RE: [GENERAL] question 2 - Mailing list pgsql-general

From Jackson, DeJuan
Subject RE: [GENERAL] question 2
Date
Msg-id F10BB1FAF801D111829B0060971D839F54EF5E@cpsmail
Whole thread Raw
List pgsql-general
The easiest way is to give it another name
  select presto.*, libro.tipo, libro.id_libro as id_libro2 from ...
and just grab id_libro2 for the second id_libro.
You can also select it by column number in most interfaces.
    -DEJ
> -----Original Message-----
> From:    ZioBudda [SMTP:michel@michel.enter.it]
> Sent:    Friday, December 04, 1998 12:06 PM
> To:    pgsql-general@postgreSQL.org
> Subject:    [GENERAL] question 2
>
> Hi again,
> I have this query :
> select prestito.*, libro.tipo,libro.id_libro from prestito,libro where
> prestito.id_libro =
>  libro.id_libro_key
> that return me (correct!):
> id_libro|id_utente|data_prestito|visto|notifica1|notifica2|notifica3|t
> ipo|id_libro
> --------+---------+-------------+-----+---------+---------+---------+-
> ---+--------
>       12|       37|   12-04-1998|f    |         |         |         |
> 1|s1
>       13|       37|   12-04-1998|f    |         |         |         |
> 2|k1
> (2 rows)
>
> but if you see the item libro.id_libro is id_libro (s1,k1) and in the
> prestito.*
>  there is another id_libro (11,12). Now I want to know how can I
> select
> the second "id_lib
> ro".
>
>
>
>
> "Il divertimento e' giusto se la scimmia ci prende gusto"
> --
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