正则表达式在第n次出现字符时剪切字符串并返回字符串的第一部分

Question

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 occurrence to remove.

If nth =3, the above will return “xcv_xcv_x”

Details in this StackOverflow post: Cutting a string at nth occurrence of a character

How to change the above regular expression to return the first part instead (ie. “qwe_fs_xczv”)?

Answer 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}$/, '') ) 

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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 .

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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] ) 

Answer 2

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]) 

Answer 3

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);