Checking user input is a number

问题

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);
    }
}

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

回答1:

You can do somehting like this :

 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

回答2:

Either catch the exception and handle it, or instead of Scanner use

(char) System.in.read();

to receive characters. In this way you can avoid handling exceptions, which takes a lot of time. You can work on chars instead of integers then or check their validity and convert them to integers in this way:

int x = Character.getNumericValue(choice);

回答3:

Just wrap it in a try catch

do{
    try{
        System.out.println(choices)
        choice = input.nextInt()
        switch(choice){
        ....
        }

    }
    catch(InputMismatchException){
        System.out.println("Please enter a valid input")
    }
}
while(whatever)

回答4:

I found the answer out. This is how I would validate the account number's.

int number;
    while(true){
        System.out.print("\nEnter account number: ");
        try{
            number = input.nextInt();
            break;
        }catch(Exception e){
               System.err.println("Error: Invalid Entry! Please try only Integers");
                input=new Scanner(System.in);
        }
    }

and this is how I would validate the menu item selected is a number and not a letter:

int choice = 0;

    do {
        while(true)
        {
            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: ");
            try{

            choice = Integer.parseInt(input.next());
            break;
            }
            catch(Exception e){
                System.err.println("Error: Invalid entry! Please Try Again!");
                input=new Scanner(System.in);
                continue;
            }
    }

来源：https://stackoverflow.com/questions/23787491/checking-user-input-is-a-number