How to group in regex

Question

I have this input string(oid) : 1.2.3.4.5.66.77.88.99.10.52

I want group each number into 3 to like this

Group 1 : 1.2.3

Group 2 : 4.5.66

Group 3 : 77.88.99

Group 4 : 10.52

It should be very dynamic depending on the input. If it has 30 numbers meaning it will return 10 groups.

I have tested using this regex : (\d+.\d+.\d+)

But the result is this

Match 1: 1.2.3

Subgroups:

1: 1.2.3

Match 2: 4.5.66

Subgroups:

1: 4.5.66

Match 3: 77.88.99

Subgroups:

1: 77.88.99

Where as still missed one more matches.

Can anyone help me to provide the Regex. Thank you

Answer 1

If you want to match up to three digits, you should try:

((?:\d+\.?){1,3})

The {1,3} part matches 1-3 of the preceding item (which is one or more digits followed by a literal .. Note that the dot is escaped so that it doesn't match any character.

Edit

Further explanation: The (?: ) part is a grouping that cannot be used for backreferences (tends to be faster), see section 4.3 here for more information. You could, of course, also just use ((\d+\.?){1,3}) if you prefer. For more information on {1,3}, see here under "Limiting Repetition".

Edit (2)

Fixed error pointed out by dtmunir. An alternative way that is a bit more explicit (and doesn't catch the extra "." at the end of the early groups) is:

((?:\d+\.){0,2}\d+)

Answer 2

Al that will not capture the 52. But this one in fact will:

((?:\d+\.?){1,3})

The only change is adding the question mark after the . This allows it to accept the last number without having a period after it

Explanation (EDIT):

The \d+ as you can imagine captures consecutive digits.

The \. captures a period

The \.? captures a period, but allows the inner group to not require a period at the end

The (?:\d+\.?) defines "one group" which in your case you want to be 3 numbers.

The {1,3} sets the limits. It requires a minimum of 1 inner group and at most 3 inner groups. These groups may or may not end with a period.

Answer 3

This is my weird code for do this without regex :-)

public static String[] getTokens(String s) {

 String[] splitted = s.split("\\.");

 //Personally I hate Double.valueOf but I don't know how to avoid it
 String[] result = new String[Double.valueOf(Math.ceil(Double.valueOf(splitted.length) / 3)).intValue()];

 for (int i = 0, j = 0; j < splitted.length; i++, j+=3) {

  //Weird concat
  result[i] = splitted[j] + ( j+1 < splitted.length ? "." + splitted[j+1] : "" ) + ( j+2 < splitted.length ? "." + splitted[j+2] : "" );

 }
 return result;
}

Answer 4

\d+(?:\.\d+){0,2}

This is basically the same as Al's final regex - ((?:\d+\.){0,2}\d+) - but I think it's clearer this way. And there's no need to put parentheses around the whole regex. Assuming you're using Matcher.find() to get the matches, you can use group() or group(0) instead of group(1) to retrieve the matched text.