Re: positive regex for inverted match

Here is a regex solution to you problem:

while (<DATA>) {
   chomp;
   print "$_ has no bar\n" if /^([^b]|b[^a]|ba[^r])+$/;
}

__DATA__
bar
foo
"bar" blah
baz "bar" blah
"test" blah
"test2"
[download]

-Mark

Comment on Re: positive regex for inverted match Download Code

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Re^2: positive regex for inverted match (classic) by tye (Sage) on Feb 26, 2004 at 07:55 UTC
Classic++, including classic mistakes. (: Your regex will match "bbar" despite it containing "bar". "bb" matches `b[^a]` and then "a" and "r" each match `[^b]`. I started exploring this idea in depth and hope to write a lengthy node on it one day. Trying to fix the problems leads you down paths similar to: `/^([^b]\|b[^a]\| ba[^r])+$/ /^([^b]+\|b+[^ab]\| b+a[^rb])ba?$/ /^([^b]+\|b+[^ab]\|(b+a)+([^rb]\|b+[^ab]))[ba]$/` [download] - tye	[reply] [d/l] [select]

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Re^2: positive regex for inverted match (classic)
by tye (Sage) on Feb 26, 2004 at 07:55 UTC

Classic++, including classic mistakes. (:

Your regex will match "bbar" despite it containing "bar". "bb" matches b[^a] and then "a" and "r" each match [^b].

I started exploring this idea in depth and hope to write a lengthy node on it one day. Trying to fix the problems leads you down paths similar to:

/^([^b]|b[^a]| ba[^r])+$/
/^([^b]+|b+[^ab]| b+a[^rb])*b*a?$/
/^([^b]+|b+[^ab]|(b+a)+([^rb]|b+[^ab]))*[ba]*$/
[download]

- tye

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