Regex Expression To Cut String At Nth Occurrence Of Character And Return First Part Of String

I have found the answer for a case which returns the second part of the string, eg: 'qwe_fs_xczv_xcv_xcv_x'.replace(/([^\_]*\_){**nth**}/, ''); - where is nth is the amount of occu

Solution 1:

You need to use end anchor($) to assert ending position.

"qwe_fs_xczv_xcv_xcv_x".replace(/(_[^_]*){nth}$/, ''); 
//            --------------------^-----------^--- here

console.log(
  "qwe_fs_xczv_xcv_xcv_x".replace(/(_[^_]*){3}$/, '')
)

---

UPDATE : In order to get the first n segments you need to use String#match method with slight variation in the regex.

"qwe_fs_xczv_xcv_xcv_x".match(/(?:(?:^|_)[^_]*){3}/)[0]

console.log(
  "qwe_fs_xczv_xcv_xcv_x".match(/(?:(?:^|_)[^_]*){3}/)[0]
)

In the above regex (?:^|_) helps to assert the start position or matching the leading _. Regex explanation here.

---

Another alternative for the regex would be, /^[^_]*(?:_[^_]*){n-1}/. Final regex would be like:

/^[^_]*(?:_[^_]*){2}/

console.log(
  "qwe_fs_xczv_xcv_xcv_x".match(/^[^_]*(?:_[^_]*){2}/)[0]
)

Solution 2:

Use String.match() to look from the start (^) of the string, for three sequences of characters without underscore, that might start with an underscore (regex101):

const result = "qwe_fs_xczv_xcv_xcv_x".match(/^(?:_?[^_]+){3}/);
console.log(result);

Solution 3:

If you want to use replace, then capture up to right before the third _ and replace with that group:

const re = /^(([^_]*_){2}[^_]*(?=_)).*$/;
console.log("qwe_fs_xczv_xcv_xcv_x".replace(re, '$1'));
console.log("qwe_fs_xczv_xcv_xcv_x_x_x".replace(re, '$1'));

But it would be nicer to use match to match the desired substring directly:

const re = /^([^_]*_){2}[^_]*(?=_)/;
console.log("qwe_fs_xczv_xcv_xcv_x".match(re)[0])
console.log("qwe_fs_xczv_xcv_xcv_x_x_x".match(re)[0])