September 9th, 2008, 08:25 PM
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Spontaneously Present
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no that wont work.
you get this
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[09-Sep-2008 19:35:45] PHP Warning: preg_match() [<a href='function.preg-match'>function.preg-match</a>]: Compilation failed: lookbehind assertion is not fixed length at offset 104
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in the link i provided
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(?=pattern)
A zero-width positive lookahead assertion. For example, /\w+(?=\t)/ matches a word followed by a tab, without including the tab in $&.
(?!pattern)
A zero-width negative lookahead assertion. For example /foo(?!bar)/ matches any occurrence of ``foo'' that isn't followed by ``bar''. Note however that lookahead and lookbehind are NOT the same thing. You cannot use this for lookbehind.
If you are looking for a ``bar'' that isn't preceded by a ``foo'', /(?!foo)bar/ will not do what you want. That's because the (?!foo) is just saying that the next thing cannot be ``foo''--and it's not, it's a ``bar'', so ``foobar'' will match. You would have to do something like /(?!foo)...bar/ for that. We say ``like'' because there's the case of your ``bar'' not having three characters before it. You could cover that this way: /(? ?!foo)...|^.{0,2})bar/. Sometimes it's still easier just to say:
if (/bar/ && $` !~ /foo$/)
For lookbehind see below.
(?<=pattern)
A zero-width positive lookbehind assertion. For example, /(?<=\t)\w+/ matches a word following a tab, without including the tab in $&. Works only for fixed-width lookbehind.
(?<!pattern)
A zero-width negative lookbehind assertion. For example /(?<!bar)foo/ matches any occurrence of ``foo'' that isn't following ``bar''. Works only for fixed-width lookbehind.
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i kind of get it, but its still sinking in. looking for a clear answer as to why i cant make it a variable length i.e. your solution
i dont understand
Quote:
cover that this way: /(??!foo)...|^.{0,2})bar/. Sometimes it's still easier just to say:
if (/bar/ && $` !~ /foo$/)
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