Java integer input validation

Question

I am trying to validate input from a user.The user puts in Y coordinate a letter between (A-J) and x coordinate a number between (1-9).I can validate the y coordinate but am having trouble validating the x coordinate. I want it so if the user puts in something other than a number between 1 and 9 it keeps asking the user for valid input.

    do {
        // inner loop checks and validates user input
        do {

            System.out.println("Enter X Co-Ord (A-J), or Q to QUIT");
            letter = input.next().toUpperCase(); // upper case this for
                                                    // comparison
            if (letter.equals("Q"))
                break; // if user enters Q then quit

            String temp = "ABCDEFGHIJ";

            while (temp.indexOf(letter) == -1) {
                validString = false;
                System.out.println("Please enter a valid input");
                letter = input.next().toUpperCase();
                col = temp.indexOf(letter);

            }

            if (temp.indexOf(letter) != -1) {
                validString = true;
                col = temp.indexOf(letter);

            }
            try {

                System.out.println("Enter Y Co-Ord (0-9)");
                row = input.nextInt();


            } catch (InputMismatchException exception) {
                validInt = false;
                System.out.println("Please enter a number between 1 -9");
            }

            catch (Exception exception) {
                exception.printStackTrace();
            }

            valuesOK = false; // only set valuesOK when the two others are
                                // true
            if (validString && validInt) {
                valuesOK = true;
            }
        } while (!valuesOK); // end inner Do loop

The output is:

Enter X Co-Ord (A-J), or Q to QUIT

d

Enter Y Co-Ord (0-9)

h

Please enter a number between 1 -9

Enter X Co-Ord (A-J), or Q to QUIT

Enter Y Co-Ord (0-9)

You'll want to align your text; one says 0-9, one says 1-9.

Dave Newton
– Dave Newton

2013-01-18 17:05:25 +00:00
Commented Jan 18, 2013 at 17:05 — Dave Newton
– Dave Newton, Commented Jan 18, 2013 at 17:05

Mike Clark · Accepted Answer · 2013-01-18 17:25:38Z

1

You just need to put a while loop around your nextInt() the same as you do when reading the letter:

  System.out.println("Enter Y Co-Ord (0-9)");
  row = -1
  while (row < 0) {
    try {
      row = input.nextInt();
      validInt = true;
    } catch (InputMismatchException exception) {
      System.out.println("Please enter a number between 1 -9");
      row = -1;
      validInt = false;
    }
  }

edited Jan 18, 2013 at 17:25

answered Jan 18, 2013 at 17:04

Mike Clark

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6 Comments

Mike Clark Over a year ago

Edited. I dropped the validInt flag by mistake.

Alex Kreutznaer Over a year ago

Please, use new syntax for try catch, it will look nicer.

user1899672 Over a year ago

What are you referring to by "validInt flag"? sorry, I'm new to programming.

Mike Clark Over a year ago

I was meaning the validInt boolean value that you had inside the catch block in your original code above.

Mike Clark Over a year ago

I'm afraid with the limited block of code here I don't see where an infinite look is coming from. I would need to see further code outside this block to get the specific class of input for example.

|

Thai Tran · Accepted Answer · 2013-01-19 07:05:24Z

0

Just want to make the validation make more sense with the requirement, since human 's eye can skip easily the line nextInt()

String value = "";
System.out.println("Enter Y Co-Ord (1-9)");

while (!(value = input.next()).matches("[1-9]+")) {
    System.out.print("Wrong input. Insert again: ");
}

System.out.println(value);

Of course, when you get the right value, then you can parse it to integer again (safely !!!)

edited Jan 19, 2013 at 7:05

answered Jan 19, 2013 at 6:56

Thai Tran

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