I have the following regex which I use to find a word:
(?<=\s)([\w\@\-]+)(?=\s)
I want to further modify this regex to exclude a list of words e.g. do not match if the word is 'cat' or 'dog'.
How do I accomodate the regex to achieve this?
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This one is similar: [I have a PHP regEx, how do add a condition for the number of characters?](http://stackoverflow.com/q/13994553/367456), even specifying a minimum length what qualifies a word. – hakre Dec 29 '12 at 16:04
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@hakre It doesn't seem that similar to me?! – ObiHill Dec 29 '12 at 16:07
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Oh it is, the answer you accepted uses the same regex features. ;) - It just does the length check as well, which you don't have here (well you have one or more characters to form a word). – hakre Dec 29 '12 at 16:08
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@hakre Ok, but the answer to that doesn't answer my question. You'll have to be a Regex ninja to draw the correlation between the two, and a regex ninja am I not, ergo my question. – ObiHill Dec 29 '12 at 16:37
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I wrote similar, not identical. It's just a link in case someone stumbles over this question and is looking for alternatives. It's not saying that your question would be wrong or anything. – hakre Dec 29 '12 at 16:56
1 Answers
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\b(?!(?:dog|cat)\b)([\w@-]+)\b
Here if you want to match word start/end with @
(?<=\s)(?!(?:dog|cat)(?=\s))([\w\@\-]+)(?=\s)
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+1, but not equivalent to @ChuckUgwuh's regex - it won't match an entire "word" if it ends in `@` or `-` (unlikely), and it *will* match at the start/end of the string (where his solution won't). – Tim Pietzcker Dec 29 '12 at 15:57
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@TimPietzcker yeah u right..he can change \b to lookaround assertion to match @ at the start/end of string – slier Dec 29 '12 at 16:02
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@slier Thanks. The lookaround assertion for @, how would I pull that off?! – ObiHill Dec 29 '12 at 16:06