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']

Possible duplicate of [Regex to find repeating numbers](https://stackoverflow.com/questions/6507982/regex-to-find-repeating-numbers) — scrappedcola, Sep 12 '18 at 17:19
`(\d)` matches one digit, `\1{3}` matches three digits, so `(\d)\1{3}` would require four digits to match. — melpomene, Sep 12 '18 at 17:26
Gotcha. Thanks! This is the winner: '451999277'.match(/(\d)\1{2}/g) — Matthew Masurka, Sep 12 '18 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, Sep 12 '18 at 17:48

score 1 · Accepted Answer · answered Sep 12 '18 at 17:38

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 '18 at 17:38

Mark

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Repeating number RegExp in javascript

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