用户输入以Java重复程序

我正在编写一个简单的猜谜游戏程序，用户将输入一个数字来尝试猜测随机生成的数字。

如果他们得到了正确的数字我想给他们再次玩的选项。

这是我的代码：

public class GuessingGame { private Random num = new Random(); private int answer = num.nextInt(10); private int guess; private String playAgain; public void inputGuess(){ System.out.println("Enter a number between 1 and 10 as your first guess: "); Scanner input = new Scanner(System.in); guess = input.nextInt(); do{ if (guess  10){ System.out.println("That is not a valid entry. Please try again: "); guess = input.nextInt(); }else if (guess > answer){ System.out.println("Too high, Try Again: "); guess = input.nextInt(); }else if (guess < answer){ System.out.println("Too low, Try Again: "); guess = input.nextInt(); } }while (guess != answer); System.out.println("Congratulations, You guessed the number!"); System.out.println("Would you like to play again? Enter Y to play or any other key to quit: "); playAgain = input.nextLine(); if(playAgain == "Y" || playAgain == "y"){ System.out.println("Enter a number between 1 and 10 as your first guess: "); guess = input.nextInt(); } } } 

游戏播放但是当用户被提示再次播放时没有任何反应？

有什么建议么？

Answer 1

这是完整的代码，完全工作和测试…不使用递归..一切都修复。

  public static void main(String[] args) { String playAgain = ""; Scanner scan = new Scanner(System.in); do { ClassName.inputGuess(); System.out.println("Would you like to play again? Enter Y to play or any other key to quit: "); playAgain = scan.nextLine(); } while(playAgain.equalsIgnoreCase("Y")); System.out.println("Thanks for playing!"); } public void inputGuess() { Random num = new Random(); int answer = num.nextInt(10)+1; Scanner input = new Scanner(System.in); int guess; System.out.println("Enter a number between 1 and 10 as your first guess: "); guess = input.nextInt(); do { if (guess < 1 || guess > 10) { System.out.println("That is not a valid entry. Please try again: "); guess = input.nextInt(); } else if (guess > answer) { System.out.println("Too high, Try Again: "); guess = input.nextInt(); } else if (guess < answer) { System.out.println("Too low, Try Again: "); guess = input.nextInt(); } input.nextLine(); } while (guess != answer); System.out.println("Congratulations, You guessed the number!"); } 

Answer 2

执行以下操作，您的代码将起作用：

1. 替换所有input.nextInt(); with Integer.parseInt(input.nextLine());
2. 用(playAgain.equalsIgnoreCase("Y"))替换(playAgain == "Y" || playAgain == "y") (playAgain.equalsIgnoreCase("Y"))
3. 在inputGuess()内部初始化answer 。
4. 用inputGuess();替换if(playAgain.equalIgnoreCase("Y"))的inputGuess();

当您通过控制台输入整数值时，它还包含\n （下一行）。 但是当你使用nextInt() ，它不会读取这个\n ，但是当你尝试使用input.nextLine()获取下一行时，它会查找已经存在于整数条目中的\n （下一行）之后什么都没有。 代码查找“Y”或“y”并中断，因为它没有找到任何一个。

这就是Integer.parseInt(input.nextLine()); 在这里工作

Answer 3

这是代码：

 private Random num = new Random(); private int answer = num.nextInt(10) +1; private int guess; private String playAgain; Scanner input = new Scanner(System.in); public void inputGuess(){ System.out.println("Enter a number between 1 and 10 as your first guess: "); guess = input.nextInt(); do{ if (guess < 1 || guess > 10){ System.out.println("That is not a valid entry. Please try again: "); guess = input.nextInt(); }else if (guess > answer){ System.out.println("Too high, Try Again: "); guess = input.nextInt(); }else if (guess < answer){ System.out.println("Too low, Try Again: "); guess = input.nextInt(); } if(guess == answer) { System.out.println("Congratulations, You guessed the number!"); System.out.println("Would you like to play again? Enter Y to play or any other key to quit: "); playAgain = input.nextLine(); } }while (!playAgain.equals("Y") && !playAgain.equals("y")); } 

你只需要在while内引入输赢逻辑，条件就是结束/继续标志。

在比较字符串以使用equals方法时，另一件事总是要记住，因为==将比较对象引用而不是String值，在某些情况下==将返回true以获得相等的字符串，因为JVM如何存储字符串，但是确保总是使用等于 。

