Java regex program not working as expected

Question

On running this program,

import java.util.regex.*;

public class PatternExample {
    public static final String numbers = "1\n22\n333\n4444\n55555\n666666\n7777777\n88888888\n999999999\n0000000000";

    public static void main(String[] args) {
        Pattern pattern = Pattern.compile("9.*", Pattern.MULTILINE);
        Matcher matcher = pattern.matcher(numbers);
        while (matcher.find()) {
            System.out.print("Start index: " + matcher.start());
            System.out.print(" End index: " + matcher.end() + " ");
            System.out.println(matcher.group());
        }
    }
}

Output is

Start index: 44 End index: 53 999999999

I expected the output include the zeros, because of Pattern.MULTILINE. What should I do to include the zeros?

assylias · Accepted Answer · 2015-05-28 15:53:05Z

3

You need to add the DOTALL flag (and don't need the MULTILINE flag which only applies to the behaviour of ^ and $):

Pattern pattern = Pattern.compile("9.*", Pattern.DOTALL);

This is stated in the javadoc

The regular expression . matches any character except a line terminator unless the DOTALL flag is specified.

answered May 28, 2015 at 15:53

assylias

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karthik manchala · Accepted Answer · 2015-05-28 15:53:20Z

2

You are looking for Pattern.DOTALL, use the following:

Pattern pattern = Pattern.compile("9.*", Pattern.DOTALL);

Also Pattern.MULTILINE is not necessary here since you are not using any start ^ and end $ anchors.

answered May 28, 2015 at 15:53

karthik manchala

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RealSkeptic · Accepted Answer · 2015-05-28 15:53:35Z

2

You need to add the flag Pattern.DOTALL. By default, . does not match line terminators.

answered May 28, 2015 at 15:53

RealSkeptic

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