Cannot get regex to be greedy in matches for items in brackets []

Question

Given this input:

[[00], [01], [02], [03]]

I'm trying to extract all the values inside the [ ].

The regex I've crafted looks like this /[^[\]]+(?=])/g But if I try in on my input, it only returns the first element

$ /[^[\]]+(?=])/g.exec(' [[00], [01], [02], [03]]');

$ [ '00', index: 3, input: ' [[00], [01], [02], [03]]' ]

What I don't understand is I've built the regex and tested it on https://regex101.com/r/QhZDMR/1, and on there it seems to return all the values just fine.

You have different regexes in the question and the regex demo. — Wiktor Stribiżew, Nov 19 '17 at 18:01
Getting multiple matches with a JS regex is a long solved problem, see the linked dupe reason. — Wiktor Stribiżew, Nov 19 '17 at 18:05
Fixed the mismatch, SO had removed my escaped `\\`. @Ec_ actually provided the answer I was looking for. — Laurier, Nov 19 '17 at 18:12
It is [already there](https://stackoverflow.com/a/43622095/3832970), no need providing it again and again. — Wiktor Stribiżew, Nov 19 '17 at 18:40

score 0 · Accepted Answer · answered Nov 19 '17 at 18:04

0

exec function finds only the first match. Instead use match like this:

"[[00], [01], [02], [03]]".match(/[^[\]]+(?=])/g)

answered Nov 19 '17 at 18:04

Ec_

66
5

Thanks, did not know this. Need more rep to upvote :) – Laurier Nov 19 '17 at 18:13

Cannot get regex to be greedy in matches for items in brackets []

1 Answers1