Java Regex Matcher get wrong group count?

Question

My code is

String debug = "1$<$2";
Matcher matcher = Pattern.compile("^[1-5]" + "\\(?\\$[^\\$]*\\$\\)?" + "([1-5])$").matcher(debug);
List<String> matches = new ArrayList<String>();
if (matcher.matches()) {
  for (var i = 0;i< matcher.groupCount();i++){
    matches.add(matcher.group(i));
  }
}
System.out.println(matcher.groupCount());
System.out.println(matches);

And the group count is only 1.

The matches is 1$<$2.

But actually the result of matcher.group(1) is 2.

How can I get the right group count?

You have only one capturing group in your regex, ([1-5]) at the end. — Mark Rotteveel
– Mark Rotteveel, Commented Aug 26, 2022 at 9:26
Also note that matcher.group(0) will always be the full match. I.e. the input string. From the javadoc: Group zero denotes the entire pattern, so the expression m.group(0) is equivalent to m.group() — Lino
– Lino, Commented Aug 26, 2022 at 9:26

assylias · Accepted Answer · 2022-08-26 09:26:41Z

4

That is because the first group (index 0) is the whole pattern and is not counted. See the javadoc:

public int groupCount()

Returns the number of capturing groups in this matcher's pattern.
Group zero denotes the entire pattern by convention. It is not included in this count.
Any non-negative integer smaller than or equal to the value returned by this method is guaranteed to be a valid group index for this matcher.

So you can change your code to this for example:

for (var i = 1; i <= matcher.groupCount(); i++){
  matches.add(matcher.group(i));
}

answered Aug 26, 2022 at 9:26

assylias

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