Regex to variable assignment fails in Perl

Question

I am wondering why the following fails to assign the value (the ip address) of the regex match to the $ipaddress variable after it is matched. I've tried a number of recommended methods (I'm new to Perl obviously), but they've all failed. When performing matches, is it not true that each match is stored in the $1 variables, which range from 1-9, so that the first is, by default, $1?

Code is:

sub block_match {

  my $line_instance_b = $_[0];
  my $ipaddress;

  if ( $line_instance_b =~ /banned_ip|Found UltraSurf Signature|ip_block / ) {
    $ipaddress = $1 if ($line_instance_b =~ /\b(?:[0-9]{1,3}\.){3}[0-9]{1,3}\b/);
    print "Found ip address: ", $ipaddress, "in the following line: \n",
    $line_instance_b;
  }
  else {
    print "Not a block line: \n", $line_instance_b, "\n"
  }
}

The line it is matching against is:

INIT: banned_ip add 208.67.219.132 for FreeGate

In list context a match returns a list of the captures, you can do my ($ipaddress) = $line_instance_b =~ /…/ instead. Additionally, the pedant in me must point out that your regex completely precludes the possibility of IPv6 addresses, which eventually will bite you. — Josh Y.
– Josh Y., Commented Sep 5, 2012 at 2:10

Jon Purdy · Accepted Answer · 2012-09-04 23:44:03Z

6

You’re using a non-capturing group, (?:...), which is never assigned to a match variable.

Wrap the (?:[0-9]{1,3}\.){3} part of the expression in () to capture into $1:

$ipaddress = $1 if $line_instance_b =~ /\b((?:[0-9]{1,3}\.){3})[0-9]{1,3}\b/;
#                                         ^~~~~~~~~~~~~~~~~~~~^

answered Sep 4, 2012 at 23:44

Jon Purdy

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1 Comment

user1647568 Over a year ago

Awesome. Thanks Jon. That did work, and now I know what to read over again as well. Thanks for taking the time :)

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