Java regex first match only

Question

How do I tell the following regex to only find the FIRST match? The following code keeps finding all possible regex within the string.

i.e. I'm looking only for the indices of the substring (200-800;50]

public static void main(String[] args) {

    String regex = "(\\[|\\().+(\\]|\\))";

    String testName=  "DCGRD_(200-800;50]MHZ_(PRE|PST)_(TESTMODE|REG_3FD)";

            Pattern pattern = 
            Pattern.compile(regex);

            Matcher matcher = 
            pattern.matcher(testName);

            boolean found = false;

            while (matcher.find()) {
                System.out.format("I found the text" +
                    " \"%s\" starting at " +
                    "index %d and ending at index %d.%n",
                    matcher.group(),
                    matcher.start(),
                    matcher.end());
                found = true;

            }

            if (!found){
                System.out.println("Sorry, no match!");
            }
}

The first match would be the first iteration of your loop. Just process one iteration and that's it. Or don't have a loop at all, and simply process the match if find returns true. — Edwin Dalorzo
– Edwin Dalorzo, Commented Sep 16, 2013 at 23:23
Your regex matches (200-800;50]MHZ_(PRE|PST)_(TESTMODE|REG_3FD). Are you in need of help with the regex? — user557597
– user557597, Commented Sep 16, 2013 at 23:56
I only needed the first occurrence. i.e (200-800;50]. Got it all squared away! :) — Mark Kennedy
– Mark Kennedy, Commented Sep 18, 2013 at 1:25

ATG · Accepted Answer · 2013-09-16 23:31:59Z

15

matcher.group(1) will return the first match.

If you mean lazy matching instead of eager matching, try adding a ? after the + in the regular expression.

Alternatively, you can consider using something more specific than .+ to match the content between the brackets. If you're only expecting letters, numbers and a few characters then maybe something like [-A-Z0-9;_.]+ would work better?

edited Sep 16, 2013 at 23:31

answered Sep 16, 2013 at 23:21

ATG

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3 Comments

Alan Moore Over a year ago

What matcher.group(1) returns is the contents of the first capturing group, (\[|\(). So it will be ( or [, and in this particular example it would always be ( because all three of the possible matches start with that. (group(2) will return ]` or ).) Anyway, the reluctant-or-more-specific advice seems to be what the OP was looking for, so +1.

Hykilpikonna Over a year ago

Wait... Shouldn't matcher.group(0) return the first match?

ATG Over a year ago

You might think that intuitively, but group 0 actually returns the whole matched expression.

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