Regexp question

davidov0009 has asked for the wisdom of the Perl Monks concerning the following question:

Dear monks,

I am trying to match a string within some HTML but seem to be having some trouble doing so. The following is part of the HTML source that I am looking to match:

name="challenge" value="26eca68705b3b0c76a6d0602937ac524"

I want to grab the string for value. I am using grouping in my regexp and expect therefore that $1 will return 26eca68...etc, essentially the string in between the quotes for the value parameter. Here is the regexp I am using:

$returned_data =~ s/\n//g;

my $c_code = $returned_data;
$c_code =~ m/name="challenge" value="(.*)"/; #"
print $1 . "\n";
[download]

When $1 is printed it does not just print the text in between the quotes but instead prints all the remaining text after the first quote of the value parameter so like this: 26eca68705b3b0c76a6d0602937ac524" /><input type="hidden" id="md5pass" name="md5pass" value="1" /><input type="hidden" id="noerror" name="noerror" value="1" /><label><span>Email:</span><input type="text" class="inputtext" name="....MORE HTML SOURCE...

Any ideas as to why $1 seems to be capturing more that it should?

use strict;
use CGI;
[download]

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Re: Regexp question by RMGir (Prior) on Dec 20, 2007 at 16:48 UTC
You've run into the "greedy" behaviour of * - it grabs the longest possible match. perlre explains it well. The workaround is to modify the * to be non-greedy, by adding a ? `$c_code =~ m/name="challenge" value="(.*?)"/;` [download] Mike	[reply] [d/l]
Re: Regexp question by kyle (Abbot) on Dec 20, 2007 at 16:50 UTC
The `` modifier in regular expressions is "greedy". That means it will match the most that it possibly can. You can get a non-greedy match by adding `?` after it like so: `m/name="challenge" value="(.?)"/` [download] Note that in general this won't cope with escaped quotes as in `"* is \"greedy\" in regex"`. Before there were non-greedy matches, I might have written the expression this way (which also doesn't cope with escaped quotes): `m/name="challenge" value="([^"]*)"/` [download] Or, given your input, this way (which will only match a hexadecimal string): `m/name="challenge" value="([0-9a-f]+)"/` [download] As an aside, let me recommend in general using something like HTML::Parser when you're parsing HTML. Getting it right with regular expressions can be really difficult, but the module does the right thing for you.	[reply] [d/l] [select]
Re: Regexp question by toolic (Bishop) on Dec 20, 2007 at 16:54 UTC
In addition to being conscious of the greedy behavior, it is also good practice to check that the match succeeded before using `$1`.	[reply] [d/l]
Re^2: Regexp question by davidov0009 (Scribe) on Dec 20, 2007 at 17:35 UTC
One last quick question. Is this valid for checking whether the match occurred, and would $1 be initialized to the matched text in this code? `if ( $html =~ m/id="$element" name="challenge" value="(.*?)"/ ) { $post{$element} = $1; }` [download] `use strict; use CGI;` [download]	[reply] [d/l] [select]
Re^3: Regexp question by toolic (Bishop) on Dec 20, 2007 at 17:42 UTC
I believe so, but you can easily prove it for yourself with a trivial testcase.	[reply]
Re^3: Regexp question by FunkyMonk (Bishop) on Dec 21, 2007 at 00:18 UTC
Yes, that'll work. As will `$post{$element} = $1 if $html =~ m/id="$element" name="challenge" value="(.?)"/;` [download] Unless you expect* `$element` to contain metacharacters, you should use `m/id="\Q$element\E" name="challenge" value="(.*?)"/;` [download] to escape any metacharacters in `$html`. See quotemeta for more.	[reply] [d/l] [select]