How can I output my numbers with commas added?
  This subroutine will add commas to your number:

  sub commify {
      local $_  = shift;
      1 while s/^([-+]?\d+)(\d{3})/$1,$2/;
      return $_;
  }

  This regex from Benjamin Goldberg will add commas to numbers:

  s/(^[-+]?\d+?(?=(?>(?:\d{3})+)(?!\d))|\G\d{3}(?=\d))/$1,/g;

  It is easier to see with comments:

    s/(
         ^[-+]?            # beginning of number.
         \d{1,3}?          # first digits before first comma
         (?=               # followed by, (but not included in the mat
+ch) :
          (?>(?:\d{3})+) # some positive multiple of three digits.
          (?!\d)         # an *exact* multiple, not x * 3 + 1 or whate
+ver.
         )
         |                  # or:
         \G\d{3}           # after the last group, get three digits
         (?=\d)            # but they have to have more digits after t
+hem.
       )/$1,/xg;
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#!/usr/bin/perl
$_ = '1234567'; 
$_ = reverse $_; 
s/(\d{3})(?=\d)/$1,/g; 
$_ = reverse $_; 
print $_ ;
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Here is another way, based on the RE you used:

The trick is to loop looking for groups of 3 digits.

while($col=~s/(\d)(\d\d\d)(,|$)/$1,$2/)
 {}
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if the number is 7 digits longs it doesnt add the comma

The reason is because you anchor your match to the end of the line with $. Your regex is doing exactly what you're asking it to do: match 4 digits at the end of the string, and insert a comma in the appropriate place.

It's messy to write a "normal" regex to do what you want, because regular expressions scan your string from left to right. The easiest way commafy a number is to actually reverse the string first, as demonstrated by borisz's earlier post.

{
    local $_ = $amount;
    /\./ ? s/(?<=\d)(?=(\d{3})+(?:\.))/,/g
         : s/(?<=\d)(?=(\d{3})+(?!\d))/,/g;
    $amount = $_;
}
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