Re: Re: regex match question

you don't need to confuse people with the extra syntax of look-ahead/behind....you just need to match exactly what you want "a non-percent sign character followed by a percent sign character followed by q", i.e. $str =~ s/[^%]%q//g;

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Re: Re: Re: regex match question by leriksen (Curate) on Aug 04, 2003 at 06:07 UTC
That doesn't work if %q starts the line - try `/^%q\|[^%]%q/` [download] or use the negative look behind /(?<!%)%q/	[reply] [d/l]
Re: Re: Re: regex match question by graff (Chancellor) on Aug 05, 2003 at 03:49 UTC
In addition to the problem leriksen pointed out, your suggestion deletes an extra character. In order to meet the OP's spec, it gets a bit uglier: `$str =~ s/^%q\|([^%])%q/$1/g;` [download] Basically, it would be less confusing to learn about zero-width assertions. Granted, it does pose a bit of a challenge to "learn" the various zero-width and non-grouping forms. I haven't memorized them yet myself -- I just scan through the "perlre" man page every time I need to use one of those "(?whatever)" thingies. I actually don't mind that. I'm used to it from always having to do the same thing for a dozen different common C library calls -- I still do that, after "knowing" C for 15 years.	[reply] [d/l]