java - Java .nextLine() repeats line

Question

Everything of my guessing game is alright, but when it gets to the part of asking the user if he/she wants to play again, it repeats the question twice. However I found out that if I change the input method from nextLine() to next(), it doesn't repeat the question. Why is that?

Here is the input and output:

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

Here is the code:(It is in Java) The last do while loop block is the part where it asks the user if he/she wants to play again.

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);
        }
    }
}

score 7 · Accepted Answer

在你的do while循环中，你不想要nextLine()，你只想要next().

所以改变这个：

answer = input.nextLine();

对此：

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当您需要至少执行一次循环时使用循环，但您不知道需要执行它的频率。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;

while只要没有输入“y”或“n”（忽略大小写），循环就会一直循环。一旦它是，循环结束。如果需要，if条件会更改 keepPlaying 值，否则不会发生任何事情，并且您的外部while循环会再次执行（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

我应该补充一下，您原来的陈述不起作用的原因是因为您的第一个do while循环。在其中，您使用：

guess = input.nextInt();

这会读取线下的数字，但不会读取线的返回值，这意味着当您使用时：

answer = input.nextLine();

nextInt()它立即从语句中检测到剩余的托架。如果你不想使用我的阅读解决方案，next()你可以这样做吞下剩下的：

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

score 5 · Accepted Answer

问题确实在于完全不同的代码段。在guess = input.nextInt();执行前一个循环时，它会在输入中留下一个换行符。然后，当answer = input.nextLine();在第二个循环中执行时，已经有一个换行符等待读取并返回一个空字符串，该字符串激活最终else并被validInput = false;执行，以重复循环（和问题）。

input.nextLine();一种解决方案是在第二个循环之前添加一个。另一种是读取guess，nextLine()然后将其解析为 int。但这会使事情复杂化，因为输入不可能是正确的int。再想一想，代码已经提出了这个问题。尝试输入非数字响应。所以，定义一个函数

public static int safeParseInt(String str) {
    int result;
    try {  
        result= Integer.parseInt(str) ;
    } catch(NumberFormatException ex) {  
      result= -1 ;  
    }  
    return result ;  
}

然后将您的第一个循环替换为：

do {
    validInput= true ;
    int guess= safeParseInt( input.nextLine() ) ;
    if( guess < 1 || guess > 10 ) {
        validInput= false ;
        System.out.print("That is not a valid input,  guess again: ");
    }
} while( !validInput );

do-whilePS：我认为循环没有任何问题。它们是语言的一部分，语法清楚地表明条件在主体至少执行一次后进行评估。我们不需要仅仅因为其他人不知道而删除语言中有用的部分（至少从实践中）。相反：如果我们确实使用它们，它们会变得更广为人知！

score 4 · Accepted Answer

    validInput = false;

    do {

        System.out.print("Do you want to play again? (Y/N) ");
        answer = input.next();

        if(answer.equalsIgnoreCase("y")){

            keepPlaying = true;
            validInput = true;

        } else if(answer.equalsIgnoreCase("n")) {

            keepPlaying = false;
            validInput = true;

        }        

    } while(!validInput);

我改变了编码风格，因为我发现这种方式更具可读性。

score 2 · Accepted Answer

您的问题是，nextInt一旦 int 结束，它将停止，但将换行符留在输入缓冲区中。为了让您的代码正确读取答案，您必须在猜测的同一行输入它，例如5SpaceYReturn.

为了使其行为超出预期，如果第一个nextLine结果仅包含空格，则忽略它，并nextLine在这种情况下再次调用而不打印消息。

score 1 · Accepted Answer

我相信的输出input.nextLine()将包括行尾的换行符，而input.next()不会（但Scanner将保持在同一行）。这意味着输出永远不等于"y"or "n"。尝试修剪结果：

answer = input.nextLine().trim();

java - Java .nextLine() repeats line

5 回答 5

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