Question

我试过在java上做一个健壮的代码，但它似乎不起作用。我正在寻找的是用户输入输入，如果输入不是必需的输入，程序将检查输入，然后用户将被选择重新输入适当的输入，直到输入匹配所需的输入，或者干脆退出这是我到目前为止所拥有的。当我运行此代码时，一切正常，除非用户输入错误的输入并想要退出。 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);

Answer 1

问题是，当您阅读新输入时，您不会更新布尔变量的值;你甚至不把它读成同一个变量。

所以：

favSeason = in.next();

应该是：

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

但请注意，这种检查有效输入的方式是糟糕。这是非常低效的（在每个检查中拉出子串），但也非常脆弱（您必须正确地获得这些索引），并且当您的需求发生变化时，您必须在多个位置更新代码。

您正在将字符串映射到整数，因此请使用Map：

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

然后您的isSeason变量变为：

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

你的条件消失了，变成了：

seasonScore = seasonScores.get(favSeasonInput);

如何使用用户输入使该稳健性工作？

1 个答案: