Question

好的，所以我编写了一个简单的石头剪刀游戏，使用用户输入进行游戏。用户键入他们的选择，随机函数选择计算机，然后生成结果。在此之后程序终止，因为它已经完成。我想使用while循环，这样当游戏结束时，它会再次启动，然后如果用户键入退出或退出，程序将停止，这可以通过简单地说playerGo != exit之类的东西轻松完成，玩游戏等等。但是，我不能让这个工作，请有人帮助我，我是一个Java菜鸟：）

import java.util.Scanner;   

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

    Scanner input = new Scanner(System.in);

    int compGoInt;
    String compGo;
    String playerGo;

    System.out.println("You can type 'Exit' to quit the game at any time.");
    System.out.print("Please enter your choice. Rock, Paper or Scissors: ");
    playerGo = input.nextLine();

    compGoInt = (int) (Math.random() * 3);

    switch (compGoInt){
    case 0:
        compGo = "Rock";
        break;
    case 1:
        compGo = "Paper";
        break;
    case 2:
        compGo = "Scissors";
        break;
    default:
        compGo = "Error";
        System.out.println("Error.");
        break;
    }

    if (playerGo.equals(compGo)){
        System.out.println("Computer chooses "+compGo);
        System.out.println("It's a draw!");
    }

    else if ((playerGo.equalsIgnoreCase("Rock") && compGo.equalsIgnoreCase("Scissors")) ||
            (playerGo.equalsIgnoreCase("Paper") && compGo.equalsIgnoreCase("Rock")) ||
            (playerGo.equalsIgnoreCase("Scissors") && compGo.equalsIgnoreCase("Paper"))){
        System.out.println("Computer chooses "+compGo);
        System.out.println("Player Wins!");
    }

    else if ((compGo.equalsIgnoreCase("Rock") && playerGo.equalsIgnoreCase("Scissors")) ||
            (compGo.equalsIgnoreCase("Paper") && playerGo.equalsIgnoreCase("Rock")) ||
            (compGo.equalsIgnoreCase("Scissors") && playerGo.equalsIgnoreCase("Paper"))){
        System.out.println("Computer chooses "+compGo);
        System.out.println("Computer Wins!");
    }

    else{
        System.out.println("Something has gone wrong!");
        System.out.println("Player chose "+playerGo);
        System.out.println("Computer chose "+compGo);
    }


}

}

Answer 1

简单地说：

while(true) {
     if(playerGo.equalsIgnoreCase("Exit")) break;
     else //GameLogic
}

但是，如果我这样说，你应该让用户选择一个数字，因为任何字母输入都容易出错。

为澄清目的：

import java.util.Scanner;   

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

        Scanner input = new Scanner(System.in);

        int compGoInt;
        String compGo;
        String playerGo;

        while(true) {
            System.out.println("You can type 'Exit' to quit the game at any time.");
            System.out.print("Please enter your choice. Rock, Paper or Scissors: ");
            playerGo = input.nextLine();

            if(playerGo.equalsIgnoreCase("Exit")) break;    //Checks for exit condition.
            else { //GameLogic

                compGoInt = (int) (Math.random() * 3);

                switch (compGoInt){
                case 0:
                    compGo = "Rock";
                    break;
                case 1:
                    compGo = "Paper";
                    break;
                case 2:
                    compGo = "Scissors";
                    break;
                default:
                    compGo = "Error";
                    System.out.println("Error.");
                    break;
                }

                if (playerGo.equals(compGo)){
                    System.out.println("Computer chooses "+compGo);
                    System.out.println("It's a draw!");
                }

                else if ((playerGo.equalsIgnoreCase("Rock") && compGo.equalsIgnoreCase("Scissors")) ||
                        (playerGo.equalsIgnoreCase("Paper") && compGo.equalsIgnoreCase("Rock")) ||
                        (playerGo.equalsIgnoreCase("Scissors") && compGo.equalsIgnoreCase("Paper"))){
                    System.out.println("Computer chooses "+compGo);
                    System.out.println("Player Wins!");
                }

                else if ((compGo.equalsIgnoreCase("Rock") && playerGo.equalsIgnoreCase("Scissors")) ||
                        (compGo.equalsIgnoreCase("Paper") && playerGo.equalsIgnoreCase("Rock")) ||
                        (compGo.equalsIgnoreCase("Scissors") && playerGo.equalsIgnoreCase("Paper"))){
                    System.out.println("Computer chooses "+compGo);
                    System.out.println("Computer Wins!");
                }

                else{
                    System.out.println("Something has gone wrong!");
                    System.out.println("Player chose "+playerGo);
                    System.out.println("Computer chose "+compGo);
                }
            }
        }
    }
}

Answer 2

您可以添加布尔变量IsGameRunning = true。将整个登录包装在一个while循环中，该循环检查IsGameRunning的状态。将一个大小写添加到检查用户输入“-1”的switch语句中，当用户输入“-1”作为输入时，大小写将IsGameRunning变量更改为false。

简单而且应该工作

Answer 3

boolean running = true;
do{
    System.out.println("You can type 'Exit' to quit the game at any time.");
    System.out.print("Please enter your choice. Rock, Paper or Scissors: ");
    playerGo = input.nextLine();
    running = !(playerGo.equalsIgnoreCase('Exit') || playerGo.equalsIgnoreCase('Quit'));
    if(running){
       //logic
       compGoInt = (int) (Math.random() * 3);
       ...
    }
}while(running)

同时 - ＆gt;输入需要至少询问一次。

和

if (playerGo.equals(compGo)){
     System.out.println("Computer chooses "+compGo);
     System.out.println("It's a draw!");
 }

playerGo.equals(compGo) - ＆gt; playerGo.equalsIgnoreCase(compGo)与代码的其余部分一样

添加while循环以使用户输入石头剪刀游戏重复，直到用户结束游戏

3 个答案: