java如何修复这个随机数猜测游戏使用dowhile循环程序？

Question

创建一个程序，随机生成1-100之间的数字，并让用户猜一猜。如果用户输入的数字为“低”或“高”，则会显示一条消息告知用户。当用户猜到随机数时，告诉用户他尝试了多少次才得到那个数字。之后，询问用户是否希望再次执行此操作，如果用户确实使用生成的新随机数重复此过程

问题是，我似乎不知道如何让用户再次这样做，当我运行程序时，它似乎在代码中显示错误。如果有人能帮我解决这个问题，那就太好了。谢谢大家!

import java.util.Scanner;
import java.util.Random;
public class RandomGuess
{
    public static void main(String [] args)
    {
        Scanner keyboard = new Scanner(System.in);
        Random randy = new Random();

        //#declaring variables
        int num, count = 0;
        final int random = randy.nextInt(100);
        String input;
        char yn;

        //#random number
        System.out.println("Num = " + random);

        //#title or header
        System.out.println("Random Number Guessing Game");
        System.out.println("===========================");

        //#asking user for input
        do
        {
            System.out.print("Guess the random number " + 
                "from 1 to 100===> ");
            num = keyboard.nextInt();

            //#if the number the user entered
            //#was less than the random number
            if(num < random)
            {
                //#display this message
                System.out.println("Your guess is too low try again...");
                System.out.println();
            }
            //#if the number the user entered
            //#was less than the random number
            if(num > random)
            {
                //#display this message
                System.out.println("Your guess is too high try again...");
                System.out.println();
            }
            count++;
            if (num == random)
            {
                System.out.println("You guessed the random number in " + 
                    count + " guesses!");
                break;
            }
            do 
            {
                System.out.print("Continue? (Y or N)==> ");
                input = keyboard.nextLine();
                yn = input.charAt(0);
            }
            while(yn == 'Y' || yn == 'y');
        }
        while (num > 1 || num > 100);

    }
}

Answer 1

您的代码有几个问题，甚至没有看到显示的错误（我在这些区域中添加了注释）：

            count++;
            if (num == random)
            {
                System.out.println("You guessed the random number in " + 
                    count + " guesses!");
                break;
            }   // You should put an else here
            do 
            {
                System.out.print("Continue? (Y or N)==> ");
                input = keyboard.nextLine();
                yn = input.charAt(0);
            }
            while(yn == 'Y' || yn == 'y');  // This will keep asking if you want to try again so long as you enter a "y"
                     // But it won't actually let you try.
                     // Why? Because if you enter a y" it will loop back to the question.   
        }
        while (num > 1 || num > 100); // This should probably be (random != num)

    }
}

这是一个修订版

            count++;
            if (num == random) {
                System.out.println("You guessed the random number in " + 
                    count + " guesses!");
            } else {
                yn = 'x';    // can be anything other than y or n
                while(yn != 'y' && yn != 'n') {
                  System.out.print("Continue? (Y or N)==> ");
                  input = keyboard.nextLine();
                  yn = input.toLowerCase().charAt(0);
                }
            }
        }
        while (num != random && yn == 'y');
    }
}

希望这足以推动你前进

此外，请发布错误消息和/或描述它做错了什么，以及描述您实际希望它做什么

至于例外情况，问题在于扫描仪。nextInt不会使用您输入的号码末尾的换行符。因此，您的“继续Y/N”问题将获得上一行的剩余内容（即新行=>；空字符串）

你可以试试这个：

    num = -1;       // Initialise the number to enable the loop
    while (num <= 1 || num >= 100) {
      System.out.print("Guess the random number from 1 to 100===> ");
      String ans = keyboard.nextline();
      try {
        num = Integer.parseInt();     // Convert the string to an integer - if possible
      } catch (NumberFormatException e) {
           // If the user's input can not be converted to an integer, we will end up here and display an error message.
        System.out.println ("Please enter an integer");
      }
    }