java - 如何创建检查 int 范围、数字类型而不是 char 的异常？

Question

我正在尝试解决异常问题，但我遇到的问题是我需要创建一个程序，要求用户输入数字 9-99。必须使用 3 种不同的异常对该数字进行错误检查。

e1: number is outside of the range (200)

e2: number is of a data type other than integer (double)

e3: input is another data type other than number (char)

我尝试在 if 结构中创建模式以使所有三个都起作用，但是我无法让它区分 e2 和 e3。它将始终默认为 e2。这就是我所拥有的，只有两个异常(exception)，但我非常感谢您帮助我弄清楚如何实现第三个异常(exception)。谢谢。

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    boolean tryAgain = true;
  do {
       try {
            System.out.println("Please enter an integer between 9 and 99: ");

            int inInt = input.nextInt();

            if (inInt >= 9 && inInt <= 99){
                System.out.println("Thank you.  Initialization completed.");
                tryAgain = false;
            }
            else if (inInt < 9 || inInt > 99){
                throw new NumberFormatException("Integer is out of range.");
            }
        }
        catch (NumberFormatException e1) { // Range check
            System.out.println("* The number you entered is not between 9 and 99.  Try again.");
            System.out.println();
            input.nextLine();
        }
       catch (InputMismatchException e2) { // Something other than a number
            System.out.println("* You did not enter an integer.  Try again.");
            System.out.println();
            input.nextLine();
        }
    } while(tryAgain);
}

}

这是我现在得到的输出:

Please enter an integer between 9 and 99: 2

• The number you entered is not between 9 and 99. Try again.

Please enter an integer between 9 and 99: f

• You did not enter an integer. Try again.

Please enter an integer between 9 and 99: 88

Thank you. Initialization completed.

https://www.ideone.com/ZiOpGH

Answer 1

在catch (InputMismatchException e2)中，测试是否input.hasNextDouble()(或input.hasNextFloat()？不确定)哪一个更普遍......)是正确的。如果是，那么您可以区分“用户输入 double ”和“用户输入非数字类型”的情况