CODEWITHSUNDEEP
javaregex
brain storm
brain storm

Reputation: 31242

regex matching pattern in java

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.

Upvotes: 2

Views: 71

Answers (2)

RP-
RP-

Reputation: 5837

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.

Upvotes: 2

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626691

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

Upvotes: 6

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