How can I remove this java exception error in my code?

When I input a string into the code below by mistake as a test, I get a red java error message in my console. However, within my if statement I added an else part which should end the program if the user doesn't input the if statement condition i.e a number between 0-100. Why this is and how can I fix it?

MY CODE

        Scanner input = new Scanner(System.in);
        System.out.println("Enter a number: ");
        int decimal = input.nextInt();
        if (decimal > 0 && decimal <= 100) {
            //code 
        }
        else {
            System.exit(0);
        }

When I input a string this message gets displayed. However, I just wanted to tell the user they exerted the wrong value and I wanted the program to quit.

Exception in thread "main" java.util.InputMismatchException
	at java.util.Scanner.throwFor(Scanner.java:864)
	at java.util.Scanner.next(Scanner.java:1485)
	at java.util.Scanner.nextInt(Scanner.java:2117)
	at java.util.Scanner.nextInt(Scanner.java:2076)
	at MainHandler.main(MainHandler.java:22)

I did try use hasNextInt at one point to try get rid of the exception error but I get an error when I use hasNextInt. https://i.stack.imgur.com/LvQYr.jpg

You would need double if you are entering decimal numbers and put your code inside try/catch block

try{
            Scanner input = new Scanner(System.in);
            System.out.println("Enter a number: ");
            double decimal = input.nextDouble();
            if (decimal > 0 && decimal <= 100) {
            }
            else {
                System.exit(0);
            }
        }catch (Exception e){
            System.out.println(e);
        }

Try binding your code inside try and catch blocks like this:

try{
 int decimal = input.nextInt();
        if (decimal > 0 && decimal <= 100) {
            //code 
        }
        else {
            System.exit(0);
        }

}
catch(Exception e){
  System.out.println("You enter wrong value");
}

Try with something like this. You surround your input inside a try-catch and as long as you get an Exception, you ask the user to put a number again. As soon as the input is valid (a number) you can proceed with your code:

    boolean canProceed = false;
    int number = -1;
    
    while (!canProceed) {
        try {
            Scanner input = new Scanner(System.in);
            System.out.println("Enter a number: ");
            number = Integer.parseInt(input.nextLine());
            canProceed = true;
        } catch (Exception e) {
            System.out.println("Invalid input.");
        }
    }
    
    if (number > 0 && number <= 100) {
        System.out.println("That's fine");
    }
    else {
        System.exit(0);
    }

The code you posted works just fine if the input is provided from the start. However given your "Enter a number:" message I suppose you want the user to provide the input interactively, in which case you'll have to implement a waiting loop :

Scanner input = new Scanner(System.in);
System.out.println("Enter a number: ");
while (!scanner.hasNextInt()) {
    if (scanner.hasNextLine()) {
        scanner.nextLine(); // we've got something but it's not a number, discard it
    }
}
int decimal = input.nextInt();
if (decimal > 0 && decimal <= 100) {
    //code 
}
else {
    System.exit(0);
}

In this case the loop will continue until it finds an integer to read from the input, discarding lines of input if they don't start with an integer.

I just thought I would throw this in here since it is another flexible alternative to accomplishing the task. It utilizes the Scanner#nextLine() method with input validation to ensure that what is supplied is actually what is expected without the worry of an exception being thrown.

Validation is done using the String#matches() method supplied with a small Regular Expression (RegEx) to insist that numerical digits are supplied and that the supplied value is within the desired inclusive unsigned numerical range of 1 to 100. If the validation fails then the User is expected to try again or enter 'Q' (or anything that starts with Q - case insensitive) to quit.

Scanner input = new Scanner(System.in);
int intValue = 0;
String numberString = "";
    
// Prompt and Input...
while (intValue == 0) {
    System.out.print("Enter a number (or 'q' to quit): --> ");
    numberString = input.nextLine().toLowerCase();
    //Quit..
    if (numberString.charAt(0) == 'q') {
        System.exit(0);  
    }
    // Integer (unsigned - Range 1 to 100)...
    else if (numberString.matches("100|[1-9][0-9]{1}?")) {
        intValue = Integer.parseInt(numberString); 
    }
    else {
        System.err.println("Invalid numerical value supplied or the supplied");
        System.err.println("value is not in the inclusive range of 1 to 100!");
        System.err.println("Try Again...");
        System.err.println();
    }
}

//Your processing code here.
System.out.println("The value you entered is: --> " + intValue);

Explanation for the RegEx used within the String#matches() method:

100|[1-9][0-9]{1}?

1st Alternative 100
    100 matches the characters 100 literally (case sensitive)

2nd Alternative [1-9][0-9]{1}?
    Match a single character present in the list below [1-9]
    1-9 a single character in the range between 1 (index 49) and 9 (index 57) (case sensitive)
    Match a single character present in the list below [0-9]{1}?
    {1}? Quantifier — Matches exactly one time (meaningless quantifier)
    0-9 a single character in the range between 0 (index 48) and 9 (index 57) (case sensitive)

If you want to change your numerical range then you may find this site useful.