multiple variable for if else statement

Question

I have written some code to check if the user has entered a number between 1 and 5, and now I would also like my code to allow the user to enter the letters A, S, D or M.

Is there a way to combine the code where I can have it identify whether the user has entered 1-5 or A, S, D, M?

How do I edit the code below so the user can enter either an Integer or a character? Do I have to write a snippet of code underneath the loop for it to identify that a user did not enter 1-5 but did enter A, S, D, or M, as in break out of the loop? Or is it a separate loop all together. I am so confused!

import java.util.InputMismatchException;
import java.util.Scanner;

public class Selection {
    Scanner readInput = new Scanner(System.in);

    int selectionOne() {
        int inputInt;
        do { //do loop will continue to run until user enters correct response
            System.out.print("Please enter a number between 1 and 5, A for Addition, S for subtraction, M for multiplication, or D for division: ");
            try { 
                inputInt = readInput.nextInt(); //user will enter a response
                if (inputInt >= 1 && inputInt <=5) {
                    System.out.print("Thank you");
                    break; //user entered a number between 1 and 5
                } else {
                    System.out.println("Sorry, you have not entered the correct number, please try again.");
                }
                continue;
            }
            catch (final InputMismatchException e) {
                System.out.println("You have entered an invalid choice. Try again.");
                readInput.nextLine(); // discard non-int input
                continue; // loop will continue until correct answer is found
            }
        } while (true);
        return inputInt;
    } 
}

Answer 1

As @MarsAtomic mentioned, first thing you should change your input to String instead of an int so you can easily handle both characters and digits.

Change:

int inputInt;

To:

String input;

Then change:

inputInt = readInput.nextInt();

To:

input = readInput.next();

To accommodate reading String instead of int.

Now you reach at 2 main cases (and 2 subcases):

1) input is a single character
   a) input is a single digit from 1-5
   b) input is a single character from the set ('A', 'S', 'D', 'M')
2) input is an error value

Also, since you are not calling Scanner.nextInt, you don't need to use the try/catch statement and can print your errors in else blocks.

Furthermore, you should have your method return a char or a String instead of an int so you can return both 1-5 or A,S,D,M. I will assume you want to return a char. If you want to return a String instead, you can return input instead of return val in the code bellow.

NOTE: The code bellow can be simplified and shortened, I just added variables and comments in an attempt to make each step clear to what is being read or converted. You can look at @mikeyaworski's answer for a more concise way of doing this.

Here is how your code could look like:

   char selectionOne() {
        String input;
        do {
            input = readInput.next();
            // check if input is a single character
            if(input.length() == 1) {
                char val = input.charAt(0);
                // check if input is a single digit from 1-5
                if(Character.isDigit(val)) {
                    int digit = Integer.parseInt(input);
                    if (digit >= 1 && digit <=5) {
                        System.out.print("Thank you");
                        return val; // no need to break, can return the correct digit right here
                    } else {
                        System.out.println("Sorry, you have not entered the correct number, please try again.");
                    }
                } else {
                    // check if input is in our valid set of characters
                    if(val == 'A' || val == 'S' || val == 'M' || val == 'D') { 
                        System.out.print("Thank you");
                        return val;  // return the correct character
                    } else {
                        System.out.println("Sorry, you have not entered the correct character, please try again.");
                    }
                }
            } else {
                System.out.println("Sorry, you have not entered the correct input format, please try again.");
            }
        } while(true);
    } 

Answer 2

I suggest instead of using an int input, just use a String input and convert it to an integer when you need to. You can use Integer.parseInt(String) to convert a String to an int.

So when you check if the input is valid, you need to check if the input is equal to "A", "S", "M" or "D", or any values from 1-5 when it is converted to an int.

So to check if it's one of the characters, you could do this:

if (input.equals("A") || input.equals("S") || input.equals("M") || input.equals("D"))

And then to test if it's an int of value 1 through 5, you could do this:

if (Integer.parseInt(input) >= 1 && Integer.parseInt(input) <= 5)

Just parse the input to an int and then check the range as you already have done.

The return type of this method will be String now, instead of int. If you need it to be an int for whatever reason, you can just parse the value to an int and then return that instead. But I just returned it as a String.

The last thing I changed was the catch block. Now, instead of an InputMismatchException (because they can enter Strings now, I changed it to NumberFormatException, which would happen if a String that could not be converted to an int was attempted to be. For example, Integer.parseInt("hello") will throw a NumberFomatException because "hello" can not be represented as an integer. But, Integer.parseInt("1") would be fine and would return 1.

Note that you should test the String equivalence first so that you don't go into your block before you have a chance to test all conditions you need to.

The method would look like this:

String selectionOne() {
    String input;
    do { //do loop will continue to run until user enters correct response
        System.out.print("Please enter a number between 1 and 5, A for Addition, S for subtraction, M for multiplication, or D for division: ");
        try { 
            input = readInput.nextLine(); //user will enter a response
            if (input.equals("A") || input.equals("S") || input.equals("M") || input.equals("D")) {
                System.out.println("Thank you");
                break; //user entered a character of A, S, M, or D
            } else if (Integer.parseInt(input) >= 1 && Integer.parseInt(input) <= 5) { 
                System.out.println("Thank you");
                break; //user entered a number between 1 and 5
            } else {
                System.out.println("Sorry, you have not entered the correct number, please try again.");
            }
            continue;
        }
        catch (final NumberFormatException e) {
            System.out.println("You have entered an invalid choice. Try again.");
            continue; // loop will continue until correct answer is found
        }
    } while (true);
    return input;
}

Answer 3

If your input can be both characters and letters, why not change to looking for a char or String? Then, you can look for "1" or "A" without any trouble.