Question

我的问题是如何将用户输入限制为Y / N或y / n（在Java中）。目前我正在使用equals（）并计划将它们更改为equalsignorecase（），这应该处理案例部分。但是，这并不能阻止用户输入其他字符（例如：H或h）。目前，当输入除y或n之外的字符时，程序直接进入＆＃34;感谢播放消息＆＃34;和游戏的结束。

我对编程比较陌生，所以请提供一些建议，最好是一个完整的例子。这对我来说真的很长。另外，如果你觉得这部分代码可以用更好的方式编写，我可以重写，但请再次提供一个完整的例子。

我意识到这个问题对于stackoverflow来说有点宽泛，但我真的可以使用更有经验的程序的洞察力。谢谢你的时间。

// creates instance of BufferedReader
// prompts user to play the game again
// places user input in a try
// if user wants to play again, call startGame()
// if user dosen't want to play again, keep asking anyways
private void showPlayAgainMessage() 
{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    System.out.println();
    System.out.println("Do you want to play again? (y/n)");

    try 
    {
        String playAgain = br.readLine();

        // Do you want to play again? Is y.
        if(playAgain.equals("y")) 
        {
            startGame();//else prompt another question with if else
        }

        // Do you want to play again? Is n.
        else if(playAgain.equals("n"))
        {
            System.out.println(); 
            System.out.println("Last chance.  Play again? (y/n)");
            playAgain = br.readLine(); 
                // Last chance.  Play again? Is y.
                if(playAgain.equals("y")) 
                {
                    startGame(); 
                }
                // Last chance.  Play again? Is n.
                else if(playAgain.equals("n")) 
                {
                    System.out.println(); 
                    System.out.println("How about Minesweeper? (y/n)");
                    playAgain = br.readLine();
                    // How about Minesweeper? Is y.
                        if(playAgain.equals("y")) 
                        {
                            System.out.println(); 
                            System.out.println("I really wish we had Minesweeper...");
                            System.out.println("Lots of Hangman though...Hangman? (y/n)");
                            playAgain = br.readLine();
                                // Lots of Hangman though...Hangman? Is y.
                                if(playAgain.equals("y")) 
                                {
                                    startGame(); 
                                }
                                // Lots of Hangman though...Hangman? Is n.
                                else if (playAgain.equals("n"))
                                {
                                    System.out.println();
                                    System.out.println("ok...");
                                }
                            }       
                        }
                    }
                }

Answer 1

这个问题看起来很烦人，但是如果必须的话，可以使用while循环和一个实现状态机的switch语句。

int state = 0;
while (state < 4) {

    switch (state) {

    case 0: System.out.println("Do you want to play again? (y/n)"); break;
    case 1: System.out.println("Last chance.  Play again? (y/n)"); break;
    case 2: System.out.println("How about Minesweeper? (y/n)"); break;
    case 3: System.out.println("I really wish we had Minesweeper...");
            System.out.println("Lots of Hangman though...Hangman? (y/n)");break;
    }

    String playAgain = br.readLine();

    if(playAgain.equals("y")) 
    {
        startGame();
        state = 0;
    }
    else if(playAgain.equals("n")) {
        state++;
    }

}
System.out.println("ok...");

每个＆＃34; n＆＃34;回答会让你进入下一个问题。非法输入将再次显示当前问题。

Answer 2

创建一个检查用户输入并在任何地方调用它的函数。

public boolean checkInput(String in){

 if(in.equalsIgnoreCase("y") || in.equalsIgnoreCase("n"))
   return true;
 else
   return false;

}

从您的代码库中调用此方法，例如

System.out.println();
System.out.println("Do you want to play again? (y/n)");

try 
{
    String playAgain = br.readLine();
    while(!checkInput){
        System.out.println("Please enter Valid Input Y/N");
        playAgain = br.readLine();
    }

    // Do you want to play again? Is y.
    if(playAgain.equals("y")) 
    {
        startGame();//else prompt another question with if else
    }

此处while循环中的代码不允许用户输入任何无效输入，如h，g等。它将提示用户输入无效，并且应输入有效选项。

Answer 3

你需要简单地这样做，即为y / n以外的字符添加其他部分。只需在else部分调用showPlayAgainMessage（）函数。

您可以为代码的其他必需部分执行此操作。

 private void showPlayAgainMessage() 
 {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    System.out.println();
    System.out.println("Do you want to play again? (y/n)");

    try 
    {
        if (playAgain.equals("y")) {
            startGame();
        } 
        else if (playAgain.equals("n")) {
            System.out.println(); 
            System.out.println("Last chance.  Play again? (y/n)");
            playAgain = br.readLine(); 

           if (playAgain.equals("y")) {
                System.out.println();
                System.out.println("I really wish we had Minesweeper...");
                System.out.println("Lots of Hangman though...Hangman? (y/n)");
                playAgain = br.readLine();
                // Lots of Hangman though...Hangman? Is y.
                if (playAgain.equals("y")) {
                    startGame();
                } // Lots of Hangman though...Hangman? Is n.
                else if (playAgain.equals("n")) {
                    System.out.println();
                    System.out.println("ok...");
                }
            }

        }
        else{
            System.out.println("Sorry, invalid input. y/n required.");
            showPlayAgainMessage();   // this will prompt the user again for y/n if any other  character is typed.
        }
     }catch(Exception e){

     }
 }

Java，BufferedReader，限制用户输入，y / n

3 个答案: