学习Java - 制作一个简单的猜谜游戏

Question

我一直在尝试自己学习Java大约一周，而且我决定尝试将我的（相当有限的）知识应用于制作一个简单的猜谜游戏。

基本上，用户输入一个介于1到10之间的数字，并根据他们是否猜测得太高，太低或正确猜测随机数字而得到反馈。

我的问题是，一旦我运行我的方法，我不知道如何允许多个猜测。基本上他们猜测一次，然后就是这样。游戏结束。

我敢肯定它就像我需要实现的循环一样简单，如果他们猜错了，可能会声明一个额外的变量？我不知道。

任何人都可以帮我吗？

这里是我的代码：

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

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

    //creates a new scanner object 
    Scanner input = new Scanner(System.in); 

    // randomly generates a number from 1-10 
    Random rand = new Random(); 
    int myNumber = rand.nextInt(10)+1; 


    String tooWarm = "Too high. Try again!"; 
    String tooCold = "Too cold. Try again!"; 
    String bingo = "Yup! Good guess!"; 
    boolean playing = true; 

    System.out.printf("What's your name? "); 
    String name = input.nextLine();  
    System.out.println("Well, " + name + ", betcha can't guess what number I'm thinking of! \nHint: it's between 1 and 10..."); 
    int value = input.nextInt(); 


     if (value > myNumber) { 
      System.out.println(tooWarm); 
     } 

     else if (value < myNumber) { 
      System.out.println(tooCold); 
     } 

     else { 
      System.out.println(bingo); 
     }   

} 
} 

Answer 1

只是这样做，

do..while让你执行，以ATLEAST执行一次。

do { 

     System.out.println("Enter : "); 
     int value = input.nextInt(); 

     if (value > myNumber) { 
      System.out.println(tooWarm); 
     } 

     else if (value < myNumber) { 
      System.out.println(tooCold); 
     } 

     else { 
      System.out.println(bingo); 
      playing = false; 
     } 
    } 
    while (playing); 

Answer 2

添加到您的代码的末尾：

... 
char yN = input.nextChar(); 
if (yN == 'y' || yN == 'Y') { 
    main(null); 
    return; 
    } 
}//end of main(). 

基本上它的作用是，它使得递归调用，使游戏可以一遍又一遍播放。 :-)

Answer 3

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

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

    //creates a new scanner object 
    Scanner input = new Scanner(System.in); 

    // randomly generates a number from 1-10 
    Random rand = new Random(); 
    int myNumber = rand.nextInt(10)+1; 


    String tooWarm = "Too high. Try again!"; 
    String tooCold = "Too cold. Try again!"; 
    String bingo = "Yup! Good guess!"; 
    boolean playing = true; 

    System.out.printf("What's your name? "); 
    String name = input.nextLine();  
    System.out.println("Well, " + name + ", betcha can't guess what number I'm thinking of! \nHint: it's between 1 and 10..."); 
    while(true){ 
     int value = input.nextInt(); 


     if (value > myNumber) { 
      System.out.println(tooWarm); 
     } 

     else if (value < myNumber) { 
      System.out.println(tooCold); 
     } 

     else { 
      System.out.println(bingo); 
      return; 
     }   

    } 
}