s/\babc\s+//;

print unless /\babc\b/;

print if /(\w+) nevermind/ && $1 ne "abc";

print if /(?<!abc) nevermind/;

I know about look-behind assertions, but I was wondering if there would be a way to do it wihtout... it looks like a common, simple thing, _not_ matching a word...

But if look-behind assertions are the only solution, then I'll use them.

I'm not interested in the other solutions, since they require a code change, and... well, it's a bit working around the problem that I wanted to know.

Thanks!

I'm not sure I'm reading your question right, but you're wondering why [^(abc)] doesn't match the string abc?

The square brackets enclose an expression that will match one character. The construct you show will match anything other than (, ), a, b, or c. I think that's what your description is trying to say.

You can get the effect you want in Perl newer than 5.6.1 with a lookbehind assertion: print if /(?<!abc) nevermind/There are other ways to do it, using $PREMATCH, but this will work for fixed strings. More information at perlre, the section about lookbehind.</code>

That is, if I read your question correctly....

Yes, that's the solution. Is there no way without look-behind assertions? It seems like a common thing to do, 'not match' a expression of grouped letters (i.e. a word).

Thanks for your answer :)

Yes, there's the negative look-a-head assertion. ;-)

    /^(?!abc)\w+ something/
[download]

But why are you afraid of look-behind assertions. Asking for a solution to X without using Y, if Y is a reasonable way of solving X should have an explaination of why Y shouldn't be used.

Abigail

print unless /\babc\b/;
[download]

print if /(?<!abc )nevermind/;
[download]

my @words = grep !/^abc\z/, split;
print "@words";
[download]

-- Mike

--
just,my${.02}

Actually, I was trying to match every sentence where 'abc' is not preceding 'nevermind' - every other word is ok.

Negative look-behind works, but I expected a more common combination to work too - something like grouping letters with () to a word, and then ^ to negate the sense. Is there no way to group these letters as a word instead of individual characters?

It's more a matter of curiosity if it can be done with just a simple regexp rather than making it work, otherwise I'd easily solve it with more code.

Thanks! ;)

Do you not want to use lookbehind because of efficiency concerns, or some other reason? It seems to be just the right tool for the job... It's doing what you were trying to do in your original code: making sure "abc" does not match before "nevermind".

As far as using [^abc] to do the job, [] represents a character class within regular expressions, and can only match a single character at a time. So unfortunately, you can't use [] to match anything more than a single character.

-- Mike

--
just,my${.02}

The Cookbook has the following solution for your problem:

    /^(?:(?!abc).)*$/
[download]

This matches a string that does not contain the text abc.

Abigail