Replace nth match of matches with regex

Question

I\'m trying to find a way to replace nth match of more matches lite this.

string = \"one two three one one\"

How do I target the se

Answer 1

Update :

To make it dynamic use this:

((?:.*?one.*?){1}.*?)one

where the value 1 means (n-1); which in your case is n=2

and replace by:

$1\(one\)

Regex101 Demo

const regex = /((?:.*?one.*?){1}.*?)one/m;
const str = `one two three one one asdfasdf one asdfasdf sdf one`;
const subst = `$1\(one\)`;
const result = str.replace(regex, subst);
console.log( result);

Answer 2

A more general approach would be to use the replacer function.

// Replace the n-th occurrence of "re" in "input" using "transform"
function replaceNth(input, re, n, transform) {
  let count = 0;

  return input.replace(
    re, 
    match => n(++count) ? transform(match) : match);
}

console.log(replaceNth(
  "one two three one one", 
  /\bone\b/gi,
  count => count ===2,
  str => `(${str})`
));

// Capitalize even-numbered words.
console.log(replaceNth(
  "Now is the time",
  /\w+/g,
  count => !(count % 2),
  str => str.toUpperCase()));