为什么我的 While 循环不会像我的 if else 语句那样循环我的 Cat？

Question

我正在尝试制作一个程序，用户将在其中输入随机整数/数字。但主要问题是，如果用户不小心输入了一个字符串，我希望他们得到通知，并且我希望他们通过再次询问来再次要求输入。但在我的代码中，我无法循环捕获。

public static void main(String[] args) {

    Scanner scan = new Scanner(System.in);

    int answer = 28;
    int attempts = 0;
    
    System.out.println("Guess a Number 1 - 50:");
    
    while(true) {
      try {
        int input = scan.nextInt();
        scan.nextLine();
        if(input > answer) {
          System.out.println("Too Big");
          System.out.println("Guess a Number 1 - 50:");
        } else if (input < answer) {
          System.out.println("Too Small");
          System.out.println("Guess a Number 1 - 50:");
        } else if (input == answer ) {
          System.out.println("Congrats!");
        } 
      } catch (InputMismatchException e) {
        System.out.println("Numbers only");
        System.out.println("Guess a Number 1 - 50:");
      } 
    }
}

Answer 1

从 try 块中删除 scan.nextLine();，然后在 catch 块中放置 scan.skip(".*");。这将跳过无效输入（.* 基本上匹配任何字符）。此外，您可以用简单的else 替换最后一个else if，因为这是唯一的可能性。话虽如此，请尝试以下操作：

public static void main(String[] args) throws IOException {
    Scanner scan = new Scanner(System.in);

    int answer = 28;
    int attempts = 0;

    System.out.println("Guess a Number 1 - 50:");

    while(true) {
        try {
            int input = scan.nextInt();
            if(input > answer) {
                System.out.println("Too Big");
                System.out.println("Guess a Number 1 - 50:");
            } else if (input < answer) {
                System.out.println("Too Small");
                System.out.println("Guess a Number 1 - 50:");
            } else {
                System.out.println("Congrats!");
            }
        } catch (InputMismatchException e) {
            System.out.println("Numbers only");
            System.out.println("Guess a Number 1 - 50:");
            scan.skip(".*");
        }
    }
}

Answer 2

你问一个原因

此代码在遇到无效输入时会无限循环，因为nextInt() 不接受无效令牌，因此无效令牌会留在扫描仪中，因为它不会因异常而被消耗，不断导致抛出异常下次您尝试读取 int 时，因为由于异常而永远无法到达 nextLine。

这可以通过将nextLine() 放在 catch 块中来解决任何导致引发异常的输入，清除输入流并允许用户再次输入。

    public static void main(String[] args) {

    Scanner scan = new Scanner(System.in);

    int answer = 28;
    int attempts = 0;
    
    System.out.println("Guess a Number 1 - 50:");
    
    while(true) {
      try {          
        int input = scan.nextInt();
        scan.nextLine();  // never reached when exception is thrown
        if(input > answer) {
          System.out.println("Too Big");
          System.out.println("Guess a Number 1 - 50:");
        } else if (input < answer) {
          System.out.println("Too Small");
          System.out.println("Guess a Number 1 - 50:");
        } else if (input == answer ) {
          System.out.println("Congrats!");
        } 
      } catch (InputMismatchException e) {
        scan.nextLine(); // This skips the invalid token 
        System.out.println("Numbers only");
        System.out.println("GuessNumber 1 - 50:");
      } 
    }
}

Answer 3

你应该试试这个代码，它会完美地满足你的要求......

     public static void main(String[] args) {
        int answer = 28;
        System.out.println("Guess a Number 1 - 50:");
        boolean condition = true;
        while (condition) {
            try {
                    Scanner scan = new Scanner(System.in);
                    int input = scan.nextInt();
                    scan.nextLine();
                    if (input > answer) {
                        System.out.println("Too Big");
                        System.out.println("Guess a Number 1 - 50:");
                    } else if (input < answer) {
                        System.out.println("Too Small");
                        System.out.println("Guess a Number 1 - 50:");
                    } else {
                        System.out.println("Congrats!");
                        condition=false;
                    }
            } catch (Exception e) {
                    System.out.println("Numbers only");
                    System.out.println("Guess a Number 1 - 50:");
            }
        }
    }