Exception in thread main java util InputMismatchException error

0 votes

I have a question on what's going on, whenever I try to compile it it keeps giving me an error like this:

Exception in thread "main" 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 Person.main(Person.java:38)

All I want is for the user to be able to input their age and name and have it stored in the "age" and "name" variables, then have it print it out in the bottom string. And if someone would like to help me clean up my code as well, it wouldn't hurt..

import java.util.*; 
import java.io.*; 
import java.util.Scanner; 

public class Person 
{ 

public static void main(String[]args) 
{ 

int age; 
int name; 
Scanner scan = new Scanner(System.in); 

System.out.println("Enter in your age."); 
age = scan.nextInt(); 

if (age < 18) 
{ 
System.out.println("So you're a kid, huh? That's fine."); } else if (age >= 18) 
{ 

System.out.println("Ah, and adult! Good."); } 

@SuppressWarnings("resource") 
Scanner in = new Scanner(System.in); 

System.out.println("Enter in your name"); 
name = in.nextInt(); 

System.out.println("So you're " + age + " years old and your name is " + name); 

}

}

Feb 18, 2022 in Java by Aditya
• 7,660 points
1,116 views

1 answer to this question.

0 votes
Problem

int name; //Name should be of type String 
... 
System.out.println("Enter in your name"); 
name = in.nextInt(); //It doesn't handle the string since your using `nextInt`

Solution

String name; 
... 
System.out.println("Enter in your name"); 
name = in.nextLine();

answered Feb 18, 2022 by Rahul
• 9,670 points

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