Question

int memberChoice;
    boolean accountLoop = true;
    while (accountLoop) {
        try {
            System.out.println("Would you like to be a... \n1.Basic Member or \n2.Premium member");
            memberChoice = input.nextInt();
            if (memberChoice == 1 || memberChoice == 2) {
                accountLoop = false;
            }
            if (memberChoice < 1 || memberChoice > 2) {
                System.out.println("Please choose either 1 or 2");
            }
            System.out.println();
        } catch (InputMismatchException e) {
            System.out.println("Invalid input");
        }
    }

每当我输入一个字符串时，它应该说无效输入会回到循环的开头并指示我再次输入一个数字。相反，它给了我一个无限循环。我觉得它的东西真的很小但我已经在这里呆了5个小时而且无法理解它。有什么建议吗？

Answer 1

对我来说，你的代码工作得很好。你的ide可能有问题。无论如何，将行if (memberChoice < 1 || memberChoice > 2)更改为if (memberChoice < 1 && memberChoice > 2)

Answer 2

将input = new Scanner(System.in);添加到catch块，因为我假设您正在使用扫描仪读取标准输入...

当出现问题时，这将重新初始化扫描仪。

当放入＆＃34;字符串＆＃34;扫描仪将尝试将其读作nextInt（），除非您再次初始化它，否则它将继续失败...

Answer 3

使用try / catch块来路由代码逻辑是不好的做法。例外情况应保留用于特殊情况，并且从不用于通过应用程序控制信息流。

相反，您可以使用（假定的）扫描程序对象的其他方法在尝试为其分配类型之前验证传入的元素。

public static void main(String[] args)  {
    int memberChoice=0;
    System.out.println("Would you like to be a... \n1.Basic Member or \n2.Premium member");

    //Use Try with Resources to ensure the Scanner is closed when we're done (requires Java 1.7+)
    try (Scanner input = new Scanner(System.in)){

        //hasNextLine tells us the user has entered a value.
        while (input.hasNextLine() && (memberChoice != 1 && memberChoice != 2)) {

            //Is that value actually an int?
            if (input.hasNextInt()) {
                //YES - Read as int and check for the value we're expecting
                int inValue = input.nextInt();
                if (inValue == 1 || inValue == 2) {
                    memberChoice = inValue;
                } else {
                    System.out.println("Invalid choice.  Try again");
                    System.out.println("Would you like to be a... \n1.Basic Member or \n2.Premium member");
                    System.out.println();
                }
            } else {

                //NO - Read the line, provide feedback and prompt the user for a correction.
                String line = input.nextLine();
                System.out.println("You have to enter an integer value. '"+ line + "' is not valid.");
                System.out.println("Would you like to be a... \n1.Basic Member or \n2.Premium member");
                System.out.println();
            }            
        }
    }
}

Answer 4

这就是我解决问题的方法：

int memberChoice = -1;
boolean accountLoop = true;
Scanner reader = new Scanner(System.in);
while (accountLoop) {
        System.out.println("Would you like to be a... \n1.Basic Member or \n2.Premium member");
        try {   
            memberChoice = reader.nextInt();
            if (memberChoice == 1 || memberChoice == 2) {
                System.out.println(" Answer was valid ");
                accountLoop = false;
            }
            if (memberChoice < 1 || memberChoice > 2) {
                System.out.println("Please choose either 1 or 2");
            }
            System.out.println();
        } catch (InputMismatchException e) {
            System.out.println("Invalid input");
            reader.next(); // this consumes the invalid token 
        }
    }
    reader.close();

使用以下链接解决了这个问题： How to handle infinite loop caused by invalid input using Scanner

这个答案的摘要：

您需要“消费”无效答案。否则，该输入将保留在缓冲区中，这将导致无限循环。

我不确定我在这个try catch块中做错了什么

4 个答案: