In Java with regular expressions, how to capture numbers from a string with unknown length?

Question

My regular expression looks like this: "[a-zA-Z]+[ \t]*(?:,[ \t]*(\\d+)[ \t]*)*"

I can match the lines with this, but I don't know how to capture the numbers,I think it has to do something with grouping.

For example: from the string "asd , 5 ,2,6 ,8", how to capture the numbers 5 2 6 and 8?

A few more examples:

sdfs6df -> no capture

fdg4dfg, 5 -> capture 5

fhhh3      ,     6,8    , 7 -> capture 6 8 and 7

asdasd1,4,2,7 -> capture 4 2 and 7

So I can continue my work with these numbers. Thanks in advance.

Just use `"\\d+"` with `Matcher#find()` in a loop. See [How to extract numbers from a string and get an array of ints?](https://stackoverflow.com/questions/2367381/how-to-extract-numbers-from-a-string-and-get-an-array-of-ints) — Wiktor Stribiżew, Apr 25 '20 at 21:22
@remmaks Do you mean like this? `(?:\w+|\G(?!^))\h*,\h*([0-9]+)` https://regex101.com/r/n14Shg/1 — The fourth bird, Apr 25 '20 at 21:55
@WiktorStribiżew I think the OP was looking for a more specific way to get the matches. Do you agree if I reopen it and post it? — The fourth bird, Apr 25 '20 at 22:06
@remmaks Does [this solution](https://stackoverflow.com/a/2367418/3832970) work for you? Does it produce the result you want with all your test cases? — Wiktor Stribiżew, Apr 25 '20 at 22:08
@WiktorStribiżew No, that is not enough for me unfortunately. But maybe my description was bad, but thanks anyway! Thefourbird's solution is what i was searching for. — remmaks, Apr 25 '20 at 22:18
[Your comment](https://stackoverflow.com/questions/61432785/in-java-with-regular-expressions-how-to-capture-numbers-from-a-string-with-unkn?noredirect=1#comment108673162_61432785) was misleading, please remove it. @Thefourthbird Please post. — Wiktor Stribiżew, Apr 25 '20 at 22:19
So yes, the problem was that the first word in the string can contain numbers too, and i do not want to capture those. Sorry for bad description. — remmaks, Apr 25 '20 at 22:20

score 1 · Accepted Answer · answered Apr 25 '20 at 22:23

You could match the leading word characters and make use of the \G anchor capturing the continuous digits after the comma.

Pattern

(?:\w+|\G(?!^))\h*,\h*([0-9]+)

Explanation

(?: Non capture group
\w+ Match 1+ word chars -| or
- \G(?!^) Assert postition at the end of previous match, not at the start
) Close non capturing group
\h*,\h* Match a comma between horizontal whitespace chars
([0-9]+) Capture group 1, match 1+ digits

Regex demo | Java demo

In Java with double escaped backslashes:

String regex = "(?:\\w+|\\G(?!^))\\h*,\\h*([0-9]+)";

Example code

String regex = "(?:\\w+|\\G(?!^))\\h*,\\h*([0-9]+)";
String string = "sdfs6df -> no capture\n\n"
     + "fdg4dfg, 5 -> capture 5\n\n"
     + "fhhh3      ,     6,8    , 7 -> capture 6 8 and 7\n\n"
     + "asdasd1,4,2,7 -> capture 4 2 and 7";

Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);

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

Output

In Java with regular expressions, how to capture numbers from a string with unknown length?

1 Answers1