java 如何创建检查 int 范围、数字类型而不是 char 的异常？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15468418/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How can I create an exception that checks for int range, number type, and not a char?
提问by user2180681
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.
我试图解决异常问题，我遇到的问题是我需要创建一个程序，要求用户输入 9-99 的数字。这个数字必须使用 3 个不同的异常进行错误检查。
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)
e1：数字超出范围 (200)
e2：数字是整数（double）以外的数据类型
e3：输入是除数字（字符）以外的另一种数据类型
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.
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 e2 和 e3。它将始终默认为 e2。这就是我所拥有的，只有两个例外，但我将不胜感激帮助弄清楚如何实现第三个。谢谢你。
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);
}
}
}
Here is the output I get right now:
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.
请输入 9 到 99 之间的整数：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
- 您没有输入整数。再试一次。
请输入 9 到 99 之间的整数：88
谢谢你。初始化完成。
回答by Patashu
In catch (InputMismatchException e2), test to see if input.hasNextDouble()(or input.hasNextFloat()? Not sure which one is more general...) is true. If it is, then you can distinguish between the case 'user entered a double' and 'user entered a non numeric type'
在 中catch (InputMismatchException e2)，测试以查看input.hasNextDouble()（或input.hasNextFloat()？不确定哪个更通用...）为真。如果是，那么您可以区分“用户输入双精度”和“用户输入非数字类型”的情况
回答by Hot Licks
If you've got to detect three circumstances, you need to have three sets of logic.
如果你必须检测三种情况，你需要有三套逻辑。
- Check if the entered characters are a valid numeric value.
- Check that the entered number is an integer.
- Check that the entered number falls between the low and high bounds.
- 检查输入的字符是否是有效的数值。
- 检查输入的数字是否为整数。
- 检查输入的数字是否介于上下限之间。
And you pretty much have to check them in that order.
而且您几乎必须按照该顺序检查它们。
Scanner conveniently has the hasXXXmethods to see whether the characters about to be read match a given pattern.
Scanner 有hasXXX多种方法可以方便地查看将要读取的字符是否与给定的模式匹配。
