From: "David A. Black" <dblack@...>
Date: 2005-04-01T06:06:51+09:00
Subject: Re: Regex parsing question

Hi --

On Fri, 1 Apr 2005, Paul Hanchett wrote:

> Why does this:
>
> 	text= "AA<X>BB<X>CC</X>DD</X>EE"
> 	regex = %r{(.*)<X>(.*)}
>
> 	t = text.sub( regex, "z" );
> 	print "$1=#{$1}\n$2=#{$2}\n$3=#{$3}\n$4=#{$4}\n"
>
> Return this:
>
> 	$1=AA<X>BB
> 	$2=CC</X>DD</X>EE
> 	$3=
> 	$4=
>
> Instead of:
>
> 	$1=AA
> 	$2=BB<X>CC</X>DD</X>EE
> 	$3=
> 	$4=

Because * is "greedy" -- meaning, it eats up as many characters as
possible, from left to right, while still allowing for a successful
match overall.

So your first .* eats up everything until it reaches as far right as
it possibly can -- namely, just before the second <X> (which it then
leaves intact so that it can be matched by the literal <X> in your
regex).  It even eats up the first <X>.

> And how would I fix it?

Use *? instead of * -- like this:

    regex = %r{(.*?)<X>(.*)}


David

-- 
David A. Black
dblack@wobblini.net