Pattern Matching

nwboy74 has asked for the wisdom of the Perl Monks concerning the following question:

Why does this match?

my $s = '<a href=\'2010-coachmen-mirada-bunk-house-w2-slides-for-sale-
+TX-i111851\' class=\'uviss-rv_inventoryListMoreInfoLink\'>';
if ($s =~ /href=(['"])([^\1>]+?for-sale[^\1>]+?)\1>/i) {
   print "$2\n";
}
[download]

Doesn't [^\1>]+? mean look for a sequence of characters not containing \1 or >? I get this result:

2010-coachmen-mirada-bunk-house-w2-slides-for-sale-TX-i111851' class='uviss-rv_inventoryListMoreInfoLink

Comment on Pattern Matching Select or Download Code

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Re: Pattern Matching by AnomalousMonk (Archbishop) on Jan 14, 2010 at 21:32 UTC
In a character set, `"\1"` is the character whose ASCII value is octal 01. It matches the same character (in a character set) as `"\x01"` does. Outside a character set, it would be a backreference to the first capture group. `>perl -wMstrict -le "print '1. no match octal 01' if qq{\1} =~ m{ [^\1] }xms; print '2. match numeral 1' if qq{1} =~ m{ [^\1] }xms; print '3. match octal 01' if qq{\1} =~ m{ [\1] }xms; print '4. no match numeral 1' if qq{1} =~ m{ [\1] }xms; " 2. match numeral 1 3. match octal 01` [download]	[reply] [d/l] [select]
Re: Pattern Matching by ahmad (Hermit) on Jan 14, 2010 at 21:22 UTC
No it doesn't, \1 is a backreference to the first match. `[^\1>]` means match anything that do not match $1 (the first match) or > Update: See the answers below, I missed the character class thing.	[reply] [d/l]
Re^2: Pattern Matching by Anonymous Monk on Jan 14, 2010 at 21:46 UTC
Backreferences don't work in character classes. Try this: `$ perl -e 'if ("aa\\"=~/(.)[^\1]/) { print "$&\n"}'` If backreferences work in character classes, we should see the output as `a\`.	[reply] [d/l] [select]