检查用户输入的数字 [英] Checking user input is a number

问题描述

我需要检查用户输入.我有一个菜单，我需要用户选择数字0-4，但是如果用户选择字母而不是数字，那么我只会得到InputMismatchException.因此，我试图验证用户输入了一个数字.这是我的代码:

I need to check users input. I have a menu and I need the user to select numbers 0-4 but if the user selects a letter instead of a number then I just get a InputMismatchException. So I am trying to validate that the user entered a number. Here is my code:

public class TestBankAccount {

static Scanner input = new Scanner(System.in);


public static void main(String[] args) throws FileNotFoundException {

    ArrayList<BankAccount> list = new ArrayList<BankAccount>();

    int choice;

    do {
        System.out.println("1. Deposit money");
        System.out.println("2. Withdraw money");
        System.out.println("3. Check balance");
        System.out.println("4. Create new account");
        System.out.print("Your choice, 0 to quit: ");
        choice = input.nextInt();

        switch (choice) {
        case 1:
            depositMoney(list);
            break;
        case 2:
            withdrawMoney(list);
            break;
        case 3:
            checkBalance(list);
            break;
        case 4:
            createNewAccount(list);
            break;
        case 0:
            System.out.println("Thank you for trusting us with your banking needs!");
            break;
        default:
            System.out.println("Invalid option is selected!");
        }
        System.out.println();
    } while (choice != 0);

    if (list.size() > 0) {
        displayResults(list);
    }
}

我正在考虑做while(选择！= 0&&选择！= input.hasNextInt())之类的事情；但我得到一个错误.有什么想法吗?

I was thinking to do something like while (choice != 0 && choice != input.hasNextInt()); but I get an error. Any ideas?

推荐答案

您可以执行以下操作:

 int choice = 0 ;
 try{
   choice = Integer.parseInt(input.next());
 }
catch(NumberFormatException e)
{
    System.out.println("invalid value enetered");
}

// Now you can check if option selected is between 1 & 4
//  and throw some custom exception

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