Re: substring result
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
|
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)--------------------------- >TIP 6: explain analyze is your friend
В списке pgsql-novice по дате отправления: