在我的while循环中遇到这个开关的问题

所以我试图复制我以前玩过的这个游戏。在这个游戏中，你会看到一个数字，并且有一个隐藏的数字。您需要猜测这个隐藏的号码是小号，大号还是与显示的号码相同。我遇到问题需要输入才能工作。我看不出让switch语句起作用。我也遇到了扫描仪问题。虽然它在while循环的外部起作用，但在它内部却不起作用。在我的while循环中遇到这个开关的问题

import java.util.Scanner; 
import java.util.Random; 
public class Jamey { 

    /** 
    * @param args 
    */ 
    public static void main(String[] args) { 

      //This will give us the first random shown number 
     Random yourRandom = new Random(); 
     int y = 1+yourRandom.nextInt(10); 

     //Here is the introduction text 
     System.out.println("Welcome to the guessing game!"); 
     System.out.println("The objective of this game is simple."); 
     System.out.println("You will be shown one of two numbers which range between one and ten."); 
     System.out.println("You have to gues if the number shown is larger, smaller, or equal to the hidden number."); 
     System.out.println("If you believe the number you see is larger enter 1."); 
     System.out.println("If you believe the number you see is smaller enter the 3."); 
     System.out.println("If you believe the number you see is the same enter the 2."); 
     System.out.println("Good luck, your number is "+y+"."); 

     boolean isDone = false; 
     while(isDone=false){    

      //This allows the user to guess 
      Scanner guess = new Scanner(System.in); 
      int g = guess.nextInt(); 

      //This will help us to keep score later. 
      int score = 0; 

      //This will give us the new random number 
      Random newRandom = new Random(); 
      int n = 1+newRandom.nextInt(10); 

      //This will give us the new hidden number 
      Random hiddenRandom = new Random(); 
      int r = 1+hiddenRandom.nextInt(10); 

      //This is to allow multiple different inputs 
      switch(score){ 
      case 1 : 
       score +=1; 
       if(y>r){ 
        System.out.println("Correct"); 
        System.out.println("Your new number is "+n+"."); 
       } 
       else{ 
        score +=1; 
        System.out.println("Inccorect, your overall score was "+score+"."); 
        isDone = true; 
       } 
       break; 

      case 2 : 
       score +=1; 
       if(y==r){ 
        System.out.println("Correct"); 
        System.out.println("Your new number is "+n+"."); 
       } 
       else{ 
        System.out.println("Inccorect, your overall score was "+score+"."); 
        isDone = true; 
       } 
       break; 

      case 3 : 
       score +=1; 
       if(y<r){ 
        System.out.println("Correct"); 
        System.out.println("Your new number is "+n+"."); 
       } 
       else{ 
        System.out.println("Inccorect, your overall score was "+score+"."); 
        isDone = true; 
       } 
       break; 
      default: 
       System.out.println("Invalid input."); 
       isDone = true; 
       break; 

      }  
     } 
    } 
}

来源

2014-01-16 user3203973

你while语句使用分配

while(isDone=false) // Incorrect 
while (isDone == false) OR while(!isDone) // Better

通知的==，而不是仅仅=。你没有做比较，每次迭代只设置isDone为假。

您的switch语句失败，因为您正在检查您的score变量而不是猜测变量。

switch(score) // Incorrect 
switch(g) // Better

此外，要创建大量Random的对象时，你只需要创建一个单一实例，然后相应地命名每个变量。例如

Random rand = new Random(); 
int yourRandom = 1 + rand.nextInt(10); 
int newRandom = 1 + rand.nextInt(10); 
int hiddenRandom = 1 + rand.nextInt(10);

来源

2014-01-16 19:36:40 csmckelvey

而（！isDone）看起来更好 – Divers

当有人写，如果（二== true），还是东西我从来没有见过像这样，每个人都写如果（b）。有时候新手会这样做。 – Divers

感谢您的帮助。我不敢相信这很简单。 – user3203973

while(isDone=false){ 应该 while(isDone==false){ 甚至更好 while(!isDone){

来源

2014-01-16 19:37:02 ren78min

谢谢，最小的东西总是让我。 – user3203973

现在我所得到的是从我的默认语句获取“无效输入”响应。对此有何帮助？ – user3203973

在我的while循环中遇到这个开关的问题

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