如何使用Java Scanner获取整数输入？无接触元素异常

Question

所以我一直在寻找如何使用扫描仪从用户那里获取输入。发生的事情是我在第一个扫描仪之后进入一个无限循环。我尝试将第二台扫描仪的名称改为read而不是in，但没有任何改变。我加入了。nextLine（）位于第一个scanner实例的末尾，然后我关闭以尝试解决问题，但这似乎不起作用。不太清楚为什么会这样。

`   private int compPlayInput(){
        int comps = -1; // Initialize 

        Scanner in = new Scanner(System.in); // Start Scanner

        //While the user input isn't between 0 and 2
        while ((comps<0) || (comps>2)){
            System.out.println("Turn: " + turnCount);
            System.out.print("How many computer controlled players will there be: ");
            //did the user input an integer?
            if (in.hasNextInt()){
                comps=in.nextInt();//change comps variable to the user's input
            }
            //The user hasn't entered an integer
            else {
                System.out.println("\n\n** ERROR: Enter an integer x that satisfies 0 <= x <= 2 **\n");
            }
        }
        in.nextLine(); //It seems like this is supposed to fix it, but it doesn't.
        in.close(); // close scanner
        return comps;
    }

    /**
     * Gets a player's guess
     * @return the player's guess (int) between 0 and 4 inclusively
     */
    private int getPlayerGuess(){
        int guess = -1; //initialize
        Scanner in = new Scanner(System.in); //start scanner

        //While the user doesn't input an int between 0 and 4
        while ((guess<0) || (guess>4)){
            System.out.println("Turn: " + turnCount);
            System.out.print("What is your guess: ");
            //If the user input an integer
            if (in.hasNextInt()){
                //make guess = to next int
                guess=in.nextInt();
            }
            else {
                System.out.println("\n\n** ERROR: Enter an integer x that satisfies 0 <= x <= 4 **\n");
            }
        }

        in.close(); // close scanner
        return guess; //return guess
    }`

这是eclipse中的输出：

    Turn: 0
How many computer controlled players will there be: 1
Turn: 1
What is your guess: 

** ERROR: Enter an integer x that satisfies 0 <= x <= 4 **

Turn: 1
What is your guess: 

** ERROR: Enter an integer x that satisfies 0 <= x <= 4 **
...

不允许用户像询问有多少电脑播放器时那样输入猜测。我不知道我应该如何解决这个问题。

我把这个放进去了。next（）转换到else语句中，但似乎出现了错误，因为扫描仪无法读取任何内容。我得到的新输出是

回合：0将有多少电脑控制的玩家：1回合：1你猜怎么着：

**错误：输入一个满足0的整数x

线程“main”java中出现异常。util。java中的NoSuchElementException。util。扫描仪。java上的throwFor（未知源）。util。扫描仪。HW2TaylorZacharyGame的下一个（未知来源）。HW2TaylorZacharyGame上的getPlayerGuess（HW2TaylorZacharyGame.java:114）。在HW2TaylorZachary转弯（HW2TaylorZacharyGame.java:52）。main（HW2TaylorZachary.java:15）

Answer 1

我找到了我的问题。似乎scanner不喜欢在单个类文件中被调用两次，所以我的解决方案是简单地创建一个新的Scanner对象，在其中我将为该对象放置我的私有变量。所以像这样：

public class HW2TaylorZacharyGame {

    private HW2TaylorZacharyPlayer player1 = new HW2TaylorZacharyPlayer();
    private HW2TaylorZacharyPlayer player2 = new HW2TaylorZacharyPlayer();
    private int winner = 0;
    private int turnCount = 0;
    Scanner in = new Scanner(System.in); // Start Scanner

对于一个让我很心痛的问题，非常简单的解决方案。不过，请注意，我还在从用户那里获取任何输入后放置了in.nextLine（）。这两件事，对于其他有相同问题的人来说，在我看来是使用Scanner的最佳方式。不过，我也从未关闭过Scanner，所以不幸的是，这有点缺点。

Answer 2

试着进来。next（）而不是in。nextLine（）；把它放在else语句中

Answer 3

您需要在您的else中使用非代码int。比如，

else {
   System.out.printf("ERROR: Enter an integer x that satisfies 0 <= x <= 2: %s "
       + "does not qualify%n", input.nextLine());

}