Thread: Getting Exception " java.util.InputMismatchException" in scanner package

Hello luke,

The reason you are getting this error is because you are using int name = sc.nextInt();

For a start, the variable 'name' is an Integer so needs to have an Integer value assigned.

It's the Scanner class that is throwing the java.util.InputMismatchException exception.

sc.nextInt(); is looking for an Integer value entered into the console.
The Scanner class has several methods such as nextBoolean(), nextDouble(), nextFloat() which all look for the corresponding variable type.

If you want to use characters or String, then you can use sc.next()

Take a look at this example:

import java.util.Scanner;
 
public class Main2 {
 
    public static void main(String[] args) {
 
        System.out.println("input");
 
        Scanner sc = new Scanner(System.in);
 
        String name = sc.next();
 
        System.out.println(name);
 
    }
 
}

We can catch this java.util.InputMismatchException exception in a try/catch block. This is error handling and will alert the user that they must use an Integer value.

import java.util.Scanner;
 
public class Main2 {
 
    public static void main(String[] args) {
 
        try{
 
        System.out.println("input");
 
        Scanner sc = new Scanner(System.in);
 
        int number = sc.nextInt();
 
        System.out.println(number);
 
        }catch(Exception e){
            System.out.println("You must enter a Integer value!!");
        }
 
    }
 
}

Originally Posted by luke

is there a way to loop this back so that it prompts the user again to enter the value?

Yes there is. You can use a while loop:

import java.util.Scanner;
 
public class Main2 {
 
    public static void main(String[] args) {
 
        System.out.println("Enter something!");
        Scanner sc = new Scanner(System.in);
 
        while(sc.hasNext()){
 
        String name = sc.next();
 
        System.out.println(name);
 
        }
 
    }
 
}

Take a look at this luke.

If the user enters a non Integer value, the exception is caught and the ScannerStuff() method is run again.

The 2 Integers myInt & yourInt are compared once a user enters an Integer value. If they match then a message is printed to the console.

I hope this is what you are looking for...

import java.util.Scanner;
 
public class Main2 {
 
    public static void ScannerStuff() {
 
        int myInt = 10;
        int yourInt;
 
        System.out.println("Enter an Integer: ");
        Scanner sc = new Scanner(System.in);
 
        try {
            while (sc.hasNext()) {
 
                yourInt = sc.nextInt();
                System.out.println(yourInt);
 
                if(yourInt == myInt){
                    System.out.println("You entered the same Integer as me!");
                }
 
            }
        } catch (Exception e) {
            System.out.println("You must enter an Integer!");
            ScannerStuff();
        }
 
    }
 
    public static void main(String[] args) {
 
        Main2 m = new Main2();
        m.ScannerStuff();
 
    }
 
}

Originally Posted by luke

alright last bit, extremely basic i know but i've forgotten.

i want it to check if the value entered is withing the boundaries.

for example i want in common language to be

if the entered value is greater than 0 and less than fourty then{
}

This can easily be done with an if statement:

                if(yourInt > 0 && yourInt <= 40){
                    System.out.println("Your value is greater than 0 and less or equal to 40");
                }

This is basically saying, if yourInt is greater than 0 and less or equal to 40 then print message.

The > && <= bits are called Operators. Here is a quick overview:

Operators (The Java™ Tutorials > Learning the Java Language > Language Basics)

Originally Posted by luke

This is an excellent an idea! thank you very much was exactly what i was looking for thank you!

Brilliant! Glad I could help Luke.

If you have no further questions, please can you mark this thread as solved. (Check the link in my signature)

Thanks!!