Re: string substitution

Here's one that works without capturing parens and backreferences.

my $ariths = '(+*\-/<>=)%^';
$string =~ s/(?:^|(?<=[$ariths]))BAD(?=[$ariths])/HELL/g;
[download]

There may be a better way to do it though. Consider \b for word boundries.

Update:Ugg, thanks Not_a_Number... corrected. I used a non-capturing paren when I meant to use a zero-width assertion.

Dave

"If I had my life to do over again, I'd be a plumber." -- Albert Einstein

Comment on Re: string substitution Download Code

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Re: Re: string substitution by Not_a_Number (Prior) on Oct 08, 2003 at 17:44 UTC
But that: 1) removes the arithmetic operators; 2) doesn't meet the 'start of line' spec. :-( dave	[reply]
Re: Re: string substitution by davido (Cardinal) on Oct 08, 2003 at 18:02 UTC
Fixed. I shouldn't post Regexp nodes before eating my breakfast cereal. For the record: 1. I mistakenly used (?: ) non-capturing parens when I meant to use (?<= ) and (?= ) zero-width assertions. 2. I didn't realize that stort meant start. The issues have been resolved. :) Dave "If I had my life to do over again, I'd be a plumber." -- Albert Einstein	[reply]