am doing wrong? > > Bob > > Select 2 into point_array ; > Select "1" into column ; Hi Bob, I think it is the double-quotes around the 1; just leave them out to get a literal integer 1: select 1 into column; If I understand correctly, the double-quotes make Postgres look for a column named "1". Ray. -- Raymond O'Donnell :: Galway :: Ireland rod@iol.ie Ray I've named columns 1 through 10 so that it will be easy to determine the next column in the loop. When I use the following it works well. Update library.compare Set "1"[2] = (select st_distance (st.............................
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