Question

I am looking to do a 2-step regular expression look-up in Perl, I have text that looks like this:

here is some text 9337 more text AA 2214 and some 1190 more BB stuff 8790 words

I also have a hash with the following values:

%my_hash = ( 9337 => 'AA', 2214 => 'BB', 8790 => 'CC' );

Here’s what I need to do:

1. Find a number
2. Look up the text code for the number using my_hash
3. Check if the text code appears within 50 characters of the identified number, and if true print the result

So the output I’m looking for is:

Found 9337, matches 'AA'
Found 2214, matches 'BB'
Found 1190, no matches
Found 8790, no matches

Here’s what I have so far:

while ( $text =~ /(\d+)(.{1,50})/g ) {
  $num = $1;
  $text_after_num = $2;
  $search_for = $my_hash{$num};
  if ( $text_after_num =~ /($search_for)/ ) {
    print "Found $num, matches $search_for\n";
  }
  else {
   print "Found $num, no matches\n";
  }

This sort of works, except that the only correct match is 9337; the code doesn’t match 2214. I think the reason is that the regular expression match on 9337 is including 50 characters after the number for the second-step match, and then when the regex engine starts again it is starting from a point after the 2214. Is there an easy way to fix this? I think the \G modifier can help me here, but I don’t quite see how.

Any suggestions or help would be great.

Answer 1

You have a problem with greediness. The 1,50 will consume as much as it can. Your regex should be /(\d+)(.+?)(?=($|\d))/

To explain, the question mark will make the multiple match non-greedy (it will stop as soon as the next pattern is matched – the next pattern gets precedence). The ?= is a lookahead operator to say “check if the next element is a digit. If so, match but do not consume.” This allows the first digit to get picked up by the beginning of the regex and be put into the next matched pattern.

[EDIT]
I added an optional end value to the lookahead so that it wouldn’t die on the last match.