Display an Error for invalid input in JAVA

Question

I have written a JAVA program that takes input from the user and check if the user has entered the correct thing or not. I have taken the input from the Scanner class. If the user enters an invalid character like a String, I want to display 'Invalid Input'.

public class Main {
    public static void main(String[] args) {
        Scanner takeInteger = new Scanner(System.in);
        System.out.println("Enter a number");
        int enteredNumber = takeInteger.nextInt();
    }
}

And what is a valid input? what did you try until now?

Filip
– Filip

2021-07-07 15:49:39 +00:00
Commented Jul 7, 2021 at 15:49 — Filip
– Filip, Commented Jul 7, 2021 at 15:49
A valid input is an integer.

Hardik
– Hardik

2021-07-07 16:00:04 +00:00
Commented Jul 7, 2021 at 16:00 — Hardik
– Hardik, Commented Jul 7, 2021 at 16:00

Holger · Accepted Answer · 2021-07-07 16:50:49Z

4

Just ask the Scanner whether the next input is a valid int value, e.g.

Scanner takeInteger = new Scanner(System.in);
System.out.println("Enter a number");
while(!takeInteger.hasNextInt()) {
    System.out.println("Invalid Input: " + takeInteger.next());
}
int enteredNumber = takeInteger.nextInt();

This will retry the operation until the user entered a number. If you just want a single attempt, use something like

Scanner takeInteger = new Scanner(System.in);
System.out.println("Enter a number");
if(!takeInteger.hasNextInt()) {
    System.out.println("Invalid Input: " + takeInteger.next());
}
else {
    int enteredNumber = takeInteger.nextInt();
    // ... proceed with the input
}

answered Jul 7, 2021 at 16:50

Holger

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Comments

Ashish Mishra · Accepted Answer · 2021-07-07 16:05:16Z

1

You will get an Exception that is InputMismatchException when an invalid input is passed.(i.e except integer value),you can use a try-catch block to hold the exception and inform the user about the invalid input. Try block , Catch block

    import java.util.*; 

    Scanner takeInteger = new Scanner(System.in);
    System.out.println("Enter a number");
    try{
       int enteredNumber = takeInteger.nextInt();
    }
    catch(InputMismatchException e) {
        System.out.println("Enter a valid input");
    }

edited Jul 7, 2021 at 16:05

answered Jul 7, 2021 at 15:52

Ashish Mishra

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2 Comments

Hardik Over a year ago

It says that Cannot resolve symbol 'InputMismatchException'

Ashish Mishra Over a year ago

@Hardik I thought you may have imported the package java.util using the wildcard import java.util.* but it looks like that you have imported only the Scanner class you can use this class -> import java.util.InputMismatchException;

jpllosa · Accepted Answer · 2021-07-08 10:43:03Z

1

You can use Exception handling for the same.

public class Main {
     public static void main(String[] args) {
         Scanner takeInteger = new Scanner(System.in);
         System.out.println("Enter a number");
         try {
             int enteredNumber = takeInteger.nextInt();
         }
         catch (Exception e) {
             System.out.println("Invalid Input");
         }
     }
}

edited Jul 8, 2021 at 10:43

jpllosa

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answered Jul 7, 2021 at 15:53

Sonam Gupta

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2 Comments

Hardik Over a year ago

When I implement the following code, it works. But let's say that I want to print out the addition of the numbers. When I take an input ( the second number ) and print it using System.out.println, for 'enteredNumber ,it says that 'Cannot resolve symbol 'enteredNumber''

Sonam Gupta Over a year ago

@Hardik It seems you are taking another input or performing addition outside 'try' block . Try doing this inside the ' try ' block only after you have taken the first input .

GURU Shreyansh · Accepted Answer · 2021-07-07 15:51:52Z

0

You can add a try-catch block in your program to check if the user's input is a number or not.

        Scanner takeInteger = new Scanner(System.in);
        System.out.println("Enter a number");
        String input = takeInteger.next();
        int enteredNumber;
        try
        {
            enteredNumber = Integer.parseInt(input); // Input is a number
        }
        catch(NumberFormatException ex)
        {
            System.out.println("Wrong Input!"); // Invalid input
        }

answered Jul 7, 2021 at 15:51

GURU Shreyansh

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Comments

mamadaliev · Accepted Answer · 2021-07-07 15:53:58Z

0

You need to call the .nextLine method of the Scanner class and then parse to the desired type.

Example:

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a number");
        String line = sc.nextLine();

        try {
            int enteredNumber = Integer.parseInt(line);
            System.out.println("You have entered: " + enteredNumber);
        } catch (Exception e) {
            System.out.println("Invalid Input");
        }
    }
}

Result with a number:

Enter a number
12
You have entered: 12

Result a text:

Enter a number
abcd
Invalid Input

answered Jul 7, 2021 at 15:53

mamadaliev

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