How to write this number detecting regexp in Javascript?

Question

This is how it looks like in Perl-compatible regexp (taken from https://stackoverflow.com/a/4824952/377920):

(?:(?<!\d)\d{1,3}(?!\d))

However, apparently Javascript lacks some regexp features so this doesn't work.

I'm trying to match 1-3 long connected digits that can have non-white characters on both ends.

Such as "Road 12A55, 10020" would match 12 and 55.

Not really the solution you asked for, but the clean way to test if a string is a number in Javascript is to use isNaN() — Guillaume Algis
– Guillaume Algis, Commented Feb 7, 2013 at 19:40
Why not just grab all substrings of digits (via the simple regex (\d+)) and then see if the matched groups have a length between 1 and 3? — user554546
– user554546, Commented Feb 7, 2013 at 19:43

Niet the Dark Absol · Accepted Answer · 2013-02-07 19:41:31Z

4

You are correct, JavaScript does not support lookbehinds.

It looks like you are trying to detect a sequence of no more than 3 digits. Depending on what the surrounding context is, you may be able to use this instead:

/(?:^|\D)\d{1,3}(?:\D|$)/

answered Feb 7, 2013 at 19:41

Niet the Dark Absol

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Ry- · Accepted Answer · 2013-02-07 19:41:49Z

4

You can rewrite the expression without lookbehinds - just make sure to get group 1:

/(?:^|\D)(\d{1,3})(?!\d)/

answered Feb 7, 2013 at 19:41

Ry-♦

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Rohit Jain · Accepted Answer · 2013-02-07 19:42:48Z

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Javascript does not support look-behinds, that is why your regex didn't work.

You can try out this alternative: -

/(?:^|\D)(\d{1,3})(?!\d)/

And get the group 1.

answered Feb 7, 2013 at 19:42

Rohit Jain

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גלעד ברקן · Accepted Answer · 2013-02-08 03:20:53Z

1

This returns 12 and 55:

var output = 'Road 12A55, 10020'.replace(/D+|\d{4,}/g, ' ').match(/\d+/g)

alert(output)

edited Feb 8, 2013 at 3:20

answered Feb 7, 2013 at 20:50

גלעד ברקן

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