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Regex to first occurrence only? [duplicate]

开发者 https://www.devze.com 2022-12-16 08:03 出处:网络
This question already has answers here开发者_如何转开发: Regular expression to stop at first match
This question already has answers here开发者_如何转开发: Regular expression to stop at first match (9 answers) Closed last year.

Let's say I have the following string:

this is a test for the sake of testing. this is only a test. The end.

and I want to select this is a test and this is only a test. What in the world do I need to do?

The following Regex I tried yields a goofy result:

this(.*)test (I also wanted to capture what was between it)

returns this is a test for the sake of testing. this is only a test

It seems like this is probably something easy I'm forgetting.


The regex is greedy meaning it will capture as many characters as it can which fall into the .* match. To make it non-greedy try:

this(.*?)test

The ? modifier will make it capture as few characters as possible in the match.


Andy E and Ipsquiggle have the right idea, but I want to point out that you might want to add a word boundary assertion, meaning you don't want to deal with words that have "this" or "test" in them-- only the words by themselves. In Perl and similar that's done with the "\b" marker.

As it is, this(.*?)test would match "thistles are the greatest", which you probably don't want.

The pattern you want is something like this: \bthis\b(.*?)\btest\b


* is a greedy quantifier. That means it matches as much as possible, i.e. what you are seeing. Depending on the specific language support for regex, you will need to find a non-greedy quantifier. Usually this is a trailing question mark, like this: *?. That means it will stop consuming letters as soon as the rest of the regex can be satisfied.

There is a good explanation of greediness here.


For me, simply remove /g worked.

See https://regex101.com/r/EaIykZ/1

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