Javascript Regex with .test()

Question

> var p = /abc/gi;
> var s = "abc";
> p.test(s);
  true
> p.test(s);
  false;

When I run this code on console of Chrome I Have this output above. Each time I call .test() I get a different value. Someone could explain to me why this happens? thanks

More dupes: [Interesting test of Javascript RegExp](http://stackoverflow.com/q/604860) and [Regular Expression test can't decide between true and false](http://stackoverflow.com/questions/2669861/regular-expression-test-cant-decide-between-true-and-false-javascript) — Andy E, Oct 05 '11 at 11:54

score 9 · Accepted Answer · edited May 23 '17 at 10:29

9

The behavior is due to the "g" modifier, i.e. matches three times, no match the fourth time:

> var p = /a/gi;
> var s = "aaa";
> p.test(s)
true
> p.test(s)
true
> p.test(s)
true
> p.test(s)
false

See similar question: Why RegExp with global flag in Javascript give wrong results?

edited May 23 '17 at 10:29

Community

1
1

answered Oct 05 '11 at 11:43

Stefan

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score 5 · Answer 2 · answered Oct 05 '11 at 11:43

The g flag causes the RegExp literal your using to track the matches LastIndex

If you were to;

print( p.test(s), p.lastIndex )
print( p.test(s), p.lastIndex )

You would see

true,3
false,0

So the 2nd test fails as there is no incremental match from the 1st.

score 2 · Answer 3 · edited May 23 '17 at 11:47

2

It's because of the /g flag. Every consecutive search starts from the character last matched in the previous search. In your case, in the second run it starts from the end of the string and returns false. The third time it starts from the beginning again. And so forth.

Also, take a look at this question: Why RegExp with global flag in Javascript give wrong results?

edited May 23 '17 at 11:47

Community

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answered Oct 05 '11 at 11:46

Dmytro Shevchenko

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Javascript Regex with .test()

3 Answers3