Sort ArrayList<String> excluding the digits on the first half of the String

Fabio :

I'm trying to sort an ArrayList which a series of String composed as well (XX and YY are numbers):

Test: XX    Genere: Maschio    Eta: YY    Protocollo: A

I would link to sort them by only considering the YY value. I found this method but it considerers all the digits of the string and I cannot remove n digits before YY because I don't know from how many digits is composed XX:

Collections.sort(strings, new Comparator<String>() {
    public int compare(String o1, String o2) {
        return extractInt(o1) - extractInt(o2);
    }

    int extractInt(String s) {
        String num = s.replaceAll("\\D", "");
        // return 0 if no digits found
        return num.isEmpty() ? 0 : Integer.parseInt(num);
    }
});

Andrew Tobilko :

You also need to come up with how you'd like to sort the strings that don't have a number in the second part.

Collections.sort(strings, new Comparator<String>() {
  public int compare(String o1, String o2) {
    return Comparator.comparingInt(this::extractInt)
        .thenComparing(Comparator.naturalOrder())
        .compare(o1, o2);
  }

  private int extractInt(String s) {
    try {
      return Integer.parseInt(s.split(":")[1].trim());
    }
    catch (NumberFormatException exception) {
      // if the given String has no number in the second part,
      // I treat such Strings equally, and compare them naturally later
      return -1;
    }
  }
});

UPDATE

If you are sure Integer.parseInt(s.split(":")[1].trim()) never fails with an exception, Comparator.comparingInt(this::extractInt) would be enough, and you could go with a shorter comparator.

Collections.sort(strings, 
  Comparator.comparingInt(s -> Integer.parseInt(s.split(":")[1].trim())));