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));
    }
}

yankee · Accepted Answer · 2011-07-28 20:42:30Z

29

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

answered Jul 28, 2011 at 20:42

yankee

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1 Comment

Tim Pietzcker Over a year ago

More precisely, change the for loop condition to i<=count.

duduamar · Accepted Answer · 2011-07-28 20:44:36Z

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, 2011 at 20:44

duduamar

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1 Comment

william.berg Over a year ago

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).

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