Re: Matching pattern in regular expression with string position

Within the if statement,

    my $pos = pos( $variable ) - 1;
[download]

should do the trick.

See perlfunc:pos for more information.

You might also look at perlvar:$+ * pervar:$-, but these are only set when captiring brackets are used.

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller
If I understand your problem, I can solve it! Of course, the same can be said for you.

Comment on Re: Matching pattern in regular expression with string position Download Code

Replies are listed 'Best First'.
Re: Re: Matching pattern in regular expression with string position by diotalevi (Canon) on Aug 09, 2003 at 13:24 UTC
Either pos() or `$-[0]`. You could also write that as `my $pos = $-[0]`.	[reply] [d/l] [select]
Re: Re: Matching pattern in regular expression with string position by sauoq (Abbot) on Aug 09, 2003 at 19:18 UTC
In order for `pos()` to work, he'll need to add a `/g` modifer. It only works on "global" matches. Also, it's not the scalars `$+` and `$-` which would be of interest but the arrays `@+` and `@-` which would be. Finally, `$+[0]` and `$-[0]` are both useful even without any capturing because they hold offsets from the entire match. In his case, `$-[0]` is exactly what he needs. -sauoq "My two cents aren't worth a dime.";	[reply]