正则表达式返回方括号（但不是括号）之间的字符

Question

I'm looking for a RegEx to return all the characters between two square brackets, and only the characters, not the brackets themselves.

For example given a string that looks like "[A][B2][1,1][ABC]"

The RegEx should return: A , B2 , 1,1 , ABC

I have tried various expressions for both String#split and Pattern & Match .

The closest I've gotten was:

String.split("[\\\\[]*[\\\\]]"); and Pattern.compile("\\\\[[^\\\\]]+(?=])");

Both of these return: [A , [B2 , [1,1 , [ABC

I've found other expressions that return the values with the enclosing brackets (ie. Pattern.compile("\\\\[[^\\\\]]*\\\\]"); ) but that is not what I'm after.

Does someone know the correct expression? or is what I'm trying to do not possible?

Answer 1

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

你需要lookaheads和lookbehinds 。或者

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

Answer 2

All you need is a capturing group to retain the part of the pattern you want.

String s  = "[A][B2][1,1][ABC]";
Pattern p = Pattern.compile("\\[([^]]*)]");
Matcher m = p.matcher(s);
while (m.find()) {
  System.out.println(m.group(1));
}

Ideone Demo

Answer 3

你可以用它。

Pattern.compile("\\[(.*?)\\]");