group counting issue with regex

Question

EDIT: The RE in the original code is irrelevant (or that it makes any sense). Let's say you're matching (X)|(Y): two patterns that are combined in the RE with an OR. How to know which pattern was actually matched??

I'm trying to extract just the text that's mattching the RE within the brackets.

The problem i'm facing is that i cannot figure out which actual group was matched, since the group index is not constant because of the OR.

Ie in the line marked XXX m.group() returns the entire pattern

pat1
abcdef2

And m.group(1) produces

pat
null

And m.group(2) produces

null
de

And m.groupCount() is just a total number of groups in the RE, so it is useless for indexing.

What I want the loop to print is

pat
de

It's a shame java doesnt have the perl's m/ operator that'll extract the stuff and put it nicely into an array ;-)

import java.io.IOException;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {

    public static void main(String[] args) throws IOException {

        Pattern p = Pattern.compile("([pat]+)1|abc([de]+)f2");
        String original = "  pat1 abcdef2555";

        Matcher m = p.matcher(original);
        boolean result = m.find();

        while (result) {
            System.out.println(m.group());        // XXX want to print only matched GROUP!!
            result = m.find();
        }

    }

}

I'm not sure I understand. Is it possible that you want your regex to be "(pat)1|pat(ter)n2"? The way you have it, I believe, would return any variation of the letters P A and T. I.e. [pat]+ would match "ppppt", "atpaptpap", and so on. Maybe I'm wrong on this. I'm kind of new to regex. — bozdoz
– bozdoz, Commented Nov 16, 2011 at 19:27
If your strings were jjjpatssjpttattpppt1 and ksjdhfabcdddddddddf2 what would be your desired output? As your expression is currently, the matches would be successful and the text would be pttattpppt and ddddddddd respectfully. I ask because like others here, I think you may have a poor understanding of what your expression is specifying and possibly regex syntax in general — Code Jockey
– Code Jockey, Commented Nov 16, 2011 at 20:57

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

I don't see that Java does support this directly.

(Perl does have a beautiful syntax for this, there you could use (?|([pat]+)1|abc([de]+)f2) and both groups would have the same number, what obviously is no problem because there can be only the one or the other.)

So to give you an answer to

What I want the loop to print is

pat
de

You have to check by your own if the group is available or not, so this loop would give you the desired output:

while (result) {
    if (m.group(1)!= null) System.out.println(m.group(1));
    if (m.group(2)!= null) System.out.println(m.group(2)); 
    result = m.find();
}

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Nov 18, 2011 at 9:36

stema

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2 Comments

Alan Moore Over a year ago

+1, but I like to use if (m.start(1) !- -1) instead, because it doesn't create a new String object each time you call it.

Saideira Over a year ago

alas, seems like i have no choice but to do post-processing after RE ;-)

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