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To: pgsql-sql@postgresql.org
From: Tim Landscheidt <tim@tim-landscheidt.de>
Subject: Re: Passing function parameters to regexp_replace
Date: Sat, 17 Sep 2011 17:07:03 +0000
Organization: <URI:http://www.tim-landscheidt.de/>
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Message-ID: <m3fwjvt7wo.fsf@passepartout.tim-landscheidt.de>
References: <201109171321.43309.leif@solumslekt.org>
	<201109171331.58663.leif@solumslekt.org>
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Leif Biberg Kristensen <leif@solumslekt.org> wrote:

>> UPDATE sources SET source_text = regexp_replace(source_text,
>> E'n="(.*?)$1(.*?)"', E'n="\\1$2\\2"', 'g') where source_text like
>> '%n="%$2%">%';

> Sorry, I pasted a literal replacement, and substituted the parameters by hand.
> The expression should of course be

> UPDATE sources SET source_text = regexp_replace(source_text,
> E'n="(.*?)$1(.*?)"', E'n="\\1$2\\2"', 'g') where source_text like
> '%n="%$1%">%'

Try:

> UPDATE sources SET source_text = regexp_replace(source_text,
> CONCAT(E'n="(.*?)', $1, E'(.*?)"'), CONCAT(E'n="\\1', $2, '\\2"', 'g') where source_text like
> CONCAT('%n="%', $1, '%">%')

If $1 and $2 (can) include meta characters, you have to es-
cape them properly.

  Please consider that regexp_replace() uses POSIX Regular
Expressions while LIKE uses a different syntax. If possible,
I would replace the LIKE expression with its "~" equivalent
so chances of confusion are minimized.

Tim