当提供的输入不是数字时，如何使代码捕获错误？

Question

程序检查用户输入并确定它是正面的还是负面的。 当用户提供无效输入（非整数）（例如“444h”或“ibi”）时，如何捕获错误。

我在想最后的 else 语句会捕获异常，但它没有。

import java.util.Scanner;
public class Demo

{

    public static void main(String[] args)
    {

        Scanner scan = new Scanner(System.in);
        System.out.print("Enter any number: ");

        int num = scan.nextInt();

        scan.close();
        if(num > 0)
        {
            System.out.println(num+" is positive");
        }
        else if(num < 0)
        {
            System.out.println(num+" is negative");
        }
        else if(num == 0)
        {
            System.out.println(num+" is neither positive nor negative");
        }
        else
        {
            System.out.println(num+" must be an integer");
        }
    }
}

我希望程序能够捕获异常并提示输入有效整数

Answer 1

只需将您的代码包装在 try..catch block 中。

完整代码：

import java.util.Scanner;
import java.util.InputMismatchException;

class Main {
    public static void main(String[] args) {
        System.out.println("Hello world!");
        Scanner scan = new Scanner(System.in);
        System.out.print("Enter any number: ");

        try {
            int num = scan.nextInt();
            if (num > 0) {
                System.out.println(num + " is positive");
            } else if (num < 0) {
                System.out.println(num + " is negative");
            } else if (num == 0) {
                System.out.println(num + " is neither positive nor negative");
            }
        } catch (InputMismatchException e) {
            System.out.println("Error: Value Must be an integer");
        }

        scan.close();
    }
}

祝你好运:)

Answer 2

int num = 0;
    try{
        num =input.nextInt();
    }catch(InputMismatchException ex) {
        System.out.println("Enter Integer Value Only");
    }

如果您从输入中获得非整数值，则必须在 try 块中编写 int num = scan.nextInt(); 这条语句，然后 InputMismatchException 将出现然后执行 catch 块。

Answer 3

首先，如果您打算在应用程序运行期间再次使用扫描仪，请不要关闭它。关闭它后，您需要重新启动应用程序才能再次使用它。

如前所述，您可以在 Scanner#nextInt() 方法上使用 try/catch：

Scanner scan = new Scanner(System.in);

int num = 0;
boolean isValid = false;
while (!isValid) {
    System.out.print("Enter any number: ");
    try {
        num = scan.nextInt();
        isValid = true;
    }
    catch (InputMismatchException ex) {
        System.out.println("You must supply a integer value");
        isValid = false;
        scan.nextLine(); // Empty buffer
    }
}

if (num > 0) {
    System.out.println(num + " is positive");
}

else if (num < 0) {
    System.out.println(num + " is negative");
}

else if (num == 0) {
    System.out.println(num + " is neither positive nor negative");
}

或者你可以使用 Scanner#nextLine() 方法来代替它接受字符串：

Scanner scan = new Scanner(System.in);

String strgNum = "";
boolean isValid = false;
while (!isValid) {
    System.out.print("Enter any Integer value: ");
    strgNum = scan.nextLine();

    /* See if a signed or unsigned integer value was supplied.
       RegEx is used with the String#matches() method for this.
       Use this RegEx: "-?\\d+(\\.\\d+)?"  if you want to allow
       the User to enter a signed or unsigned Integer, Long, 
       float, or double values.         */
    if (!strgNum.matches("-?\\d+")) {
        System.out.println("You must supply a numerical value (no alpha characters allowed)!");
        continue;
    }
    isValid = true; // set to true so as to exit loop
}
int num = Integer.parseInt(strgNum); // Convert string numerical value to Integer
/* Uncomment the below line if you use the other RegEx. You would have to also 
   comment out the above line as well.   */
// double num = Double.parseDouble(strgNum); // Convert string numerical value to double
if (num > 0) {
    System.out.println(num + " is positive");
}
else if (num < 0) {
    System.out.println(num + " is negative");
}
else if (num == 0) {
    System.out.println(num + " is neither positive nor negative");
}

Answer 4

使用正则表达式检查输入是否为数字

import java.util.regex.*;
Pattern.matches("[0-9]+", "123"); // this will return a boolean