Re: Regex boundary match (updated demo code)

Did you try a negative lookahead assertion to exclude the dash ?

update

 DB<21> $test = qq|Some test. This is a test-with a dash. We want to m
+atch test here|

 DB<22> x $test =~ /\b(test)(?!-)\b/g
0  'test'
1  'test'

 DB<23> print pos($test),":$1\n" while $test =~ /\b(test)(?!-)\b/g
9:test
60:test
[download]

Cheers Rolf
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Comment on Re: Regex boundary match (updated demo code) Download Code

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Re^2: Regex boundary match (updated demo code) by ultranerds (Hermit) on Feb 08, 2020 at 14:08 UTC
Thanks. I ended up with `([^\-])`. Is your version quicker? Or negligible difference? (its going to be run quite a lot of times)	[reply] [d/l]
Re^3: Regex boundary match (updated demo code) by AnomalousMonk (Archbishop) on Feb 08, 2020 at 15:55 UTC
I ended up with `([^\-])`. ... difference? It depends on exactly what you want. `([^\-])` requires a character match and consumes a character. As LanX pointed out, `([^\-]\|$)` consumes a character if it can. Neither seem to do what you want: `c:\@Work\Perl\monks>perl -wMstrict -le "my $s = 'test atest atested tested test- -test test .test test. test' +; ;; $s =~ s/\b(test)([^-])/DONE/g; print qq{'$s'}; " 'DONEatest atested DONEd test- -DONEDONE.DONEDONE test' c:\@Work\Perl\monks>perl -wMstrict -le "my $s = 'test atest atested tested test- -test test .test test. test' +; ;; $s =~ s/\b(test)([^-]\|$)/DONE/g; print qq{'$s'}; " 'DONEatest atested DONEd test- -DONEDONE.DONEDONE DONE'` [download] Maybe this does: `c:\@Work\Perl\monks>perl -wMstrict -le "my $s = 'test atest atested tested test- -test test .test test. test' +; ;; $s =~ s/\b(test)(?!-)\b/DONE/g; print qq{'$s'}; " 'DONE atest atested tested test- -DONE DONE .DONE DONE. DONE'` [download] Update: Just noticed from the timestamps that ultranerds already decided to go with LanX's suggestion. :) Give a man a fish: `<%-{-{-{-<`	[reply] [d/l] [select]


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