User put String instead of integer

Question

i am begginer with java and i made a code which the user puts a number but if he writes a string instead of integer it's a error, how can i control the error to show a message to the user? Here's the error:

Exception in thread "main" java.util.InputMismatchException
    at java.base/java.util.Scanner.throwFor(Scanner.java:939)
    at java.base/java.util.Scanner.next(Scanner.java:1594)
    at java.base/java.util.Scanner.nextInt(Scanner.java:2258)
    at java.base/java.util.Scanner.nextInt(Scanner.java:2212)
    at Main.main(Main.java:21)

Process finished with exit code 1

And here's my code:

import java.util.Scanner;
import java.util.ArrayList;

public class Main extends MyZoo {

    public static void main(String[] args) {

        System.out.println("1.View all available zoo animals");

        Scanner input = new Scanner(System.in);
        System.out.print("Write Here: ");
        int UserInput = input.nextInt();

        while (true) {

            try
            {
                return input.nextInt();
            }
            catch (InputMismatchException e)
            {
                input.next();
                System.out.print("That’s not an integer. Try again: ");
            }
       }
    }
}

With the `try`-`catch` construct. I see you have already written one, so what's stopping you from writing a second one? — MC Emperor, Apr 25 '20 at 22:50
whats's try-catch construct? And how did i write it? (as i said i am begginer so i need you to be more spesific) — Tonia Giannoulaki, Apr 25 '20 at 22:56
also the intellij checks the command **return input.nextInt();** with red, i don't know why — Tonia Giannoulaki, Apr 25 '20 at 22:58
You cannot return an `int` from a method of which the return type is `void`. So `return input.nextInt()` cannot possibly work within `public static void main(String[] args)`. What exacly is it you want to achieve? — MC Emperor, Apr 25 '20 at 23:00
when i run the program if the user writes a string instead of integer , i want to tell the user that message: That’s not an integer. Try again! — Tonia Giannoulaki, Apr 25 '20 at 23:05
And i understand now that you told me about the return, thank you — Tonia Giannoulaki, Apr 25 '20 at 23:06
I suggest you study the [Java Tutorials](https://docs.oracle.com/javase/tutorial/java/index.html). It'll provide you with valuable details of the basics. — MC Emperor, Apr 25 '20 at 23:07

score 0 · Answer 1 · answered Apr 25 '20 at 23:01

Your try catch looks good, but you should place the first int UserInput = input.nextLine() in the try of course.

Probably you want something like this:

System.out.println("1.View all available zoo animals");

Scanner input = new Scanner(System.in);
while (true) {

    try
    {
        System.out.print("Write Here: ");
        int UserInput = input.nextInt();
        System.out.println("Your input: " + UserInput);
    }
    catch (InputMismatchException e)
    {
        input.next();
        System.out.print("That’s not an integer. Try again: ");
    }
}

score 0 · Answer 2 · answered Apr 25 '20 at 23:03

Your code has some flaws:

a main method doesn't (shouldn't) have a return statement
you need to import java.util.InputMismatchException
you don't assign the user input to your variable UserInput

Here is your code with the above issues removed. As is the code does nothing except getting input and printing it out. Try to add some condition to terminate your programm or add some logic to do something with the input.

import java.util.InputMismatchException;
import java.util.Scanner;

public class NewClass1 {
    public static void main(String[] args) {
        System.out.println("1.View all available zoo animals");

        Scanner input = new Scanner(System.in);

        int UserInput;

        while (true) {
            try
            {
                System.out.println("Write Here: ");
                UserInput = input.nextInt();
                System.out.println("Inputed value: " + UserInput);
            }
            catch (InputMismatchException e)
            {
                input.next();
                System.out.println("That’s not an integer. Try again: ");
            }
        }
    }
}

score 0 · Answer 3 · answered Apr 25 '20 at 23:10

I think the problem is in the catch part. It should be NumberFormatException instead of InputMismatchException.

The following code block works for you.

public static void main(String[] args) {
    System.out.println("1.View all available zoo animals");
    System.out.print("Write Here: ");

    Scanner scanner = new Scanner(System.in);
    String input = scanner.nextLine();

    while (true) {
        try {
            int userInput = Integer.parseInt(input);
        } catch (NumberFormatException e) {
            System.out.println("Wrong input.");
            System.out.print("Write Here: ");
            input = scanner.nextLine();
        }
    }
}

User put String instead of integer

3 Answers3