Regular expression takes bounds into group

Question

I want to group some characters between special symbol bounds, but this bounds are involved into group. It is not connected with greedy and reluctant quantifier.

String str1 = "#162xd14a#";
Matcher matcher = Pattern.compile("#(.*?)#").matcher(str1);
if (matcher.matches()) {
    for (int i = 0; i < matcher.groupCount(); i++) {
        System.out.println(matcher.group(i));
    }
}

Result:

#162xd14a#

But i expected:

162xd14a

you can also use lookbehinds and lookaheads like `(?<=#)(.*?)(?=#)` — rock321987, Jul 10 '16 at 08:37

score 5 · Accepted Answer · answered Jul 10 '16 at 08:46

From the official tutorial about Capturing Groups:

There is also a special group, group 0, which always represents the entire expression. This group is not included in the total reported by groupCount.

In the context of your example:

Group 0 contains the entire matched string, #162xd14a#
Group 1 contains the first (and only) matched group: 162xd14a
You have one group in your regex, therefore groupCount() returns 1
The loop stops before reaching group 1, because of the < condition

You probably want to change the loop condition to <= instead of <:

for (int i = 0; i <= matcher.groupCount(); i++) {
    System.out.println(matcher.group(i));
}

Regular expression takes bounds into group

1 Answers1