Once replaced, skip that part of string - Regex Javascript

Question

I expect a string of just numbers from HTML inputbox but i am looking for a regex so that i can capture some part of string and replace it with something and now what i want is, for next iteration, regex should skip part which has been processed.

Take a look at my regex

string.replace(/([\d]{10})/gm, "$1,")

Expected results for iterations

895645784578457845784578457845 source
8956457845,7845784578,4578457845,9089 more data is coming

But problem is result

8956457845,7845784578,4578457845
8956457845,,,7845784578,,,4578457845,,,9089

Works for me. "895645784578457845784578457845".replace(/([\d]{10})/gm, "$1,") returns "8956457845,7845784578,4578457845," as one would expect. — T.J. Crowder
– T.J. Crowder, Commented Sep 4, 2018 at 16:51
probably you;d like to use negative lookahead /(\d{10}(?!\,))/gm, like this fiddle — Sphinx
– Sphinx, Commented Sep 4, 2018 at 17:06

Sphinx · Accepted Answer · 2018-09-04 17:15:40Z

3

If I understood correctly, you'd like to apply regex-replace for one numerical string recursely like string.replace(\regex\, '$1,').replace(\regex\, '$1,').replace(\regex\, '$1,'), but ignored the parts which already replaced.

Below is one solution which uses negative lookahead.

let test = '895645784578457845784578457845' //org string
let test1 = test.replace(/(\d{10}(?!\,))/gm, "$1,")
                .replace(/(\d{10}(?!\,))/gm, "$1,")
                .replace(/(\d{10}(?!\,))/gm, "$1,") 
                // simulate recurse-replace three times
console.log(test1)

test1 += '1234567890123' //new string came
let test2 = test1.replace(/(\d{10}(?!\,))/gm, "$1,")
console.log(test2)

test2 += '1234567890123' //new string came
let test3 = test2.replace(/(\d{10}(?!\,))/gm, "$1,")
console.log(test3)

edited Sep 4, 2018 at 17:15

answered Sep 4, 2018 at 17:09

Sphinx

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