如何使用正则表达式匹配此字符串中的“ aas”？

Question

String str = "aa(aaq(aas)ppp)eeews";

How can I use a regular expression to match "aas" ?

want to match the contents of the brackets, but not sure how many brackets,may be more and more

Answer 1

(?<=\\()[^()]*(?=\\))

• (?<=\\\\() - positive lookbehind to check for the existence of a opening bracket
• [^()] - a character, which is not a closing or opening bracket. You don't want closing brackets so you don't match more than you need. You don't want opening brackets, because you wanted the inner most.
• * - repeat the above zero or more times
• (?=\\\\)) - positive lookahead to check for the existence of a closing bracket

Answer 2

Since you need to match the text inside parentheses that has no ( or ) inside, you can use

String str = "aa(aaq(aas)ppp)eeews";
String rx = "\\(([^()]*)\\)";
Pattern ptrn = Pattern.compile(rx);
Matcher m = ptrn.matcher(str);
while (m.find()) {
   System.out.println(m.group(1)); // => aas
}

See regex demo

The text you need is in Capture group 1. See IDEONE demo . To only get the first occurrence, use if instead of while .

Another way to get the first occurrence is with "aa(aaq(aas)ppp)eeews".replaceAll("(?s).*?\\\\(([^()]*)\\\\).*$", "$1") ( demo ).

Why not use lookarounds? You can, but they always involve some additional overhead to the regex engine. Certainly, with this basic pattern as here, performance is not the key, however, it is best practice to only use lookarounds when you have to.

The regex details:

• \\( - matches an opening (
• ([^()]*) - Capture group 1 matching zero or more characters other than ( or )
• \\) - matches closing round bracket.