How to match, but not capture, part of a regex?
I have a list of strings. Some of them are of the form 123-...456 . The variable portion "..." may be:
123-apple-456 123-banana-456 123-456 (note there's only one hyphen) Any word other than "apple" or "banana" is invalid.
For these three cases, I would like to match "apple", "banana", and "", respectively. Note that I never want capture the hyphen, but I always want to match it. If the string is not of the form 123-...456 as described above, then there is no match at all.
How do I write a regular expression to do this? Assume I have a flavor that allows lookahead, lookbehind, lookaround, and non-capturing groups.
The key observation here is that when you have either "apple" or "banana", you must also have the trailing hyphen, but you don't want to match it. And when you're matching the blank string, you must not have the trailing hyphen. A regex that encapsulates this assertion will be the right one, I think.
The only way not to capture something is using look-around assertions:
(?<=123-)((apple|banana)(?=-456)|(?=456))
Because even with non-capturing groups (?:…) the whole regular expression captures their matched contents. But this regular expression matches only apple or banana if it's preceded by 123- and followed by -456 , or it matches the empty string if it's preceded by 123- and followed by 456 .
Update: Thanks to Germán Rodríguez Herrera!
In javascript try: /123-(apple(?=-)|banana(?=-)|(?!-))-?456/
Remember that the result is in group 1
Debuggex Demo
Try:
123-(?:(apple|banana|)-|)456
That will match apple , banana , or a blank string, and following it there will be a 0 or 1 hyphens. I was wrong about not having a need for a capturing group. Silly me.
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