in reply to Re: Re: Re: regex (?{CODE}) question
in thread regex (?{CODE}) question

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;
}

[download]
Thanks,
Christos

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Re: Re: Re: Re: Re: regex (?{CODE}) question
by elusion (Curate) on Aug 01, 2002 at 19:59 UTC
    Try using the /e modifier. What it does is make the second half of the regex executed as code. In your case you'd want something like this: 
    $str = "(<ref links=\" cit314-1 cit314-2 cit314-3 cit314-4 cit314-5\"/
+>)";

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

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

    [download]
    Your $str changed from your previous posts, so I changed it back to the way it was in the other posts. Notice how I quoted the tags and concatenated them with the return value from the subroutine. I also did some extra formatting. Hope this helps.

    elusion : http://matt.diephouse.com

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