剪刀石头布Java [英] Rock, Paper, Scissors Game Java

问题描述

我是编程的新手，我正尝试用Java编写一个非常简单的Rock，Paper，Scissors游戏.它将编译并运行良好，但是我想说些类似无效的举动.再试一次".或当用户(personPlay)未输入正确的字符(r，p或s)时沿这些行显示的内容.这样做的最佳方法是什么?例如，如果输入"q"，则应打印"Invalid move".提前非常感谢您！

I'm new to programming and I'm trying to write a very simple Rock, Paper, Scissors game in Java. It will compile and run fine, but I am looking to say something like "Invalid move. Try again." or something along those lines for when the user (personPlay) does not enter a correct character (r, p, or s). What would be the best way to do so? For example, if you enter a "q", it should print "Invalid move." Thank you so much in advance!

// *************
// Rock.java 
// ************* 

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


public class Rock 
{ 
public static void main(String[] args) 
{ 
    String personPlay; //User's play -- "R", "P", or "S" 
    String computerPlay = ""; //Computer's play -- "R", "P", or "S" 
    int computerInt; //Randomly generated number used to determine 
                     //computer's play 
    String response; 


    Scanner scan = new Scanner(System.in); 
    Random generator = new Random(); 

    System.out.println("Hey, let's play Rock, Paper, Scissors!\n" + 
                       "Please enter a move.\n" + "Rock = R, Paper" + 
                       "= P, and Scissors = S.");

    System.out.println();

    //Generate computer's play (0,1,2) 
    computerInt = generator.nextInt(3)+1; 

    //Translate computer's randomly generated play to 
    //string using if //statements 

    if (computerInt == 1) 
       computerPlay = "R"; 
    else if (computerInt == 2) 
       computerPlay = "P"; 
    else if (computerInt == 3) 
       computerPlay = "S"; 


    //Get player's play from input-- note that this is 
    // stored as a string 
    System.out.println("Enter your play: "); 
    personPlay = scan.next();

    //Make player's play uppercase for ease of comparison 
    personPlay = personPlay.toUpperCase(); 

    //Print computer's play 
    System.out.println("Computer play is: " + computerPlay); 


    //See who won. Use nested ifs 

    if (personPlay.equals(computerPlay)) 
       System.out.println("It's a tie!"); 
    else if (personPlay.equals("R")) 
       if (computerPlay.equals("S")) 
          System.out.println("Rock crushes scissors. You win!!");
    else if (computerPlay.equals("P")) 
            System.out.println("Paper eats rock. You lose!!"); 
    else if (personPlay.equals("P")) 
       if (computerPlay.equals("S")) 
       System.out.println("Scissor cuts paper. You lose!!"); 
    else if (computerPlay.equals("R")) 
            System.out.println("Paper eats rock. You win!!"); 
    else if (personPlay.equals("S")) 
         if (computerPlay.equals("P")) 
         System.out.println("Scissor cuts paper. You win!!"); 
    else if (computerPlay.equals("R")) 
            System.out.println("Rock breaks scissors. You lose!!"); 
    else 
         System.out.println("Invalid user input."); 
}

}

推荐答案

我建议制作Rock，Paper和Scissors对象.这些对象既可以翻译成字符串也可以翻译成字符串，也可以知道"什么胜过什么. Java枚举是完美的选择.

I would recommend making Rock, Paper and Scissors objects. The objects would have the logic of both translating to/from Strings and also "knowing" what beats what. The Java enum is perfect for this.

public enum Type{

  ROCK, PAPER, SCISSOR;

  public static Type parseType(String value){
     //if /else logic here to return either ROCK, PAPER or SCISSOR

     //if value is not either, you can return null
  }
}

如果String不是有效的类型，则parseType方法可以返回null.并且您的代码可以检查该值是否为null，如果是，则打印无效的重试"并循环返回以重新读取扫描仪.

The parseType method can return null if the String is not a valid type. And you code can check if the value is null and if so, print "invalid try again" and loop back to re-read the Scanner.

Type person=null;

 while(person==null){
      System.out.println("Enter your play: "); 
      person= Type.parseType(scan.next());
      if(person ==null){
         System.out.println("invalid try again");
      }
 }

此外，您的类型枚举可以通过让每个Type对象知道来确定什么胜过了什么

Furthermore, your type enum can determine what beats what by having each Type object know:

public enum Type{

    //...

    //each type will implement this method differently
    public abstract boolean beats(Type other);


}

每种类型将以不同的方式实现此方法，以了解什么胜过什么:

each type will implement this method differently to see what beats what:

ROCK{

   @Override
   public boolean beats(Type other){            
        return other ==  SCISSOR;

   }
}

 ...

然后在您的代码中

 Type person, computer;
   if (person.equals(computer)) 
   System.out.println("It's a tie!");
  }else if(person.beats(computer)){
     System.out.println(person+ " beats " + computer + "You win!!"); 
  }else{
     System.out.println(computer + " beats " + person+ "You lose!!");
  }

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