如何创建检查int范围，数字类型而不是char的异常？ [英] How can I create an exception that checks for int range, number type, and not a char?

问题描述

我正试图绕过异常，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个不同的例外对此数字进行错误检查。

I'm trying to wrap my head around exceptions and the problem I'm running into is that I'm being required to create a program that asks for user input for a number 9-99. This number must be error-checked using 3 different exceptions.

e1：数字超出范围（200）

e1: number is outside of the range (200)

e2：number是整数以外的数据类型（double）

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

e3：输入是数字以外的其他数据类型（ char）

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

我试图在我的if结构中创建模式以使所有三个工作，但是我无法区分它在e2和e3之间。它将始终默认为e2。这是我所拥有的只有两个例外，但我非常感谢帮助找出如何实现第三个。谢谢。

I have tried to create patterns in my if structure to make all three work, however I can't get it to differentiate between e2 and e3. It will always default to e2. This is what I have with only two exceptions, but I would greatly appreciate help with figuring out how to implement the third. Thank you.

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：

Here is the output I get right now:

请输入9到99之间的整数：
2

Please enter an integer between 9 and 99: 2

• 您输入的数字不在9到99之间。请重试。

请输入9到99之间的整数：
f

Please enter an integer between 9 and 99: f

• 您没有输入整数。再试一次。

请输入9到99之间的整数：
88

Please enter an integer between 9 and 99: 88

谢谢。初始化完成。

https://www.ideone.com/ZiOpGH

推荐答案

如果你必须检测三种情况，你需要有三组逻辑。

If you've got to detect three circumstances, you need to have three sets of logic.

1. 检查输入的字符是否是有效的数值。


2. 检查输入的数字是否为整数。


3. 检查输入的数字是否介于低限和高限之间。

你几乎必须按顺序检查它们。

And you pretty much have to check them in that order.

扫描仪方便地拥有 hasXXX 查看要读取的字符是否与给定模式匹配的方法。

Scanner conveniently has the hasXXX methods to see whether the characters about to be read match a given pattern.

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