Java 如何使用用户输入使健壮性发挥作用？

我曾尝试在java上编写一个健壮的代码，但它似乎不起作用。我要找的是用户输入一个输入，如果它不是必需的输入，程序将检查输入，然后给用户一个选项重新输入适当的输入，直到输入与必需的输入匹配，或者干脆退出。这是我到目前为止所拥有的。当我运行这段代码时，除了用户输入错误并想要退出外，其他一切都正常工作。while循环保持运行，即使用户重新输入正确的输入或退出，也不会停止。我怎样才能做到这一点

    //question
    System.out.println("Summer, Winter, Fall, or Spring");
    System.out.print("Which season is your favarite? ");
    String favSeason = in.next();
    System.out.println();

    //Control the inputs by converting them to Upper Case
    String favSeasonInput = favSeason.toUpperCase();

    //required answers of the question
    String seasons = "SUMMER, WINTER, FALL, SPRING?";
    String quit = "QUIT!";

    boolean isSeasons = (favSeasonInput.equals(seasons.substring(0, 6)) ||
            favSeasonInput.equals(seasons.substring(8, 14)) ||
            favSeasonInput.equals(seasons.substring(16, 20)) ||
            favSeasonInput.equals(seasons.substring(22, 28)));
    boolean isQuit = favSeasonInput.equals(quit.substring(0, 4));
    //inialize variables that will compute scores
    int favSeasonScore = 0;

    //if user enters an input otherthan seasons
    while (!isSeasons){

        favSeason = in.next();

        if(isQuit){
            System.exit(0); 
        }

    }


    //Conditions to set up scores for seasons
    if(favSeasonInput.equals(seasons.substring(0, 6))){
        favSeasonScore = 6;
        System.out.println("Summer is " + favSeasonScore + " points");
    }
    else if(favSeasonInput.equals(seasons.substring(8, 14))){
        favSeasonScore = 14;
        System.out.println("Winter is " + favSeasonScore + " points");
    }
    else if(favSeasonInput.equals(seasons.substring(16, 20))){
        favSeasonScore = 20;
        System.out.println("Fall is " + favSeasonScore + " points");
    }
    else if(favSeasonInput.equals(seasons.substring(22, 28))){
        favSeasonScore = 28;
        System.out.println("Spring is " + favSeasonScore + " points");
    }

    System.out.println(favSeasonScore);

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问题是，当您读取新输入时，您没有更新布尔变量的值；你甚至没有把它读入同一个变量

因此：

应该是：

favSeasonInput = in.next().toUpperCase();
isSeaons = ...;
isQuit = ...;

---

但请注意，这种检查有效输入的方法非常糟糕。它效率很低（每次检查时都会拉出子字符串），但也非常脆弱（必须正确获取这些索引），并且当需求发生变化时，必须在多个位置更新代码

您正在将字符串映射为整数，因此请使用映射：

Map<String, Integer> seasonScores = new HashMap<>();
seasonScores.put("SPRING", 28);
// Etc.

你的条件句消失了，变成：

seasonScore = seasonScores.get(favSeasonInput);

---

请勿将代码作为图像粘贴到此处请将代码粘贴到此处，并解释您收到的错误您的现有代码有什么问题？

isSeason = seasonScores.keySet().containsKey(favSeasonInput);

seasonScore = seasonScores.get(favSeasonInput);