java - 如果 Scanner.nextInt() 无效，则显示 else block 中的错误输入

Question

我知道有很多代码，但在为扫描仪定义速度后，重点主要集中在 block 中的空白 else 空间内。如果用户没有输入整数，我希望 else 打印代码循环，并说“再试一次!输入无效。” ETC;但我不知道该去哪里。我听说过“catch”和“Exception”，但是有人可以向我解释一下，或者为我发布一个更新的代码块并进行解释。谢谢!

import java.util.Scanner;

public class M2A1
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        String Birthday;
        int Bday = 2;
        int speed;
        int penalty;
        System.out.println("You were pulled by a police officer believing you were speeding in a 55mph zone.");
        System.out.println("You hand over your license and registration. Is it possibly your birthday today?");
        System.out.println("Use yes(y) or no(n) to answer the above!");
        Birthday = input.nextLine();
        if (Birthday.equalsIgnoreCase("yes") || Birthday.equalsIgnoreCase("y"))
        {
            System.out.println("Great, your bad driving may have happened on a good day.");
            Bday = 1;
        }
        else if (Birthday.equalsIgnoreCase("n") || Birthday.equalsIgnoreCase("no"))
        {
            System.out.println("That's ok. Maybe you drove under the speed limit.");
            Bday = 0;
        }
        else 
        {
            while (!(Birthday.equalsIgnoreCase("y") || Birthday.equalsIgnoreCase("yes") || Birthday.equalsIgnoreCase("n") || Birthday.equalsIgnoreCase("no"))) {
                System.out.println("Invalid response. Try again!");
                System.out.println("Use yes(y) or no(n) to answer the above!");
                Birthday = input.nextLine();
                if (Birthday.equalsIgnoreCase("y") || Birthday.equalsIgnoreCase("yes"))
                {
                    System.out.println("Great, your bad driving may have happened on a good day.");
                }
                else if (Birthday.equalsIgnoreCase("n") || Birthday.equalsIgnoreCase("no"))
                {
                    System.out.println("That's ok.");
                }
            }
        }
        System.out.println("How fast were you going?");
        speed = input.nextInt();

        if (Bday == 1) {
            if (speed <= 65)
            {
                penalty = 0;
                System.out.println("Great job driving, and the officer wishes you a happy birthday! You drove at " +speed+ "mph! You lost " +penalty+ " points from your license.");
            }
            else if (speed > 65 && speed <= 85)
            {
                penalty = 1;
                System.out.println("You are driving a little fast, but the officer wishes you a happy birthday! You drove at " +speed+ "mph! You lost " +penalty+ " points from your license.");
            }
            else if (speed > 85)
            {
                penalty = 2;
                System.out.println("You are driving way too fast, even for your birthday. You drove at " +speed+ "mph! You lost " +penalty+ " points from your license.");
            }
            else
            {

            }
        }
        if (Bday == 0) {
            if (speed <= 60)
            {
                penalty = 0;
                System.out.println("Great job driving! You drove at " +speed+ "mph! You lost " +penalty+ " points from your license.");
            }
            else if (speed > 60 && speed <= 80)
            {
                penalty = 1;
                System.out.println("You are driving a little fast! You drove at " +speed+ "mph! You lost " +penalty+ " points from your license.");
            }
            else if (speed > 80)
            {
                penalty = 2;
                System.out.println("You are driving WAY too fast! You drove at " +speed+ "mph! You lost " +penalty+ " points from your license.");
            }
            else
            {

            }
        }
    }
}

Answer 1

扫描仪实际上有一个 hasNextInt() 方法，这可能就是您正在寻找的方法。

在您的示例中使用此方法的示例可能类似于:

while(!input.hasNextInt()){
    System.out.println("Try again! Invalid Input");
    input.nextLine(); //This clears the buffer and will allow the user to enter again.
}
speed = input.nextInt();