Repeating number RegExp in javascript

Question

How can I get ['999'] out of this string? '451999277'? I only want repetitions of the same character.

This is what I've tried:

'451999277'.match(/(\d{3})/g) // === ["451", "999", "277"]
'451999277'.match(/(\d){3}/g) // === ["451", "999", "277"]
'451999277'.match(/([0-9]){3}/g) // === ["451", "999", "277"]
'451999277'.match(/(\d)\1{3}/g) // === null

.......

[EDIT]

solution:

'451999277'.match(/(\d)\1{2}/g) // ===  ['999']

(\d) matches one digit, \1{3} matches three digits, so (\d)\1{3} would require four digits to match. — melpomene
– melpomene, Commented Sep 12, 2018 at 17:26
Gotcha. Thanks! This is the winner: '451999277'.match(/(\d)\1{2}/g) — Matthew Masurka
– Matthew Masurka, Commented Sep 12, 2018 at 17:40
As a nit-pick, I object to your use of === in the comment. I'd like to point out that even ['999'] === ['999'] is false. We get what you meant though. — Wyck
– Wyck, Commented Sep 12, 2018 at 17:48

Mark · Accepted Answer · 2018-09-12 17:38:08Z

1

You're almost there with your last example. If you just want groups of 3 use {2} instead of {3}:

console.log('4519992277'.match(/(\d)\1{2}/g))

console.log('455519992277'.match(/(\d)\1{2}/g))

answered Sep 12, 2018 at 17:38

Mark

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