Question

我似乎无法让我的程序询问用户是否想再玩一次。该程序在询问是否愿意继续时结束。我添加了一个休息时间，因为在程序不断循环正确的答案和尝试次数之前。

import java.util.Random;
import java.util.Scanner;

public class GuessTheNumber {
      public static void main(String[] args) {

              boolean play = true;
              Scanner input = new Scanner(System.in);
              Random Number = new Random();
              int GuessNumber = 1+ Number.nextInt(1000);
              int guess = 0;
              int Tries = 0;

              while(play) {
                      while(guess != GuessNumber) {
                              System.out.println("Please Enter a number between 1 and 1000:");
                              guess = input.nextInt();
                              Tries++;

                              if(guess == GuessNumber) {
                                      System.out.println("You Win!");
                                      break;
                              }
                              else if(guess < GuessNumber) {
                                      System.out.println("Guess is too low");
                              }
                              else if(guess > GuessNumber) {
                                      System.out.println("Guess is too high");
                              }
                      }
                      System.out.printf("Number was: %d",GuessNumber);
                      System.out.println("");
                      System.out.printf("Number of tries was: %d ",Tries);
                      System.out.println("");
                      System.out.println("Would you like to play again?(Yes/No)");
                      String playagain = input.nextLine();
                      if("Yes".equals(playagain))
                              play = true;
                      else
                              play = false;
              }
      }
}

Answer 1

您的计划的一些内容。

正如评论所指出的，您需要在使用nextInt()后刷新换行符。

但是，在用户声明他们想要再次播放之后，你还必须设置猜测值和尝试计数回到0 - 否则程序将继续“你赢了！”环。

我建议将变量声明移动到外部循环：

while(play) { 
  int guess = 0;
  int guessNumber = 1 + Number.nextInt(1000);
  int tries = 0;
  ...
}

Play Again功能无效

1 个答案: