如何管理 Wordle Game (JavaFx) 中的重复字母

You have to take the number of occurrences into account. So, process the exact matches first, recording the letters not found so far. For each other occurrence in the guess, remove the letter from the recorded letters, so that you will not match it again. E.g.

ArrayList<Character> missing = new ArrayList<>();
for(int i = 0; i < 5; i++) {
    if(secret.charAt(i) == prop.charAt(i)) {
        status[i] = LetterStatus.IN;
    }
    else missing.add(secret.charAt(i));
}
if(!missing.isEmpty()) {
  for(int i = 0; i < 5; i++) {
      if(secret.charAt(i) != prop.charAt(i)) {
        Object ch = prop.charAt(i); // ensure to use remove(Object) not remove(int)
        status[i] = missing.remove(ch)? LetterStatus.OK: LetterStatus.NOTIN;
      }
  }
}

For large data, you’d use something like a Map<Character,Integer> to record the number of occurrences for each element, but since we have at most five elements here, using an ArrayList is simpler while providing reasonable performance. It’s just important to keep in mind that this approach would not scale to other problems.