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

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

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Re: Re: Re: Re: regex (?{CODE}) question
by ctgIT (Initiate) on Aug 01, 2002 at 19:53 UTC
    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
      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; }
      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