CODEWITHSUNDEEP
java
user121582
user121582

Reputation: 49

How to check the type of a user input in Java?

I was wondering how you can do type checking for a user input. Here I have a simple test to check if the user input is between 1 and 10. I want to set it up so that the user can also enter a letter, primarily so I can use the input 'q' to quit the program.

Is there a part of the scanner that can type check? My thought was to have an if statement: if user inputs type int continue, if it is not type int, check if it is q to quit the program, else output this is an error. Below is what I have so far, it throws an expression when a number is put in since the types do not match.

public static void main(String[] args) {
    //Create new scanner named Input
    Scanner Input = new Scanner(System.in);
    //Initialize Number as 0 to reset while loop
    int Number = 0;

    do
    {
        //Ask user to input a number between 1 and 10
        System.out.println("At anytime please press 'q' to quit the program.");
        System.out.println();
        System.out.print("Please enter a number between 1 and 10:");
        Number = Input.nextInt();

        if (Number == 'Q'|| Number == 'q')
        {
            System.out.println("You are now exiting the program");
        }
        else if (Number >= 1 && Number <= 10)
        {
            System.out.println("Your number is between 1 and 10");
        }   
        else
        {
            System.out.println("Error: The number you have entered is not between"
                    + " 1 and 10, try again");
        }
    }
    //Continue the loop while Number is not equal to Q
   while (Number != 'Q' & Number != 'q');
}

}

Thanks everyone for the responses. I am a bit new so the try statement is new to me but looks like it will work (seems somewhat self explanatory of what it does). I will look into its use more and implementing it correctly.

Upvotes: 4

Views: 13662

Answers (6)

Soham Hazra
Soham Hazra

Reputation: 1

Found this piece of code check it out, it worked for me.

          Scanner input = new Scanner (System.in);
          if (input.hasNextInt()) {
              int a = input.nextInt();
              System.out.println("This input is of type Integer."+a);
          }   
                   
          else if (input.hasNextLong())
            System.out.println("This input is of type Long.");
       
          else if (input.hasNextFloat())
          System.out.println("This input is of type Float.");
          
          else if (input.hasNextDouble()) 
          System.out.println("This input is of type Double."); 
           
          else if (input.hasNextBoolean())
           System.out.println("This input is of type Boolean.");  
          
          else if (input.hasNextLine())
              System.out.println("This input is of type string.");

Upvotes: 0

rRadwan
rRadwan

Reputation: 11

try this

    Scanner Input = new Scanner(System.in);
    //Initialize Number as 0 to reset while loop
    String Number = "0";

    do {
        try {
            //Ask user to input a number between 1 and 10
            System.out.println("At anytime please press 'q' to quit the program.");
            System.out.println();
            System.out.print("Please enter a number between 1 and 10:");
            Number = Input.next();

            if (Number.equalsIgnoreCase("q")) {
                System.out.println("You are now exiting the program");
            } else if (Integer.valueOf(Number) >= 1 && Integer.valueOf(Number) <= 10) {
                System.out.println("Your number is " + Number);
            } else {
                System.out.println("Error: The (" + Number + ") is not between 1 and 10, try again");
            }
        } catch (NumberFormatException ne) {
            System.out.println("Error: The (" + Number + ") is not between 1 and 10, try again");
        } catch (Exception e) {
            System.out.println(e.getMessage());
        }
    } //Continue the loop while Number is not equal to Q
    while (!Number.equalsIgnoreCase("q"));

Upvotes: 1

Harry
Harry

Reputation: 1472

You could get the input as a String and then use regular expression matching on the said String.

String txt = Input.nextLine();
if(txt.matches("[0-9]+") // We have a number
{
   // Safe to parse to number
}
else if(txt.matches("[A-Za-z]+") // We have alphabeticals
{
   // Check for various characters and act accordingly (here 'q')
}

Upvotes: -1

carcaret
carcaret

Reputation: 3358

You could check if it's an int this way:

String inputArg = Input.next();

private boolean isInt(String inputArg){
    boolean isInt = true;
    try {
         Integer.parseInt(inputArg);
    } catch(NumberFormatException e) {
         isInt = false;
    }
    return isInt;
}

Other way is using regular expressions

Upvotes: 0

brso05
brso05

Reputation: 13222

I would use nextLine() and parseInt() to see if it is an int:

    int Number;
    String test = "";
    boolean isNumber = false;
    ......

    test = Input.nextLine();
    try
    {
        Number = Integer.parseInt(test);
        isNumber = true;
    }
    catch(NumberFormatException e)
    {
         isNumber = false;
    }
    if(isNumber)
    {
        if (Number >= 1 && Number <= 10)
        {
            System.out.println("Your number is between 1 and 10");
        }   
        else
        {
            System.out.println("Error: The number you have entered is not between"
                + " 1 and 10, try again");
        }
    }
    else
    {
        if (test.equalsIgnoreCase("q"))
        {
            System.out.println("You are now exiting the program");
        }
    }

.........
while (!test.equalsIgnoreCase("q"));

Upvotes: 1

user3864935
user3864935

Reputation:

Have your input read in as a string, to check for the input of Q/q. Then after that you can parseInt from input string to integer in a way as follows, catching potential NumberFormatExceptions.

String input = "5"
try {
int number = Integer.parseInt(input);
} catch (NumberFormatException e) {
System.out.println("Not a number!") 
}

Or something similar to that.

Upvotes: 0

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