bittondb has asked for the wisdom of the Perl Monks concerning the following question:

I want to replace all instances of a regular expression with a value and an incrementing number. So, the replacements would be foo.1, foo.2, etc. How would I do this?

Replies are listed 'Best First'.
Re: Increase number per regex replacement
by runrig (Abbot) on Jan 08, 2002 at 10:00 UTC
[reply]
Re: Increase number per regex replacement
by theorbtwo (Prior) on Jan 08, 2002 at 10:04 UTC

    The answer lies in the /e option to s///; read perlop.

    Thanks,
    James Mastros,
    Just Another Perl Scribe

[reply]
Re: Increase number per regex replacement
by belg4mit (Prior) on Jan 08, 2002 at 10:23 UTC
    Just for fun, here's an alternative version without /e ;-) 
    $i = 0;
@no = qw(one two three four five);
foreach ( @no ){
  s/e/foo.$i/ && $i++ while /e/;
}
print "@no\n";
__END__
onfoo.0 two thrfoo.1foo.2 four fivfoo.3

    [download]

    --
    perl -pe "s/\b;([st])/'\1/mg"

[reply]
[d/l]
      Ah, but what if you needed to use "fee.$i" as the replacement string instead of "foo.$i"?

      Fee fi foo fum, I smell the infinite loop of an eratum...

      -Blake

[reply]
        True true, but this was meant to be a whimscial/goldbergian solution. You'd have to be wary of the same issue with /e though not /ge ;-) Of course this fixes that: 
        $i = 0;
@no = qw(one two three four five);
foreach ( @no ){
  my $str = '';
  while( s/([^e]*)e// ){
    $str .= $1 . "fee" . $i++;
  }
  $_ = $str . $_; #UPDATED: Append addresses Re: to this node
}
print "@no\n";

        [download]

        --
        perl -pe "s/\b;([st])/'\1/mg"

[reply]
[d/l]

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