Re: substring result

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От Luis Silva
Тема Re: substring result
Дата
Msg-id BAY18-F21A610453467C5A2E42C00B5070@phx.gbl
обсуждение исходный текст
Ответ на Re: substring result  ("A. Kretschmer" <andreas.kretschmer@schollglas.com>)
Ответы Re: substring result  ("Luis Silva" <lfs12@hotmail.com>)
Список pgsql-novice
HI!!! that worked fine, but is it possible to get 'name' and  'address'
using the regexp_replace too? TKS A LOT again

>From: "A. Kretschmer" <andreas.kretschmer@schollglas.com>
>To: pgsql-novice@postgresql.org
>Subject: Re: [NOVICE] substring result
>Date: Mon, 13 Feb 2006 13:06:28 +0100
>
>am  13.02.2006, um 10:53:48 +0000 mailte Luis Silva folgendes:
> >
> > I there!! I'm trying to use regular expressions with postgresql. My
> > objective is to get from a long string the information that I need.
> > For example
> >
> > "name='joe' , address='portugal' " and I need to get 'joe' and
>'portugal'.
> > can I do it with select substring()? if I can,how? tks a lot
>
>test=# select * from foo;
>              string
>---------------------------------
>  name='joe' , address='portugal'
>(1 row)
>
>
>Write a function:
>- count the fields separeted by ',' and then for every field:
>
>
>test=# select regexp_replace(split_part(string,',',1), '\\m.*=', '') from
>foo;
>  regexp_replace
>----------------
>  'joe'
>(1 row)
>
>test=# select regexp_replace(split_part(string,',',2), '\\m.*=', '') from
>foo;
>  regexp_replace
>----------------
>   'portugal'
>(1 row)
>
>
>You can with trim() remove spaces if you need.
>
>
>
>HTH, Andreas
>--
>Andreas Kretschmer    (Kontakt: siehe Header)
>Heynitz:  035242/47215,      D1: 0160/7141639
>GnuPG-ID 0x3FFF606C http://wwwkeys.de.pgp.net
>  ===    Schollglas Unternehmensgruppe    ===
>
>---------------------------(end of broadcast)---------------------------
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