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I've *kinda* found a **very** limited, not so elegant, far from ideal "solution", using [`replaceAll`](https://docs.oracle.com/javase/8/docs/api/java/lang/String.html#replaceAll-java.lang.String-java.lang.String-):

```java
String regex = "(?:([aeiou]+)[0-9]+|([123]+)[a-z]+)\\W+";
System.out.println("ae123.".replaceAll(regex, "$1$2"));
System.out.println("111abc!!".replaceAll(regex, "$1$2"));
```

This prints:

> ae<br>
> 111

The trick is in the second argument: `"$1$2"` means that I'm concatenating groups 1 (`$1`) and 2 (`$2`). Because of the alternation (`|`), only one of the groups is captured and the other will be empty. And when I concatenate them, the result is always the contents of the captured group.

---
### Limitations

But as I said, this solution is very limited. Let's suppose the regex is a little bit more complicated with lots of different groups. Something like that:

```none
(1) | (2) (3) (4) | (5) (6) | (7) | (8)
```

In that case, I can have only group 1, **or** only groups 2, 3 and 4, **or** only groups 5 and 6, **or** only group 7, **or** only group 8.

Of course I could still use `replaceAll` with `"$1$2$3$4$5$6$7$8"`, but if the regex matches groups 2, 3 and 4, they will be concatenated and I wouldn't know each group's value individually. Unless I use some separator, such as `"$1,$2,$3,$4,$5,$6,$7,$8"` and then `split` the result, but that would be too "ugly" IMO (not to mention that the separator itself can't be part of the group's value, etc).

**If** Java supported *branch reset*, the groups numbering would be like this:

```none
(?| (1) | (1) (2) (3) | (1) (2) | (1) | (1) )
```

And I'd just need to loop through them, always starting with 1, until `m.groupCount()`.

*Which means I'm still waiting for better solutions* 😉