JavaScript Regex: Make ungreedy - Stack Overflow

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I have this regex which looks for %{any charactering including new lines}%:

/[%][{]\s*((.|\n|\r)*)\s*[}][%]/gm

If I test the regex on a string like "%{hey}%", the regex returns "hey" as a match.

However, if I give it "%{hey}%%{there}%", it doesn't match both "hey" and "there" seperately, it has one match—"hey}%%{there".

How do I make it ungreedy to so it returns a match for each %{}%?

I have this regex which looks for %{any charactering including new lines}%:

/[%][{]\s*((.|\n|\r)*)\s*[}][%]/gm

If I test the regex on a string like "%{hey}%", the regex returns "hey" as a match.

However, if I give it "%{hey}%%{there}%", it doesn't match both "hey" and "there" seperately, it has one match—"hey}%%{there".

How do I make it ungreedy to so it returns a match for each %{}%?

• javascript
• regex

Share Improve this question Follow asked Feb 15, 2010 at 1:40 JamesBrownIsDeadJamesBrownIsDead 93522 gold badges99 silver badges1515 bronze badges 1

• As I always mention on Regular expression questions, check out Regexr, a cool Flash Based Regex tool, by gSkinner Link: gskinner./RegExr There is also the AS3 Regular Expression tester: idsklijnsma.nl/regexps – Moshe Commented Feb 15, 2010 at 1:49

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3 Answers 3

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/[%][{]\s*((.|\n|\r)*?)\s*[}][%]/gm

Firstly, to make a wildcard match non-greedy, just append it with ? (so *? instead of * and +? instead of +).

Secondly, your pattern can be simplified in a number of ways.

/%\{\s*([\s\S]*?)\s*\}%/gm

There's no need to put a single character in square brackets.

Lastly the expression in the middle you want to capture, you'll note I put [\s\S]. That es from Matching newlines in JavaScript as a replacement for the DOTALL behaviour.

Shorter and faster working:

/%\{([^}]*)\}%/gm

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