Overlapping group capturing

问题

Please take a look at the following code:

public static void main(String[] args) {
    String s = "a < b > c > d";
    String regex = "(\\w\\s*[<>]\\s*\\w)";
    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(s);
    int i = 0;
    while (m.find()) System.out.println(m.group(i++));
}

The output of the above program is: a < b, c > d

But I actually expect a < b, b > c, c > d.

Anything wrong with my regexp here?

回答1:

Try this.

    String s = "a < b > c > d";
    String regex = "(?=(\\w{1}\\s{1}[<>]{1}\\s{1}\\w{1})).";
    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(s);
    while(m.find()) {
        System.out.println(m.group(1));
    }

Updated(Based on green's solution):

    String s = " something.js > /some/path/to/x19-v1.0.js < y < z < a > b > c > d";
    String regex = "(?=[\\s,;]+|(?<![\\w\\/\\-\\.])([\\w\\/\\-\\.]+\\s*[<>]\\s*[\\w\\/\\-\\.]+))";

    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(s);
    while (m.find()) {
        String d = m.group(1);
        if(d != null) {
            System.out.println(d);
        }
    }

回答2:

You're right in your thinking that b > c matches the regex because it does.

But when you call Matcher::find(), it returns the next substring of the input which matches the regex and is disjoint from previous find() matches. Since "b > c" begins with the 'b' which was part of the "a > b" match returned by the previous invocation, it won't be returned by find().

回答3:

Based on John's solution and adding some boundary matchers, this works finally.

    String s = " something.js > /some/path/to/x19-v1.0.js < y < z < a > b > c > d";
    String regex = "(?=[\\s,;]+([\\w\\/\\-\\.]+\\s*[<>]\\s*[\\w\\/\\-\\.]+)[\\s,;$]*).";
    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(s);
    while(m.find()) {
        System.out.println(m.group(1));
    }

来源：https://stackoverflow.com/questions/5508965/overlapping-group-capturing