RegEx: find() and group() issue...

Question

Hi,
In the code below, I use the find() method to search the first occurence of a match and then group(1) to retrieve that match if the find() method returns true.
The problem is that the find() method returns true and the group(1) throws an ArrayOutOfBoundException :(
Here is the RegEx code I use:
  String value = "${local.production}\\makeresults\\repository";
  Matcher matcher = Pattern.compile("\\$\\{.*\\}").matcher(value);
  if (matcher.find()) {
    System.out.println(matcher.group(1));   // throws an ArrayOutOfBoundException
Got this output:
Exception in thread "main" java.lang.IndexOutOfBoundsException: No group 1
  at java.util.regex.Matcher.group(Matcher.java:463)
  at Test2.main(Test2.java:28)Have I made any mistakes?
Thanks in advance for your help.
Best regards,
Serge.

Answer 1

You shall use group() without parameter to get complete sequence.
Only capturing groups (between parenthesis) are indexed (from left to right, starting at one.)
Here you don't have capturing group (and that's why it says No group 1), and it seems that you need the entire pattern, so you need the group() method.
Matcher
Tim - Actually, you can also use group(0) ...