/^\d{1,3}(,\d{3})*$/
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jdporter
The 6th Rule of Perl Club is -- There is no Rule #6.

thank you.
how would I store the matched number, without the commas, in a variable. $num = 23564, instead of $num = 23,678.

The following is off the cuff, but it might work for you. Note that it also allows for negative numbers. Unlike Popcorn Dave's solution, I combine the test and the number extraction into one step. You may or may not like this. (and I just cleaned up the code)

#!/usr/bin/perl -l
use strict;
use warnings;

my @numbers = ( 23, '1,234', '2,2', '12,234,567', -1, '-12,234',
   '-23,23', '999,999', 0 );

foreach my $number (@numbers) {
   if (my @sections = $number =~ /^(-?\d{1,3})(?:,(\d\d\d))*$/) {
       print join '', grep {defined} @sections;
   }
   else {
       # not a valid number
   }
}
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Cheers,
Ovid

New address of my CGI Course.
Silence is Evil (feel free to copy and distribute widely - note copyright text)

Well, once you've gotten a match with the above regex, then all you need to do is delete the commas.

  $num = '654,321';
  if ( $num =~ /^\d{1,3}(,\d{3})*$/ ) {
    $num =~ y/,//d; # delete commas
  }
  else {
    # not a conforming number
  }
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jdporter
The 6th Rule of Perl Club is -- There is no Rule #6.

Using a substitution does the trick for you.

$num =~ s/\,//g
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There is no emoticon for what I'm feeling now.

$_ = 123456789;
s/ # begin substitution operator
(\d{1,3}) # match and capture one-to-three digits
(?= # if they are followed by:
(?:\d\d\d)+ # one-or-more groups of three digits
(?!\d) # that are not followed by a digit
) # end lookahead assertion
/$1,/gx; # /g = perform substitution repeatedly
print; #prints: 123,456,789
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(_8(|)

    use Regexp::Common;
    if ($tag =~ /^$RE{num}{int}{-sep => ','}$/) {
        ...
    }
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Abigail