Question

所以我已经使用Java几周了，想尝试制作我的第一个小游戏。在尝试添加再次播放功能之前，我所做的一切都工作正常。看起来好像只是跳过了底部的一小部分代码，我不知道为什么。

我尝试过移动代码的不同部分，但无法弄清楚为什么它不起作用。

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

    System.out.println("You have 10 tries to correctly guess a random number between 1 and 10. Good luck! \n"
            + "After you have entered a number, press ENTER to see if you won.");

    boolean play;
       play = true;

   do 
   {     
       int tries = 10, triesLeft = 0, triesUsed = 0;

       String YorN; 
    do
    {

        System.out.print("\nEnter A Number: ");

        int numInp = scan.nextInt();    
        int randomNumber = (int)(Math.random()*9) + 1;

        System.out.println("The number was " + randomNumber);

        if(numInp == randomNumber) {
            System.out.println("\nYou Win! Great job.");
            triesLeft = --tries;
            triesUsed = 10 - triesLeft;
            tries = -1;
        } else {
            System.out.println("Try Again...");
            tries --;
        }
    }
    while( tries > 0);

    String playAgain = (" Feel free to play again soon!");

    if(tries == 0) {
        System.out.println("\nOut of tries :(   You lose loser  " + playAgain);
    }
    if(tries == -1) {
    System.out.println("\nYou used " + triesUsed + " tries and had "
            + triesLeft + " tries left! :)" + playAgain);
    }

    System.out.print("\nDo you want to play again? Y or N?");

    YorN = scan.nextLine();

    if (YorN == ("Y") || YorN == ("y"))
    {
        play = true;
    }
    else if (YorN == ("N") || YorN == ("n"))
    {
        play = false;
    }
    else {}

   } while(play = true);

   if (play = false){
       System.out.println("\nGood Bye"); 
   }

}

当我运行代码时，该系统似乎没有执行任何操作：

YorN = scan.nextLine();

    if (YorN == ("Y") || YorN == ("y"))
    {
        play = true;
    }
    else if (YorN == ("N") || YorN == ("n"))
    {
        play = false;
    }
    else {}

这是我运行它时看到的内容：

您赢了！干得好。

您使用了5次尝试，还剩下5次尝试！ :)随时可以再次玩！

您想再次玩吗？是或否？输入数字：

它并没有给我输入y或n的位置。如果有人知道为什么我会非常感谢您的帮助

Answer 1

nextInt（）或next（）方法不读取enter。通常在编号之后您必须按Enter，但上述方法尚未准备好。一种为了避免这种情况，请避免进行1次额外的sc.nextLine（）方法调用。它将读取并丢弃回车。

scan.nextLine();
YorN = scan.nextLine();

    if (YorN == ("Y") || YorN == ("y"))
    {
        play = true;
    }
    else if (YorN == ("N") || YorN == ("n"))
    {
        play = false;
    }
    else {}

我不知道为什么我的代码中的“重播”部分不起作用

1 个答案: