Regex to first occurence only?

0 votes
asked Jan 15, 2010 by ben-lesh

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.

4 Answers

0 votes
answered Jan 15, 2010 by andy-e

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:


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

0 votes
answered Jan 15, 2010 by ipsquiggle

* 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.

0 votes
answered Jan 15, 2010 by platinum-azure

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

0 votes
answered Sep 15, 2017 by meloman

For me, simply remove /g worked.


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