Java .nextLine() 重复行

Question

我的猜谜游戏一切正常，但是当涉及到询问用户是否想再玩的部分时，它会重复这个问题两次。但是我发现，如果我将输入法从 nextLine() 更改为 next()，它不会重复问题。这是为什么？

这里是输入和输出:

I'm guessing a number between 1-10
What is your guess? 5
You were wrong. It was 3
Do you want to play again? (Y/N) Do you want to play again? (Y/N) n

这是代码:(它是用Java编写的)最后一个 do while 循环 block 是询问用户是否想再次玩的部分。

import java.util.Scanner;

public class GuessingGame 
{   
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        boolean keepPlaying = true;

        System.out.println("Welcome to the Guessing Game!");

        while (keepPlaying) {
            boolean validInput = true;
            int guess, number;
            String answer;

            number = (int) (Math.random() * 10) + 1;
            System.out.println("I'm guessing a number between 1-10");
            System.out.print("What is your guess? ");
            do {
                validInput = true;
                guess = input.nextInt();
                if (guess < 1 || guess > 10) {
                    validInput = false;
                    System.out.print("That is not a valid input, " +
                            "guess again: ");
                }
            } while(!validInput);
            if (guess == number)
                System.out.println("You guessed correct!");
            if (guess != number)
                System.out.println("You were wrong. It was " + number);
            do {
                validInput = true;
                System.out.print("Do you want to play again? (Y/N) ");
                answer = input.nextLine();
                if (answer.equalsIgnoreCase("y"))
                    keepPlaying = true;
                else if (answer.equalsIgnoreCase("n"))
                    keepPlaying = false;
                else
                    validInput = false;
            } while (!validInput);
        }
    }
}

Answer 1

在您的 do while 循环中，您不需要 nextLine()，您只需要 next()。

所以改变这个:

answer = input.nextLine();

为此:

answer = input.next();

请注意，正如其他人所建议的那样，您可以将其转换为 while 循环。这样做的原因是，do while 循环用于需要至少执行一次循环，但不知道需要多久执行一次的情况。虽然在这种情况下它肯定是可行的，但这样的事情就足够了:

System.out.println("Do you want to play again? (Y/N) ");
answer = input.next();
while (!answer.equalsIgnoreCase("y") && !answer.equalsIgnoreCase("n")) {
    System.out.println("That is not valid input. Please enter again");
    answer = input.next();
}

if (answer.equalsIgnoreCase("n"))
    keepPlaying = false;

只要未输入“y”或“n”(忽略大小写)，while 循环就会一直循环。只要它是，循环就结束了。 if 条件会在必要时更改 keepPlaying 值，否则什么也不会发生，您的外部 while 循环将再次执行(从而重新启动程序)。

编辑:这解释了为什么您的原始代码不起作用

我应该补充一点，您的原始语句不起作用的原因是您的第一个 do while 循环。在其中，您使用:

guess = input.nextInt();

这会读取行外的数字，但不会返回行，这意味着当您使用:

answer = input.nextLine();

它会立即从 nextInt() 语句中检测到剩余的回车符。如果你不想使用我的解决方案，只阅读 next()，你可以通过这样做吞下剩下的:

guess = input.nextInt();
input.nextLine();
rest of code as normal...