Answer 4

尝试这样的事情：

 public void inputGuess(){ System.out.println("Enter a number between 1 and 10 as your first guess: "); Scanner input = new Scanner(System.in); guess = input.nextInt(); playAgain = "Y"; do{ if (guess < 1 || guess > 10){ System.out.println("That is not a valid entry. Please try again: "); guess = input.nextInt(); }else if (guess > answer){ System.out.println("Too high, Try Again: "); guess = input.nextInt(); }else if (guess < answer){ System.out.println("Too low, Try Again: "); guess = input.nextInt(); } if(guess == answer) { System.out.println("Congratulations, You guessed the number!"); System.out.println("Would you like to play again? Enter Y to play or N to quit: "); input.nextLine(); playAgain = input.next(); answer = num.nextInt(10); guess = -1; if(!playAgain.equalsIgnoreCase("N")) { System.out.println("Enter a number between 1 and 10 as your first guess: "); guess = input.nextInt(); } } }while (!playAgain.equalsIgnoreCase("N")); } 

您需要使用代码来检查是否要在循环内再次播放。 这样你就等到他们正确猜到了这个号码，然后问他们是否想再玩一次。 如果他们这样做，如果他们不退出循环，则重新启动该过程。

Answer 5

我上面看到的一些解决方案是不正确的。 随机数，你需要加1来得到1到10之间，你还需要与equals进行比较。 我在这里使用不区分大小写。

以下代码可以根据需要使用。

 import java.util.Random; import java.util.Scanner; public class Test2 { private static Random num = new Random(); private static int answer = 0; private static int guess; private static String playAgain; public static void main(String[] args) { inputGuess(); } // Guess Method. public static void inputGuess(){ // create answer. answer = 1+ num.nextInt(10); System.out.println("Enter a number between 1 and 10 as your first guess: "); Scanner input = new Scanner(System.in); guess = input.nextInt(); do{ if (guess < 1 || guess > 10){ System.out.println("That is not a valid entry. Please try again: "); guess = input.nextInt(); }else if (guess > answer){ System.out.println("Too high, Try Again: "); guess = input.nextInt(); }else if (guess < answer){ System.out.println("Too low, Try Again: "); guess = input.nextInt(); } }while (guess != answer); System.out.println("Congratulations, You guessed the number!"); System.out.println("Would you like to play again? Enter Y to play or any other key to quit: "); playAgain = input.nextLine(); if( playAgain.equalsIgnoreCase("Y") ){ inputGuess(); } } } 

Answer 6

到现在为止，你已经猜到了正确的方法。 以下是我将如何处理它

 public class Test { public static void main (String...a) { inputGuess(); } public static void inputGuess() { Scanner input = new Scanner(System.in); String playAgain = "Y"; int guess; Random ran = new Random(); int answer = ran.nextInt(10) + 1; while (playAgain.equalsIgnoreCase("Y")) { System.out.println("Enter a number between 1 and 10 as your first guess: " + answer); guess = input.nextInt(); do { if (guess < 1 || guess > 10) { System.out.println("That is not a valid entry. Please try again: "); guess = input.nextInt(); } else if (guess > answer) { System.out.println("Too high, Try Again: "); guess = input.nextInt(); } else if (guess < answer) { System.out.println("Too low, Try Again: "); guess = input.nextInt(); } } while (guess != answer); System.out.println("Congratulations, You guessed the number!"); System.out.println("Would you like to play again? Enter Y to play or any other key to quit: "); input.nextLine(); playAgain = input.nextLine(); answer = ran.nextInt(10) + 1 } } } 

Answer 7

此代码可以满足您的目的……

 import java.util.Random; import java.util.Scanner; public class GuessingGame { private Random num = new Random(); private int answer ; private int guess; private String playAgain; public void inputGuess() { answer = num.nextInt(11); System.out.println("Enter a number between 1 and 10 as your first guess: "); Scanner input = new Scanner(System.in); guess = input.nextInt(); do { if (guess < 1 || guess > 10) { System.out .println("That is not a valid entry. Please try again: "); guess = input.nextInt(); } else if (guess > answer) { System.out.println("Too high, Try Again: "); guess = input.nextInt(); } else if (guess < answer) { System.out.println("Too low, Try Again: "); guess = input.nextInt(); } } while (guess != answer); System.out.println("Congratulations, You guessed the number!"); System.out.println("Would you like to play again? Enter Y to play or any other key to quit: "); do { playAgain = input.nextLine(); }while(playAgain.length()<1); if (playAgain.trim().equalsIgnoreCase("y")) { inputGuess(); } else { System.out.println("Good Bye!!!"); } } public static void main(String[] args) { new GuessingGame().inputGuess(); } } 

Answer 8

一个问题 ：

不要用==来比较两个字符串的内容，你应该使用equals()

想象一下，在这种情况下，正确答案就是一个

请将此作为蓝图样本

  int answer = 0; String yes = ""; Scanner input = new Scanner(System.in); do{ do{ System.out.println("Enter your number"); answer = input.nextInt(); } while ( answer != 1); System.out.println("Would you like to play again?"); yes = input.next(); } while ( yes.equalsIgnoreCase("yes")); 

输出：