java regex capture group

Question

I am trying to capture the right part after the : using java expr, but in the following code, the printed capture group is the whole string, what's wrong?

String s ="xyz: 123a-45";   
String patternStr="xyz:[ \\t]*([\\S ]+)";
Pattern p = Pattern.compile(patternStr);
Matcher m = p.matcher(s);
//System.err.println(s);
if(m.find()){
    int count = m.groupCount();
    System.out.println("group count is "+count);
    for(int i=0;i<count;i++){
        System.out.println(m.group(i));
    }
}

score 29 · Accepted Answer · answered Jul 28 '11 at 20:42

29

Numbering of subgroups starts with 1, 0 is the full text. Just go till count+1 with your loop.

answered Jul 28 '11 at 20:42

yankee

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8

More precisely, change the `for` loop condition to `i<=count`. – Tim Pietzcker Jul 28 '11 at 20:49

score 1 · Answer 2 · answered Jul 28 '11 at 20:44

1

This is because group's indices are starting with 1. Group 0 is the entire pattern.

From the JavaDoc: "Capturing groups are indexed from left to right, starting at one. Group zero denotes the entire pattern, so the expression m.group(0) is equivalent to m.group()." See more here

answered Jul 28 '11 at 20:44

duduamar

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I think the confusion over this stems from the fact that the doc has muddled "indexing" (always from 0) with "numbering" (from whatever value you like). – william.berg Mar 22 '13 at 16:39

java regex capture group

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