yckart yckart - 1 month ago 7
Javascript Question

Get duplicate characters in string

I try to match/get all repetitions in a string. This is what I've done so far:

var str = 'abcabc123123';
var REPEATED_CHARS_REGEX = /(.).*\1/gi;

console.log( str.match(REPEATED_CHARS_REGEX) ); // => ['abca', '1231']


As you can see the matching result is
['abca', '1231']
, but I excpect to get
['abc', '123']
. Any ideas to accomplish that?

2nd question:

Another thing I excpect, is to make it possible to change the duration how often a char needs to be in the string to get matched...

For example if the string is
abcabcabc
and the repetation-time is set to
2
it should result in
['abcabc']
. If set to
3
it should be
['abc']
.

Update

A non-
RegExp
solution is perfectly alright!

Answer

Well, I think falsetru had a good idea with a zero-width look-ahead.

'abcabc123123'.match(/(.+)(?=\1)/g)
// ["abc", "123"]

This allows it to match just the initial substring while ensuring at least 1 repetition follows.

For M42's follow-up example, it could be modified with a .*? to allow for gaps between repetitions.

'abc123ab12'.match(/(.+)(?=.*?\1)/g)
// ["ab", "12"]

Then, to find where the repetition starts with multiple uses together, a quantifier ({n}) can be added for the capture group:

'abcabc1234abc'.match(/(.+){2}(?=.*?\1)/g)
// ["abcabc"]

Or, to match just the initial with a number of repetitions following, add the quantifier within the look-ahead.

'abc123ab12ab'.match(/(.+)(?=(.*?\1){2})/g)
// ["ab"]

It can also match a minimum number of repetitions with a range quantifier without a max -- {2,}

'abcd1234ab12cd34bcd234'.match(/(.+)(?=(.*?\1){2,})/g)
// ["b", "cd", "2", "34"]