javascript regex split match at char once, not multiple

Question

Having the following example:

var s = '#hello \n##1234 \n #hello2';
var splits = s.split(/#[^##]/);

produces: ["", "ello ↵#", "234 ↵ ", "ello2"]

But want it to produce: ["hello ↵##1234 ↵ ", "hello2"]

How to match single occurrence of character (#) but NOT if its occurring multiple times, and also not the last occurrence of the multiple occurred one.

was not able to edit again: Shure I mean - Wanting: ["", "hello ↵##1234 ↵ ", "hello2"] — svens
– svens, Commented Mar 14, 2017 at 10:04
[^#] produces: ["", "ello ↵#", "234 ↵ ", "ello2"] ~ so yes / [^##] == [^#] // but how to Exclude? — svens
– svens, Commented Mar 14, 2017 at 10:08

Wiktor Stribiżew · Accepted Answer · 2017-03-14 10:03:38Z

5

With split, you would need a negative lookbehind to prevent matching # that comes after another #, but JavaScript does not support lookbehinds.

Match the chunks rather than split with .match(/(?:##+|[^#])+/g) that will match 1 or more occurrences of 2+ # or any char other than #:

var s = '#hello \n##1234 \n #hello2';
var splits = s.match(/(?:##+|[^#])+/g);
console.log(splits);

Pattern details:

(?:##+|[^#])+ - 1 or more occurrences (+) of:
- ##+ - 2 or more # symbols, as many as possible
- [^#] - any character other than # (a [^...] is a negated character class that matches any single char that is not inside the defined set/ranges).

NOTE: [^##] is the same as [^#], you cannot negate a sequence with a negated character class, only specific chars or ranges of chars.

answered Mar 14, 2017 at 10:03

Wiktor Stribiżew

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