正则表达式 - 如何删除括号内的字符串？

Question

I want to remove all words inside brackets and square brackets. I'm using this regex, but it only removes words inside brackets. It does not work with square brackets...

var str = 'hey [xx] (xhini) rexhin (zzz)';
var r = str.replace(/ *\([^)]*\)*\] */g, '');

r should be hey rexhin

Any suggestions?

Answer 1

You can use this regex:

var str = 'hey [xx] (xhini) rexhin (zzz)';
var r = str.replace(/(\[.*?\]|\(.*?\)) */g, "");
//=> hey rexhin

• \\[.*?\\] will find square brackets and string inside them
• \\(.*?\\) will find round brackets and string inside them

You can also use (if square and round brackets are not nested):

var r = str.replace(/[(\[].*?[)\]] */g, "");

• [(\\[] is a character class that finds ( or [
• [)\\]] is a character class that finds ) or ]

You can call trim() to trim trailing space also.

Answer 2

You could try the below regex.

> var str = "[xxx] hey [xx] (xhini) rexhin (zzz)";
undefined
> str.replace(/^(?:\[[^\]]*\]|\([^()]*\))\s*|\s*(?:\[[^\]]*\]|\([^()]*\))/g, "")
'hey rexhin'

DEMO

• ^(?:\\[[^\\]]*\\]|\\([^()]*\\))\\s* would match the brackets as well as the following spaces which are present at the start ( start of the line ).

• | OR

• \\s*(?:\\[[^\\]]*\\]|\\([^()]*\\)) Matches all the remaining brackets along with their preceding spaces.