Can you help me to understand / convert follwoing .net regex into java
@"(?<!\\)(?'M'[^|%])" ?
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Daniel Hilgarth
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Rajender Saini
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There ain't no such thing as Java or .NET Regex. – Damian Leszczyński - Vash Mar 09 '11 at 13:33
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2@Vash: For all pratical purposes, there is. The .NET and Java regex implementations differ in some aspects. – Jens Mar 09 '11 at 13:36
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@Vash: Actually, it could be said that there are - considering each have their own engine implementation. Ultimately, however, I agree that the expressions themselves ought to be universal; this doesn't alter the fact that languages use different constructs to achieve a single goal. – Grant Thomas Mar 09 '11 at 13:37
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@Jens: Interesting, do you know some page where this is described ? – Damian Leszczyński - Vash Mar 09 '11 at 13:38
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1@Vash, http://www.regular-expressions.info/lookaround.html in the section "Important Notes About Lookbehind", for one – jswolf19 Mar 09 '11 at 13:41
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Are you sure there isn't a typo in this? I'm looking at the `(?'M'` part. I think there should be something between the `?` and `'` perhaps... – jswolf19 Mar 09 '11 at 13:44
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@jswolf19: (?'M'...) is the same as (?
...): a named capturing group. – Jens Mar 09 '11 at 13:47 -
@Vash: Here's a comprehensive list: http://www.regular-expressions.info/refflavors.html – Jens Mar 09 '11 at 13:49
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@Jeans & Mr. Disappointment : Thanks. – Damian Leszczyński - Vash Mar 09 '11 at 13:54
1 Answers
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This regex consists of two groups.
The first one (?<!\\) is a lookbehind assertion. It will match only if the previous letter is not a backslash. The second one (?'M'[^|%]) is a named capturing group (called M), that matches any character except "|" and "%".
I.e. the regex will match "a", and not match "\a" or "%".
Java does not support the named capture, but
(?<!\\)([^|%])
should work fine for you. You'd reference the first group by number, instead of by name then.
Note that you may have to escape backslashes leading to (?<!\\\\) for the first part.
Jens
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