regex matching pattern in java

Question

I am struggling to understand a simple regex. I googled around. somehow it is not striking me.

Here is the method:

public static void testMethod(){
        String line = "This order was placed for QT3000! OK?";
        String pattern = "(.*)(\\d+)(.*)";

        // Create a Pattern object
        Pattern r = Pattern.compile(pattern);

        // Now create matcher object.
        Matcher m = r.matcher(line);
        if (m.find( )) {
           System.out.println("Found value: " + m.group(0) );
           System.out.println("Found value: " + m.group(1) );
           System.out.println("Found value: " + m.group(2) );
           System.out.println("Found value: " + m.group(3) );
        }
    }

Here is the output:

I was expecting group(2) to print 3000. but why it prints only 0.

Wiktor Stribiżew · Accepted Answer · 2015-09-11 23:14:54Z

6

Group 2 captured text only contains 0 because of the first greedy .*. It matched up to the last digit, and let \d+ have only the last digit. See demo of your regex.

To fix it, use lazy dot matching:

(.*?)(\d+)(.*)
   ^

See another demo

answered Sep 11, 2015 at 23:14

Wiktor Stribiżew

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1 Comment

Wiktor Stribiżew Over a year ago

See Watch Out for The Greediness! and The Greedy Trap for a comprehensive description of how backtracking works with greedy and lazy quantifiers. I just tried to shortly explain the root of the problem, and 1 possible way of fixing it. Surely, depending on what you need, you might use a negated character class (e.g. [^0-9]*), or tempered greedy token.

RP- · Accepted Answer · 2015-09-11 23:15:30Z

2

You need ([^0-9.]*)(\\d+)(.*).

The first group matching everything until last zero as you have + in second group. You need to escape numbers from the first group.

answered Sep 11, 2015 at 23:15

RP-

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