Why doesn't String.match() produce expected results when a global flag is present?

Question

I was trying this match

'/links/51f5382e7b7993e335000015'.match(/^\/links\/([0-9a-f]{24})$/g)

and got:

['/links/51f5382e7b7993e335000015']

while I was expecting:

['/links/51f5382e7b7993e335000015', '51f5382e7b7993e335000015']

I had no luck until I removed the global flag, which I did not think would impact results my results!

With the global flag removed,

'/links/51f5382e7b7993e335000015'.match(/^\/links\/([0-9a-f]{24})$/)

produced:

[ '/links/51f5382e7b7993e335000015',
  '51f5382e7b7993e335000015',
  index: 0,
  input: '/links/51f5382e7b7993e335000015' ]

which is cool, but reading the docs I can't figure out:

Why the first form didn't work
Why the global flag interfered with the () matching
How to get my expected result without the index and input properties

On JavaScript Regex and Submatches the top answer says:

Using String's match() function won't return captured groups if the global modifier is set, as you found out.

However,

> 'fofoofooofoooo'.match(/f(o+)/g) 
["fo", "foo", "fooo", "foooo"]

seems to produce captured groups just fine.

Thank you.

I edited the question to address the possible duplicate issue. — Dmitry Minkovsky, Jul 28 '13 at 21:09
It's a duplicate, but +1 for "I tried..., I got..., I was expecting...". I wish more questions were like this. — Brigand, Jul 28 '13 at 21:09
Hah! Thanks... I will look again, and more carefully, at the linked page. — Dmitry Minkovsky, Jul 28 '13 at 21:10
@PavelAnossov My bad, I meant to include the (). Please see now. Works the same! Is that not a captured group, despite the ()? — Dmitry Minkovsky, Jul 28 '13 at 21:12
The edited last example isn't putting captured groups in the output, it is including different matches of the whole pattern. — nnnnnn, Jul 28 '13 at 21:14
@dimadima: the thing is, without `g` String.match produces "first match and its captured groups", with `g` it produces "all matches of the entire pattern". — Pavel Anossov, Jul 28 '13 at 21:15
@PavelAnossov, ah, I understand, thank you: The resulting array can only represent one or the other: either (i) the first match and its captured groups, OR (ii) all matches of the pattern. So `g` necessitates that the result array is of the latter form. — Dmitry Minkovsky, Jul 28 '13 at 21:21

Rohit Jain · Accepted Answer · 2013-07-28T21:32:34.823

From this msdn documentation for match method:

If the global flag (g) is not set, Element zero of the array contains the entire match, while elements 1 through n contain any submatches. This behavior is the same as the behavior of the exec Method (Regular Expression) (JavaScript) when the global flag is not set. If the global flag is set, elements 0 through n contain all matches that occurred.

Emphasis mine.

So, in your 1st case:

'/links/51f5382e7b7993e335000015'.match(/^\/links\/([0-9a-f]{24})$/g)

Since /g modifier is set, it will return just the complete matches that occurred, and not the submatches. That is why you just got an array with single element. As there is only 1 match for that regex.

2nd case:

'/links/51f5382e7b7993e335000015'.match(/^\/links\/([0-9a-f]{24})$/)

/g modifier is not set. So the array contains the complete match at 0th index. And further elements (1st index) in array are submatches - in this case, 1st capture group.

As for your last example:

'fofoofooofoooo'.match(/f(o+)/g)

Again, since /g modifier is set, it will return all the matches from the string, and not the submatches. So, in the string, the regex f(o+) matches 4 times:

fo    - 1st complete match (sub-match 'o' in 1st captured group ignored)
foo   - 2nd complete match (sub-match 'oo' ignored)
fooo  - 3rd complete match (sub-match 'ooo' ignored)
foooo - 4th complete match (sub-match 'oooo' ignored)

If you use the last regex without /g modifier, you would get each submatch as separate element, for the first match. Try:

'fofoofooofoooo'.match(/f(o+)/)

You will get:

["fo", "o"]  // With index and input element of course.

Without /g it just stops after first match (fo), and returns the entire match and sub-matches.

Thank you very much for this Rohit! Wow, really appreciate this answer. — Dmitry Minkovsky, Jul 28 '13 at 21:25
Rohit: Actually, `'fofoofooofoooo'.match(/f(o+)/)` produces `['fo', 'o', index: 0, input: 'fofoofooofoooo']`. I believe this is because there is no backtracking that goes on. — Dmitry Minkovsky, Jul 28 '13 at 21:29
@dimadima. Oh! So, it just returns the first match and the submatch. I just guessed that output. Hmm, it's clear, since global flag is not set, it will not try to match more than once. — Rohit Jain, Jul 28 '13 at 21:31
Yep, which is actually simpler and makes sense. I'm glad that's the real result :). — Dmitry Minkovsky, Jul 28 '13 at 21:31

score 1 · Answer 2 · answered Jul 28 '13 at 21:17

According to MDN, if the g flag is not specified it returns the same results as RegExp.exec(), which returns an array including items for capturing parentheses.

If the g flag is specified it returns an array containing all matches.

This description is consistent with the examples in the question, but your two examples are sort of apples with oranges:

Your "links" regex matches the entire input string once, possibly with one captured group depending on the flag.
Your /f(o+)/g regex matches several substrings in the input.

Thanks for this answer and your clarification of my example – Dmitry Minkovsky Jul 28 '13 at 21:24 — Dmitry Minkovsky, Jul 28 '13 at 21:24

Why doesn't String.match() produce expected results when a global flag is present?

2 Answers2