Trensett wrote at Wed, 16 Jul 2003 18:29:51 -0500:
>> The next will work:
>> my @variable = $string =~ /\((.*?)\)/g;
>
> what's the exact implication of ".*?"?
> Why wouldn't just .* in parenthesis work for this case?
A good answer can also be found in
perldoc perlre
and
perldoc -q greedy
Gree
trensett wrote:
-- Forwarded message --
Date: Wed, 16 Jul 2003 18:23:35 -0500 (EST)
From: trensett <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED]
Subject: Re: Regular Expression
On Wed, 16 Jul 2003, Janek Schleicher wrote:
Nick Diel wrote at Tue, 15 Jul 2003 11:12:18 -0600:
, with the question mark it matches up to the
first, which is what you want (I assume).
no ? = greedy
? = on a diet
Cheers,
Nigel
> -Original Message-
> From: trensett [mailto:[EMAIL PROTECTED]
> Sent: 17 July 2003 00:30
> To: [EMAIL PROTECTED]
> Subject: Re: Regular
-- Forwarded message --
Date: Wed, 16 Jul 2003 18:23:35 -0500 (EST)
From: trensett <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED]
Subject: Re: Regular Expression
On Wed, 16 Jul 2003, Janek Schleicher wrote:
> Nick Diel wrote at Tue, 15 Jul 2003 11:12:18 -0600:
>
> > I am having a h
Sorry!! again, failed to send to the list
-- Forwarded message --
Date: Thu, 22 Aug 2002 00:29:11 +0530 (IST)
From: Sudarshan Raghavan <[EMAIL PROTECTED]>
To: Javeed SAR <[EMAIL PROTECTED]>
Subject: Re: regular expression
On Wed, 21 Aug 2002, Javeed SAR wrote:
> # I want to gre
