look behind with multiple whitespace characters

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

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Re: look behind with multiple whitespace characters by AnomalousMonk (Archbishop) on Apr 05, 2015 at 17:46 UTC
The regex `qr{(?<!foo)\s+bar}` will not match `'foo bar'` `^^` `\|\|` `no match --++-- can match here` at the first position because it's preceded by `'foo'`, but at the second position it's not preceded by `'foo'` but by a whitespace character, so it can match! Update: If you have Perl version 5.10+, here's one way to simulate a variable-width negative look-behind: `c:\@Work\Perl>perl -wMstrict -le "use 5.010; ;; my $s = 'foo 1 foo 2 foo 3 4 x 567 foo foo 8 foo9'; my @matches = $s =~ m{ (?> foo \s+ \d+ (SKIP)(FAIL))? \s+ \d+}xmsg; printf qq{'$_' } for @matches; " ' 4' ' 567'` [download] See Special Backtracking Control Verbs in perlre for `(SKIP) (FAIL)` and friends. See also Look Behind issues and Another Look behind for further recent discussion of the variable-width negative look-behind issue. (Update: Variable-width positive look-behind is nicely handled by the `\K` operator.) Give a man a fish: `<%-(-(-(-<`	[reply] [d/l] [select]
Re: look behind with multiple whitespace characters by Anonymous Monk on Apr 05, 2015 at 17:49 UTC
Run it through rxrx and see what happens, or even $ perl -Mre=debug -e" $_ =qq{foo bar \ntest}; m/(?<!foo)\s+bar\s+t +est/g " Compiling REx "(?<!foo)\s+bar\s+test" Final program: 1: UNLESSM[-3] (7) 3: EXACT <foo> (5) 5: SUCCEED (0) 6: TAIL (7) 7: PLUS (9) 8: SPACE (0) 9: EXACT <bar> (11) 11: PLUS (13) 12: SPACE (0) 13: EXACT <test> (15) 15: END (0) floating "test" at 5..2147483647 (checking floating) minlen 9 Guessing start of match in sv for REx "(?<!foo)\s+bar\s+test" against +"foo bar %ntest" Found floating substr "test" at offset 12... Guessed: match at offset 0 Matching REx "(?<!foo)\s+bar\s+test" against "foo bar %ntest" 0 <> <foo bar> \| 1:UNLESSM[-3](7) 0 <> <foo bar> \| 7:PLUS(9) SPACE can match 0 times out of 21474 +83647... failed... 1 <f> <oo bar > \| 1:UNLESSM[-3](7) 1 <f> <oo bar > \| 7:PLUS(9) SPACE can match 0 times out of 21474 +83647... failed... 2 <fo> <o bar > \| 1:UNLESSM[-3](7) 2 <fo> <o bar > \| 7:PLUS(9) SPACE can match 0 times out of 21474 +83647... failed... 3 <foo> < bar %n> \| 1:UNLESSM[-3](7) 0 <> <foo bar> \| 3: EXACT <foo>(5) 3 <foo> < bar %n> \| 5: SUCCEED(0) subpattern success... failed... 4 <foo > < bar %nt> \| 1:UNLESSM[-3](7) 1 <f> <oo bar > \| 3: EXACT <foo>(5) failed... 4 <foo > < bar %nt> \| 7:PLUS(9) SPACE can match 3 times out of 21474 +83647... 7 <o > <bar %ntest> \| 9: EXACT <bar>(11) 10 < bar> < %ntest> \| 11: PLUS(13) SPACE can match 2 times out of 214 +7483647... 12 < bar %n> <test> \| 13: EXACT <test>(15) 16 < bar %ntest> <> \| 15: END(0) Match successful! Freeing REx: "(?<!foo)\s+bar\s+test" [download] First it matches "foo" which isn't allowed, but then it matches "foo " which is allowed since its not "foo" Basically, you need to rethink what you're trying to achieve, see Look Behind issues	[reply] [d/l]
Re: look behind with multiple whitespace characters by Marshall (Canon) on Apr 07, 2015 at 05:18 UTC
This look behind regex stuff is complex. There might be an easier way to code this. If you could give say 10 example records, this might work out better. `As you say, this doesn't match: foo bar test == But I guess that this should?: bar test == And this match?: foo bar test` [download]	[reply] [d/l]