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Re: Re: quick question about parenthesis and regular expressions

by cranberry13 (Beadle)
on Nov 04, 2003 at 12:31 UTC ( #304393=note: print w/replies, xml ) Need Help??


in reply to Re: quick question about parenthesis and regular expressions
in thread quick question about parenthesis and regular expressions

Enlil--
Thanks for the reg expression -- it works :)

But here's an interesting problem, it seems if i'm using a regular expression with the substitution operator, the value of $1 comes up blank.

Why is that?

What I really want to do is this:

if ($line =~ s/($word(?:s|ed))/<b>$1<\/b>/igm)

assuming that:

$line= "She was very absorbed in her homework."; $word = "absorb";

I want it to match absorbed and but store the value of it in $1 so that I can use it later.

Replies are listed 'Best First'.
Re: Re: Re: quick question about parenthesis and regular expressions
by jreades (Friar) on Nov 04, 2003 at 13:32 UTC

    Ummm, I have no problem doing the substitution using the code supplied:

    $line = "She was very absorbed in her homework."; $word = "absorb"; print STDOUT "Line: $line\n"; $line =~ s/($word(?:s|ed))/<b>$1<\/b>/igm; print STDOUT "New Line: $line\n"; print STDOUT "\$1 contains: $1\n";

    Of course, I might consider changing it slightly to read:

    $line =~ s/\b(${word}(?:s|ed)?)\b/<b>$1<\/b>/igm;

    However, I did just discover that there is something more going on here, because while the original (without the \b) does set something in $1, my change doesn't (despite doing the substitution properly).

    And finally, if you're just doing a match why are you using s///?

      You're right: the \bs in the pattern do seem to sabotage $1. In particular, any char in front of the () group kills the $1, if the g and i switches are both set.
      my $line = 'She was very absorbed in her homework.'; my $word = 'absorb'; $_=$line; s/ ($word)//g; print "\$1 is $1\n"; $_=$line; s/ ($word)//i; print "\$1 is $1\n"; $_=$line; s/ ($word)//ig; print "\$1 is $1\n";
      One other note:
      #This also fails s/(\b$word)//ig; #Although this is ok s/( $word)//ig; #And this is fine, too! my $pat = qr/\b($word)/; s/$pat//gi; #or even my $pat = qr/($word)/; s/\b$pat//gi; #or EVEN THIS! my $pat = qr/$word/; s/\b($word)//gi;
      I think we have a perlbug. Frenzy of updates completed. Really.
      It looks like the $1 is being lost because of the g modifier. That is, it matches, it replaces, it tries again. It starts the matching process again, finds the start of a possible match at the word "in", wipes out $1 etc, finds that the possible match isn't really, and then when the substitution loop finishes, you have lost $1.

      This appears to be a bug. (Report with perlbug if you like.) However I would also point out that any code which relies on the correct behaviour is likely to be buggy anyways - if the word appears multiple times then you won't catch all of the substitutions. If you really want to have fine access to all of the substitution information after the fact then you either need to write your own substitution loop (using matching with /g, pos and substr) or you need to embed code in the substitution. Like this:

      my @matches; $line =~ s/\b($word(?:s|ed))/ push @matches, $1; "<b>$1<\/b>" /iegm;

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