regex | match first pattern | and stop if not matched

Question

String to be matched:

id=abc somethingelse value=1 value=2 value=3 value=1

String not to be matched:

id=abc somethingelse value=2 value=2 value=3 value=1

I would like to:

skip the somethingelse string, i.e., can't predict what this string is, so must use wildcard
match the first occurrence value=1
if the first value does not equal to 1, ignore any other occurrences and abort. So the second example should not be matched.

I tried id=abc.*?value=1 in regex101, but obviously it doesn't stop at the first value...

Many thanks.

So for the first string, you want two separate matches of `id=abc` and `value=1` ("skip the somethingelse"?), and for the second, you want no matches at all, is that right? — CertainPerformance, Mar 07 '19 at 08:40

score 0 · Accepted Answer · answered Mar 07 '19 at 08:40

0

You may use this regex with negative lookahead:

^id=abc(?:(?!value=).)*value=1

RegEx Demo

Key part is use of (?:(?!value=).)* which means match any character as long as it is not followed by value= (enforced by negative lookahead (?!value=)).

answered Mar 07 '19 at 08:40

anubhava

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This is awesome. Works like a charm!! Noted! Thank you so much – DBSQUARED Mar 07 '19 at 08:48

regex | match first pattern | and stop if not matched

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