Re: regex (?{CODE}) question

[elusion]

The print command isn't executed because it's not in the regex. The first portion of the s/// (between the first two //'s) is a regex; the other portion could be considered a string. So you need to place the (?{ print $1 }) at the end of the regex part, like so:

$str = "(<ref links=\" cit314-1 cit314-2 cit314-3 cit314-4 cit314-5\"/
+>)";

if( $str =~ s/<ref links="(.*?)"(.*?)\/>(?{ print $1, "\n" })/<ref lin
+ks="$1"$2><\/ref>/ ){
    print "Match was found." . "\n";
}
print $str . "\n";

Notice I also didn't escape the quotes -- this isn't necessary and, I think, it makes things look a little messier.

elusion : http://matt.diephouse.com

[Ferret]

And don't forget, since you're referring to a capture while still in the regex, it hasn't set $1 yet, so you have to use \1 instead:

 (?{ print $1, "\n" })

would need to be

 (?{ print \1, "\n" })

unless there's something special about the codeblock I'm not aware of?

Update: Strike that, reverse it, elusion's correct. Make sure you don't do anything silly like using \1 instead of $1 ;-)

[elusion]

While in a code block, $1 is correct. This is necessary because it's executed as actual code. I'm not sure how \1 would actually parse, though I assume it'd be a reference to 1.

elusion : http://matt.diephouse.com

[ctgIT]

Thanks for your reply. Actually, my objective is not to print $1 but to call a subroutine from inside the replace part of the regex which will use the value of $1 to produce a string that will be used in the replace part. For example, something like:


$str = "(<citationref linkend=\" cit314-1 cit314-2 cit314-3 cit314-4 c
+it314-5\"/>)";
if( $str =~ s/<ref links=\"(.*?)\"(.*?)\/>/<ref links=\"$1\"$2>(?{ &ge
+tString($1) })<\/ref>/ ){
    print "Match was found." . "\n";
}
print $str . "\n";

sub getString($s){    
    my $s = shift;
    ...more processing of $s...
    return $s;
}

Thanks,
Christos