CODEWITHSUNDEEP
javaregex
user1653240
user1653240

Reputation: 291

Java Regex group 0

Can someone please help me to explain the following code? Thanks. I am a little bit confused about the regex grouping.

public static String evaluate(String s) {
    if (s == null) return "";

    Matcher m = Pattern.compile("[0-9]*").matcher(s);

    if (m.find()) {
        MatchResult mr = m.toMatchResult();
        return mr.group(0);
    }
    return "";
}

Upvotes: 12

Views: 17257

Answers (2)

ollo
ollo

Reputation: 25350

Hope this makes the group 0 more clear:

Example:

    String str = "start123456end"; // Your input String


    //                                 Group#1 Group#2
    //                                   |      |  
    Pattern p = Pattern.compile("start([0-9]*)(end)");
    //                           |<--- Group#0 --->|


    Matcher m = p.matcher(str); // Create a matcher for regex and input

    while( m.find() ) // As long as your regex matches something
    {
        System.out.println("group#0:\t" + m.group()); // Or: m.group(0)
        System.out.println("group#1:\t" + m.group(1));
        System.out.println("group#2:\t" + m.group(2));
    }

Output:

group#0:    start123456end
group#1:    123456
group#2:    end

You can "store" some parts of your regex into groups. in my example you have 3 of them (groups are between (and )):

  • Group 1: numbers between start and end words.
  • Group 2: the end word only
  • Group 0: thats the whole thing that matches your pattern - group 0 is reserved and will always return the whole match, while all others are optional and defined by you.

According to your code:

Example:

Matcher m = Pattern.compile("[0-9]*").matcher("123456end"); // Matches all numbers

if( m.find() )
{
    System.out.println(m.group(0)); // m.group() possible too
}

There only one group: 0!

Output: 123456 (= group 0)

now lets put some more groups into the pattern:

Code:

    //                            Group#1      Group#2
    //                              |            |
    Matcher m = Pattern.compile("([0-9])[0-9]([0-9])*").matcher(str); // Matches all numbers
    //                           |<---- Group#0 ---->|

    if( m.find() )
    {
        System.out.println("group#0:\t" + m.group(0)); // m.group() possible too
        System.out.println("group#1:\t" + m.group(1)); // 1st digit
        System.out.println("group#2:\t" + m.group(2)); // 3rd digit
    }

There are two more groups now.

Output:

group#0:    123456
group#1:    1
group#2:    6

I recommend you this documentation: Lesson: Regular Expressions. Realy start from first chapter and try examples by your own.

Additional:

Upvotes: 25

NPE
NPE

Reputation: 500327

From the documentation:

Group zero denotes the entire pattern, so the expression m.group(0) is equivalent to m.group().

In other words, mr.group(0) is the entire match.

Upvotes: 13

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