How to loop user input until an integer is inputted?

Question

I'm new to Java and I wanted to keep on asking for user input until the user enters an integer, so that there's no InputMismatchException. I've tried this code, but I still get the exception when I enter a non-integer value.

int getInt(String prompt){
        System.out.print(prompt);
        Scanner sc = new Scanner(System.in);
        while(!sc.hasNextInt()){
            System.out.println("Enter a whole number.");
            sc.nextInt();
        }
        return sc.nextInt();
}

Thanks for your time!

Juned Ahsan · Accepted Answer · 2013-10-02 05:14:40Z

11

Take the input using next instead of nextInt. Put a try catch to parse the input using parseInt method. If parsing is successful break the while loop, otherwise continue. Try this:

        System.out.print("input");
        Scanner sc = new Scanner(System.in);
        while (true) {
            System.out.println("Enter a whole number.");
            String input = sc.next();
            int intInputValue = 0;
            try {
                intInputValue = Integer.parseInt(input);
                System.out.println("Correct input, exit");
                break;
            } catch (NumberFormatException ne) {
                System.out.println("Input is not a number, continue");
            }
        }

edited Oct 2, 2013 at 5:14

answered Oct 2, 2013 at 4:58

Juned Ahsan

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1 Comment

Juned Ahsan Over a year ago

@Shail There may be better solutions available but this is the one i can create using my limited abilities. Accept the answer in case it helped. Thanks!

boxfish · Accepted Answer · 2013-10-02 05:51:58Z

7

Shorter solution. Just take input in sc.next()

 public int getInt(String prompt) {
    Scanner sc = new Scanner(System.in);
    System.out.print(prompt);
    while (!sc.hasNextInt()) {
        System.out.println("Enter a whole number");
        sc.next();
    }
    return sc.nextInt();

}

edited Oct 2, 2013 at 5:51

answered Oct 2, 2013 at 5:28

boxfish

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Comments

Shail · Accepted Answer · 2013-10-02 06:03:59Z

6

Working on Juned's code, I was able to make it shorter.

int getInt(String prompt) {
    System.out.print(prompt);
    while(true){
        try {
            return Integer.parseInt(new Scanner(System.in).next());
        } catch(NumberFormatException ne) {
            System.out.print("That's not a whole number.\n"+prompt);
        }
    }
}

answered Oct 2, 2013 at 6:03

Shail

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Comments

Ava_Katushka · Accepted Answer · 2020-09-28 15:39:29Z

Keep gently scanning while you still have input, and check if it's indeed integer, as you need:

String s = "This is not yet number 10"; 
  
        // create a new scanner 
        // with the specified String Object 
        Scanner scanner = new Scanner(s); 
  
        while (scanner.hasNext()) { 
  
            // if the next is a Int, 
            // print found and the Int 
            if (scanner.hasNextInt()) { 
                System.out.println("Found Int value :"
                                   + scanner.nextInt()); 
            } 
  
            // if no Int is found, 
            // print "Not Found:" and the token 
            else { 
                System.out.println("Not found Int value :"
                                   + scanner.next()); 
            } 
        } 
        scanner.close();

Niki · Accepted Answer · 2013-10-02 05:27:38Z

0

As an alternative, if it is just a single digit integer [0-9], then you can check its ASCII code. It should be between 48-57 to be an integer.

Building up on Juned's code, you can replace try block with an if condition:

    System.out.print("input");
    Scanner sc = new Scanner(System.in);
    while (true) {
            System.out.println("Enter a whole number.");
            String input = sc.next();
            int intInputValue = 0;
            if(input.charAt(0) >= 48 && input.charAt(0) <= 57){
                System.out.println("Correct input, exit");
                    break;
            }
            System.out.println("Input is not a number, continue");
    }

answered Oct 2, 2013 at 5:27

Niki

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