From: | Oliver Elphick <olly(at)lfix(dot)co(dot)uk> |
---|---|
To: | Nabil Sayegh <postgresql(at)e-trolley(dot)de> |
Cc: | pgsql-novice(at)postgresql(dot)org |
Subject: | Re: perlsub |
Date: | 2003-10-04 06:20:10 |
Message-ID: | 1065248410.4459.95.camel@linda.lfix.co.uk |
Views: | Raw Message | Whole Thread | Download mbox | Resend email |
Thread: | |
Lists: | pgsql-novice |
On Sat, 2003-10-04 at 00:29, Nabil Sayegh wrote:
> Hello again,
>
> now that perlsub works i need to know how backreferences work with
> plperl. $1 - $9 doesn't work. Any idea ?
>
> perlsub:
> -----------------------------------------------
> CREATE FUNCTION perlsub(text, text, text) RETURNS text AS '
> my ($data, $pat, $repl) = @_;
> $data =~ s/$pat/$repl/;
> return $data
> ' LANGUAGE 'plperl';
> -----------------------------------------------
>
> SELECT perlsub('a=b','([^=]+)=(.+)','key:$1;val:$2');
>
> Any idea ? Just a matter of quoting ?
It's a problem with Perl itself rather than with PL/Perl. I tried that
in a Perl script and it produces the same result. I couldn't find any
way of including $ in the replacement string as a metacharacter.
You need to consult a Perl guru.
When you do get an example that works in a script, remember to double
any single quotes and backslashes when you create the function.
> Also this example doesn't work with * instead of + :(
I don't see any difference:
junk=# SELECT perlsub('a=b','([^=]*)=(.*)','key:$1;val:$2');
perlsub
---------------
key:$1;val:$2
(1 row)
junk=# SELECT perlsub('a=b','([^=]+)=(.+)','key:$1;val:$2');
perlsub
---------------
key:$1;val:$2
(1 row)
--
Oliver Elphick Oliver(dot)Elphick(at)lfix(dot)co(dot)uk
Isle of Wight, UK http://www.lfix.co.uk/oliver
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========================================
"For the word of God is quick, and powerful, and
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dividing asunder of soul and spirit, and of the joints
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