处理 InputMismatchException 异常，但未捕获它 Java

Handle InputMismatchException exception but doesn't catch it Java

提问人：Tonia Giannoulaki 提问时间：4/2/2022 最后编辑：MureinikTonia Giannoulaki 更新时间：4/3/2022 访问量：42

问：

由于某种未知原因，当用户输入是字符串或特殊字符时，它不会捕获我的异常（扫描程序只接受整数和 1 到 3）。代码如下：

package mainpackage;
import java.util.InputMismatchException;
import java.util.Scanner;

public class CreateUsers {
    
    public static void main(String[] args) {
        
        Scanner chooseNumber = new Scanner(System.in);
        int number;
        
        System.out.println("Welcome to UniStudents Application!\n"
                + "Please select how to Log in: \n"
                + "1. Log in as a Student \n"
                + "2. Log in as a Professor \n"
                + "3. Log in as a Secretary \n"
                + "To choose an action, please write the number of the action: ");  
        number = chooseNumber.nextInt();
        
        while(true) {
            if(number > 3 || number < 1) {
                System.out.println("The number of the action does not exist.\n"
                        +"Please try again: ");
                chooseNumber.nextInt();
            }
            else {
                System.out.println("ok");
                chooseNumber.close();
                break;
            }   
        }
        try{
            chooseNumber.nextInt();
        }
        catch(InputMismatchException e) {
             
            System.out.println("Please write the number of the action you want to choose \n"
                    + "Please select how to Log in: \n"
                    + "1. Log in as a Student \n"
                    + "2. Log in as a Professor\n"
                    + "3. Log in as a Secretary\n");
        }

代码的结果是：

1. Log in as a Student 
2. Log in as a Professor 
3. Log in as a Secretary 
To choose an action, please write the number of the action: 
$^&*
Exception in thread "main" java.util.InputMismatchException
    at java.base/java.util.Scanner.throwFor(Scanner.java:939)
    at java.base/java.util.Scanner.next(Scanner.java:1594)
    at java.base/java.util.Scanner.nextInt(Scanner.java:2258)
    at java.base/java.util.Scanner.nextInt(Scanner.java:2212)
    at JavaBasics/mainpackage.CreateUsers.main(CreateUsers.java:18)

我想不通

java 异常 数 try-catch java.util.scanner

评论

0赞 geocodezip 4/3/2022
除了不通过 try/catch 异常处理包围第一个实例之外，您也不会保存第二个实例的结果，因此循环永远无法退出。number = chooseNumber.nextInt();

答：

1赞 Mureinik 4/2/2022 #1

第一个调用不在块中，因此不会捕获异常。将块的开头移到它上面，你应该没问题。number = chooseNumber.nextInt();trytry

1赞 Dylech30th 4/2/2022 #2

由于此异常是在上面的第 18 行抛出的，并且 try-catch 块未涵盖它，因此请尝试：number = chooseNumber.nextInt();while (true)

package mainpackage;
import java.util.InputMismatchException;
import java.util.Scanner;

public class CreateUsers {
    
    public static void main(String[] args) {
        
        Scanner chooseNumber = new Scanner(System.in);
        int number;
        
        System.out.println("Welcome to UniStudents Application!\n"
                + "Please select how to Log in: \n"
                + "1. Log in as a Student \n"
                + "2. Log in as a Professor \n"
                + "3. Log in as a Secretary \n"
                + "To choose an action, please write the number of the action: ");  
        try {
            number = chooseNumber.nextInt();
        
            while(true) {
                if(number > 3 || number < 1) {
                    System.out.println("The number of the action does not exist.\n"
                            +"Please try again: ");
                    chooseNumber.nextInt();
                }
                else {
                    System.out.println("ok");
                    chooseNumber.close();
                    break;
                }   
            }
            chooseNumber.nextInt();
        } catch (InputMismatchException e) {
             
            System.out.println("Please write the number of the action you want to choose \n"
                    + "Please select how to Log in: \n"
                    + "1. Log in as a Student \n"
                    + "2. Log in as a Professor\n"
                    + "3. Log in as a Secretary\n");
        }

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