Javascript Regex back-reference not populating all capturing groups

Question

Strange one here (or maybe not), I am attempting to retrieve two capturing groups via Javascript regex, first group: one or more digits (0-9), second group: one or more word characters or hyphens (A-Z, 0-9, -) but for some reason I never can retrieve the latter group.

Please note: I have purposely included the alternation (|) character as I wish to potentially receive one or the other)

This is the code I am using:

var subject = '#/34/test-data'
var myregexp = /#\/(\d+)|\/([\w-]+)/;
var match = myregexp.exec(subject);
if (match != null && match.length > 1) {
  console.log(match[1]); // returns '34' successfully
  console.log(match[2]); // undefined? should return 'test-data'
}

Funny thing is Regex Buddy tells me I do have two capturing groups and actually highlights them correctly on the test phrase.

Is this a problem in my JavaScript syntax?

Michael Low · Accepted Answer · 2014-01-17 08:41:17Z

1

If you change:

var myregexp = /#\/(\d+)|\/([\w-]+)/;

by removing the | alternation meta-character to just:

var myregexp = /#\/(\d+)\/([\w-]+)/;

it will then match both groups. At present, your regex is looking for either \d+ or [\w-]+ so once it matches the first group it stops and the second will be empty. If you remove |, it's looking for \d+ followed by /, followed by [\w-]+ so it will always match either both or none.

Edit: To match on all of #/34/test-data, #/test-data or #/34, you can use #(?:\/(\d+))?\/([\w-]+) instead.

edited Jan 17, 2014 at 8:41

answered Oct 11, 2011 at 1:44

Michael Low

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4 Comments

GONeale Over a year ago

Sorry I should have mentioned, I would like to match on #/test-data OR #/34 that's why the alternation character was there, how would I do this?

GONeale Over a year ago

Oh I think I just had a thought, I can use your version, but change + to * to match 0 or many times!

GONeale Over a year ago

Amazing, thanks. Works well. Probably better than my idea too :)

GONeale Over a year ago

Just to let you know I advanced on this a bit, just to keep the capturing groups at the correct index positions. This is what I settled on: #(?:/(\d+))?(?:/([\w-]+))? it will keep a match of #/34 in group 1 and #/test-data in group 2 and then both in group 1 & 2 respectively.

Deleteman · Accepted Answer · 2011-10-11 01:43:19Z

1

If you remove the "|" you get the result you want... does this help?

var subject = '#/34/test-data'
var myregexp = /#\/(\d+)\/([\w-]+)/;
var match = myregexp.exec(subject);
if (match != null && match.length > 1) {
  console.log(match[1]); // returns '34' successfully
  console.log(match[2]); // undefined? should return 'test-data'
}

Happy coding!

Edit

I think your problem was, that since you were using the "|", you were telling JS to catch either the first group or the second one, and since JS eval is lazy, when it found the first group, it stopped there... By removing the OR operand from the RegExp, you get both results...(something like an AND).

answered Oct 11, 2011 at 1:43

Deleteman

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2 Comments

GONeale Over a year ago

Please see the response I wrote to mike, I actually need the or test.

GONeale Over a year ago

Hi Deleteman, I had to accept Mike's answer because his met my requirement I forgot to mention, my fault, but thanks for the effort though!

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