Regex - How to remove strings inside brackets?

问题

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?

回答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.

回答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.

来源：https://stackoverflow.com/questions/28517941/regex-how-to-remove-strings-inside-brackets