Simple java Regex doesn't

Question

When i use this code, i don't have the expected result :

    pattern = Pattern.compile("create\\stable\\s(\\w*)\\s\\(", Pattern.CASE_INSENSITIVE);
    matcher = pattern.matcher("create table CONTACT (");
    if(matcher.matches()) {
        for(int i =0; i<matcher.groupCount();i++) {
            System.out.println("table : " + matcher.group(i) + matcher.start(i) + " - " + matcher.end(i));
        }
    }
}

I expect to catch CONTACT but the regex catch the whole expression "create table CONTACT (". Has someone an idea of the problem ?

Thanks

http://regex101.com/r/lS5tT3/66 it does capture `contact` in group1 — vks, Sep 27 '14 at 14:39
Thank you so much guys. I though that if regex matched groupCount will return 2 (the whole expression and the group captured). — Sylvain LGenka, Sep 27 '14 at 14:41

score 1 · Accepted Answer · answered Sep 27 '14 at 14:42

The regex engine actually counts the entire regex as a group. The first group in your regex is actually the second group returned by the match, which is at index 1.

If you ignore the first group, then you should find what you're looking for in the second.

The reason that the group isn't printed by your code is that groupCount doesn't count the entire regex as a group, so you're only getting 1 group in your loop.

Group zero denotes the entire pattern by convention. It is not included in this count.

You probably don't need a loop, and you can just extract the desired string directly with group(1).

score 0 · Answer 2 · answered Sep 27 '14 at 14:41

0

Group number starts from 1, not from 0.

Following expression:

matcher.group(i)

should be replaced with:

matcher.group(i+1)

Or simply print group 1 if you want print only one group:

System.out.println("table: " + matcher.group(1));

answered Sep 27 '14 at 14:41

falsetru

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Simple java Regex doesn't

2 Answers2