Consider how   m!(\d+).+to.+(\d+)!i matches against   123to456 Since you're using greedy matching, each + will match as much as it can. The regexp matches like this: 
  (12)3to45(6)
   $1       $2

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That's not what you intend. A first cut at fixing this is to rewrite the regex so that it it won't gobble up extra digits on either side of "to".   m!(\d+)\D+to\D+(\d+)!i; This will match target strings that have non-digit substrings surrounding the "to", but won't match "123to456", since there are no non-digit characters surrounding "to". If that's a problem, you can take the regex a step further, and write   m!(\d+)\D*?to\D*?(\d+)!i; which will accept zero or more non-digit characters on either side of "to".

In reply to Re: $1 and regex by dws
in thread $1 and regex by Anonymous Monk

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