Regular expression with variable number of groups?

According to the documentation, Java regular expressions can't do this:

The captured input associated with a group is always the subsequence that the group most recently matched. If a group is evaluated a second time because of quantification then its previously-captured value, if any, will be retained if the second evaluation fails. Matching the string "aba" against the expression (a(b)?)+, for example, leaves group two set to "b". All captured input is discarded at the beginning of each match.

(emphasis added)

You can use split to get the fields you need into an array and loop through that.

http://download.oracle.com/javase/1,5.0/docs/api/java/lang/String.html#split(java.lang.String)

I have not used java regex, but for many languages the answer is: No.

Capturing groups seem to be created when the regex is parsed, and filled when it matches the string. The expression (a)|(b)(c) has three capturing groups, only if either one, or two of them can be filled. (a)* has just one group, the parser leaves the last match in the group after matching.

Pattern p = Pattern.compile("ab(?:(c)|(d))*ef");
Matcher m = p.matcher("abcdef");
m.matches();

should do what you want.

EDIT:

@aioobe, I understand now. You want to be able to do something like the grammar

A    ::== <Foo> <Bars> <Baz>
Foo  ::== "foo"
Baz  ::== "baz"
Bars ::== <Bar> <Bars>
        | ε
Bar  ::== "A"
        | "B"

and pull out all the individual matches of Bar.

No, there is no way to do that using java.util.regex. You can recurse and use a regex on the match of Bars or use a parser generator like ANTLR and attach a side-effect to Bar.