[Solved]java.util.InputMismatchException


java.util.InputMismatchException which throw by a Scanner to indicate that the token retrieved does not match the pattern for the expected type, or that the token is out of range for the expected type.

  • InputMismatchException is runtime unchecked exception.
  • InputMismatchException is sub class of NoSuchElementException.

Constructors

  • InputMismatchException() : Constructs an InputMismatchException with null as its error message string.
  • InputMismatchException(String s): Constructs an InputMismatchException, saving a reference to the error message string s for later retrieval by the getMessage method.

Example

In below example create this java.util.InputMismatchException by passing some others type values other than expectation. For Example : age expected as integer value from console while passing as decimal value.

import java.util.Scanner;

public class JavaScannerExample {

	public static void main(String[] args) {
		// Create a Scanner object
		Scanner scanner = new Scanner(System.in); 

		try
		{
		//User different type of input from console
		System.out.println("Please enter user name:");
		String userName = scanner.nextLine();
		System.out.println("Please enter age:");
		int age = scanner.nextInt();
		System.out.println("Please enter salary:");
	    double salary = scanner.nextDouble();

	    //Output input by user
	    System.out.println("User Name: " + userName);
	    System.out.println("Age: " + age);
	    System.out.println("Salary: " + salary);
		}
		catch(Exception ex)
		{
			System.err.println("Entered user input are not match with required type:");
			ex.printStackTrace();
		}

	}

}
Please enter user name:
Saurabh Gupta
Please enter age:
13.5
Entered user input are not match with required type:
java.util.InputMismatchException
	at java.util.Scanner.throwFor(Unknown Source)
	at java.util.Scanner.next(Unknown Source)
	at java.util.Scanner.nextInt(Unknown Source)
	at java.util.Scanner.nextInt(Unknown Source)
	at com.userinput.JavaScannerExample.main(JavaScannerExample.java:28)

Solution

  • In such type scenarios where user interaction required always write code and ask input as scanner.nextLine();
  • Always validate values type before assign to variable.
  • If any mismatch found and throw above exception show message to insert value again as required format.

References

https://docs.oracle.com/javase/8/docs/api/java/util/InputMismatchException.html

Know More

To know more about Java Exception Hierarchy, in-built exception , checked exception, unchecked exceptions and solutions. You can learn about Exception Handling in override methods and lots more. You can follow below links:  s

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s