Question

Java Code:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegExpTest {
    public static void main(String[] args) {

        String str = "X-Value = -0.525108, Y-Value = 7.746691, Z-Value = 5.863008, Timestamp(milliseconds) = 23001";
        String p = "Value = (.*?), ";
        Pattern pattern = Pattern.compile(p);
        Matcher matcher = pattern.matcher(str);
        if (matcher.find()){
             System.out.println(matcher.group(1));
             System.out.println(matcher.group(2));
             System.out.println(matcher.group(3));
        }
    }
}

Java code’s output:

$ java RegExpTest 
-0.525108
Exception in thread "main" java.lang.IndexOutOfBoundsException: No group 2
        at java.util.regex.Matcher.group(Matcher.java:487)
        at RegExpTest.main(RegExpTest.java:15)
$ 

Python code (in Interpreter):

>>> import re
>>> re.findall("Value = (.*?), ", 'X-Value = -0.525108, Y-Value = 7.746691, Z-Value = 5.863008, Timestamp(milliseconds) = 23001;')
['-0.525108', '7.746691', '5.863008']
>>> 

So, why is Java unable to match all the occurrences of the match?

Answer 1

It’s because a group for a java match is a capturing bracket.

Your regex only has one set of non-escaped (ie capturing) brackets, the (.*?).

Group 1 contains the value that gets matched.

There is no group 2 because there is no second set of brackets in your regex.

In the java example, you want to loop through all matches, and print matcher.group(1).

while ( matcher.find() ) {
    System.out.println(matcher.group(1));
}

Note the while, which will loop through all matches and tell you group 1 from each.