Java regex - overlapping matches

Question

In the following code:

public static void main(String[] args) {
    List<String> allMatches = new ArrayList<String>();
    Matcher m = Pattern.compile("\\d+\\D+\\d+").matcher("2abc3abc4abc5");
    while (m.find()) {
        allMatches.add(m.group());
    }

    String[] res = allMatches.toArray(new String[0]);
    System.out.println(Arrays.toString(res));
}

The result is:

[2abc3, 4abc5]

I'd like it to be

[2abc3, 3abc4, 4abc5]

How can it be achieved?

You would need to search starting at every index; use the find(int startingIndex) method and search starting at every character position. Of course, then you're likely to find too many matches... Assuming you want to start at every number, you might try combining an iteration over Matcher.find(String.indexOf(digits, index)) for all matching indices. — user1676075
– user1676075, Commented Jul 31, 2013 at 13:17
I suppose if it's single digits, you could back up from the match starting position and find from there for the next match. — user1676075
– user1676075, Commented Jul 31, 2013 at 13:20
For input "12abc13abc14abc15", do you want [12abc13, 2abc13, 13abc14, 3abc14, 14abc15, 4abc15] or [12abc13, 13abc14, 14abc15]? — johnchen902
– johnchen902, Commented Jul 31, 2013 at 13:22

johnchen902 · Accepted Answer · 2013-07-31 13:26:10Z

18

Make the matcher attempt to start its next scan from the latter \d+.

Matcher m = Pattern.compile("\\d+\\D+(\\d+)").matcher("2abc3abc4abc5");
if (m.find()) {
    do {
        allMatches.add(m.group());
    } while (m.find(m.start(1)));
}

answered Jul 31, 2013 at 13:26

johnchen902

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1 Comment

johnchen902 Over a year ago

To the first two up-voter: the ordinary version contains a bug that if nothing matches, an IllegalStateException will be thrown.

Community · Accepted Answer · 2017-05-23 12:02:38Z

16

Not sure if this is possible in Java, but in PCRE you could do the following:
(?=(\d+\D+\d+)).

Explanation
The technique is to use a matching group in a lookahead, and then "eat" one character to move forward.

(?= : start of positive lookahead
- ( : start matching group 1
  - \d+ : match a digit one or more times
  - \D+ : match a non-digit character one or more times
  - \d+ : match a digit one or more times
- ) : end of group 1
) : end of lookahead
. : match anything, this is to "move forward".

Online demo

Thanks to Casimir et Hippolyte it really seems to work in Java. You just need to add backslashes and display the first capturing group: (?=(\\d+\\D+\\d+)).. Tested on www.regexplanet.com:

enter image description here

edited May 23, 2017 at 12:02

CommunityBot

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answered Jul 31, 2013 at 13:24

HamZa

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9 Comments

HamZa Over a year ago

@anubhava It gives the right results for PCRE. That's what I stated anyways.

anubhava Over a year ago

Yes I just meant that in Java it didn't give expected results.

johnchen902 Over a year ago

Not really working. Use 12abc13abc14abc15 as input and the result is [12abc13, 2abc13, 13abc14, 3abc14, 14abc15, 4abc15] instead of [12abc13, 13abc14, 14abc15]. See my and OP's comments under the question.

HamZa Over a year ago

@johnchen902 overlapping matches ! That's what he wanted to begin with ... If that's not the case then (?=(\d+\D+\d+))\d+ would do the job

Casimir et Hippolyte Over a year ago

@johnchen902: right you must replace the pattern by (?=((?<!\\d)\\d+\\D+\\d+)) if you don't want overlapped results in your overlapped results. :)

|

bric3 · Accepted Answer · 2022-12-13 12:17:16Z

3

The above solution of HamZa works perfectly in Java. If you want to find a specific pattern in a text all you have to do is:

String regex = "\\d+\\D+\\d+";

String updatedRegex = "(?=(" + regex + ")).";

Where the regex is the pattern you are looking for and to be overlapping you need to surround it with (?=(" at the start and ")). at the end.

edited Dec 13, 2022 at 12:17

bric3

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answered Oct 1, 2017 at 23:25

Fluch

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