regex 匹配括号内包含（或不包含）空格的字符串

Question

How do you capture a string with regex, that is a) inside round brackets, including the brackets, and b) includes a space (or alternatively doesn't include a space) inside the brackets, so to capture (test test), ( test), (test ) but not (test) .

I've found a similar problem here that uses the expression

\((?!\s)[^()]+(?<!\s)\)

but with this lookahead/lookbehind rule just looks for spaces that are attached to the brackets. How do you modify that to look for every space inside the bracket?

Thank you in advance!

Answer 1

You can use /\\([^)]* [^)]*\\)/ . This matches 0 or more non-closing parentheses characters followed by a space followed by 0 or more non-closing parenthesis characters. So it will match "( )" and "( )" but not "()" .

 const s = "() ( test) (test) (test ) ( ) ( ) ( test ) (t est)"; console.log([...s.matchAll(/\\([^)]* [^)]*\\)/g)]);

Answer 2

Simply \\([^()]+\\) matches everything between a pair of parentheses. That includes spaces, and any other character except (newline or) parentheses.

(Assuming a regex dialect where literal parens need to be escaped, and plus matches one or more repetitions. Whether a negated character class matches a newline is also dialect-specific.)

If (as you say in a comment) you require at least one space, that's easy to add. Rephrase: match an opening parenthesis character, zero or more non-space, non-parenthesis characters, one space, zero or more non-parenthesis characters (spaces are okay too) until the closing parenthesis.

\([^ ()]* [^())*\)