java regex - capture repeated groups

Question

I'm using Java's regex library. I want to validate a string against the following format:

31,5,46,7,86(...)

The amount of numbers is not known. I want to make sure there is at least one number in that string, and that every two numbers are separated by a comma. I also want to get the numbers from the string.

(Note: this is just a simplified example, string.split will not solve my actual problem)

I wrote the following regex:

({[0-9]++)((?:,[0-9]++)*+)

The validation part works. However, when I try to extract the numbers, I get 2 groups:

Group1: 31
Group2: ,5,46,7,86

regex101 version: https://regex101.com/r/xJ5oQ6/3

Is there a way I can get each number separately? i.e. to end up with the collection:

[31, 5, 46, 7, 86]

Thanks in advance.

can you get it recursively?

morels
– morels

2016-06-10 14:09:04 +00:00
Commented Jun 10, 2016 at 14:09 — morels
– morels, Commented Jun 10, 2016 at 14:09

Community · Accepted Answer · 2017-05-23 10:34:01Z

Java does not allow you to access the individual matches of a repeated capturing group. For more information look at this question: Regular Expression - Capturing all repeating groups

The code provided by Tim Pietzcker can help you as well. If you rework it a bit and add a special case for the first number you can use something like this:

String target = "31,5,46,7,86";

Pattern compileFirst = Pattern.compile("(?<number>[0-9]+)(,([0-9])+)*");
Pattern compileFollowing = Pattern.compile(",(?<number>[0-9]+)");

Matcher matcherFirst = compileFirst.matcher(target);
Matcher matcherFollowing = compileFollowing.matcher(target);

System.out.println("matches: " + matcherFirst.matches());
System.out.println("first: " + matcherFirst.group("number"));

int start = 0;
while (matcherFollowing.find(start)) {
    String group = matcherFollowing.group("number");

    System.out.println("following: " + start + " - " + group);
    start = matcherFollowing.end();
}

This outputs:

matches: true
first: 31
following: 0 - 5
following: 4 - 46
following: 7 - 7
following: 9 - 86

Andreas Louv · Accepted Answer · 2016-06-10 14:46:55Z

0

This might work for you:

/(?=[0-9,]+$)((?<=,|^)[0-9]{1,2})(?=,|$)/g

The above captures one or two digits followed by , or end of input.

Note that I used the global modifier.

Try it online

edited Jun 10, 2016 at 14:46

answered Jun 10, 2016 at 14:22

Andreas Louv

47.3k14 gold badges109 silver badges126 bronze badges

3 Comments

Roy Millo Over a year ago

Now the captures are working, but the validation isn't: I tried to add a letter in the middle and instead of 'no match' i got this. Your suggestion is the closest I've got to what I wanted though, any ideas how to tweak it?

Andreas Louv Over a year ago

@RoyMillo I updated the regex, but what you want is not possible to put in one PCRE regex, since dynamic length lookbehinds are not supported. I don't know what regex flavor Java is using.

Andreas Louv Over a year ago

Your validation regex could however look something like this: /^[0-9,]+$/

Collectives™ on Stack Overflow

java regex - capture repeated groups

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related