Re: How to find the outermost pair of square brackets with a regex?

With difficulty.

our $re;
local $re = qr{
   \[
   (?:
      (?> [^\[\]]+ )
   |
      (??{ $re })
   )*
   \]
}x;

my $s = <<'__EOI__';
blah blah blah blah
blah
blah blah blah [blah
[blah blah] [blah blah blah
blah] blah]
blah blah
__EOI__

$s =~ s/$re/moo/g;

print($s);
[download]

Partial credits to perlre.

Comment on Re: How to find the outermost pair of square brackets with a regex? Download Code

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Re^2: How to find the outermost pair of square brackets with a regex? by lokiloki (Beadle) on Jan 17, 2007 at 01:18 UTC
that's beautiful... so, if i have this right... match the first bracket, IF, non-backtracking, we find an additional or , then recurse, match the last bracket? what is the purpose of the `^`?	[reply] [d/l]
Re^3: How to find the outermost pair of square brackets with a regex? by lokiloki (Beadle) on Jan 17, 2007 at 01:20 UTC
sorry, i meant to say... that's beautiful... so, if i have this right... match the first bracket; IF, non-backtracking, we find an additional `[` or `]`, then recurse; match the last bracket? what is the purpose of the ^?	[reply] [d/l] [select]
Re^4: How to find the outermost pair of square brackets with a regex? by lokiloki (Beadle) on Jan 17, 2007 at 02:26 UTC
oops, caret is inverted character class	[reply]
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