CODEWITHSUNDEEP
java
user3817065
user3817065

Reputation: 11

Exception in thread "main" java.lang.NumberFormatException: For input string: PLEASE TRY IT TO YOUR PC

This program will solve the numbers that you will input but I want it to show ERROR if I will input some letters. Thanks for helping

This is my code

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;

class jedz
{
    public static void main (String args[])
    {
       BufferedReader dataln = new BufferedReader(new InputStreamReader(System.in));
       int a, b, c;
       String num1 =" ";
       String num2 =" ";
       System.out.print("Enter FN: ");
       try {
          num1 = dataln.readLine();
       }
       catch (IOException e) 
       {
          System.out.println("Error");
       }
       a = Integer.parseInt(num1);
       System.out.print("Enter SN: ");
       try
       {
           num2 = dataln.readLine();
       }
       catch (IOException e) 
       {
           System.out.println("Error");    
       }
       b = Integer.parseInt(num2);
       c = a + b;
       System.out.print("The answer is "+ c) ;
    }
}

This is the output when I put some letters.

D:\>java jedz
Enter FN: s
    Exception in thread "main" java.lang.NumberFormatException: For input string: "s "
        at java.lang.NumberFormatException.forInputString(NumberFormatException. java:65)
        at java.lang.Integer.parseInt(Integer.java:492)
        at java.lang.Integer.parseInt(Integer.java:527)
        at jedz.main(jedz.java:21)

Upvotes: 0

Views: 5277

Answers (7)

kajacx
kajacx

Reputation: 12939

change

a = Integer.parseInt(num1);

to

try {
  a = Integer.parseInt(num1);
} catch(NumberFormatException ex) {
  System.out.println("Error, "+num1+" isn't a whole number!");
}

Upvotes: 0

JBires
JBires

Reputation: 441

Try using a method to convert your characters into their ASCII representations.

Upvotes: 0

BackSlash
BackSlash

Reputation: 22233

I want it to show ERROR if I will input some letters

You can use the Scanner class to check if there is an int available:

Scanner dataln = new Scanner(System.in);

if(dataln.hasNextInt()) {
    a = dataln.nextInt();
} else {
    //no ints, display error
}

Otherwise you can use your current solution and add a try-catch block

try{
    a = Integer.parseInt(num1);
}
catch (NumberFormatException e) 
{
    System.out.println("Error");
}

The first solution gives you better performance. See this question

Upvotes: 2

Stultuske
Stultuske

Reputation: 9437

Well ... "s" is a valid String, but it is not a numerical value, which is why you can not cast it to a number.

So, if you run this:

a = Integer.parseInt(num1);

And the value of num1 = "5", then there's no problem, because 5 is a numerical value within the boundaries of an int. An "s" however, will throw the exception you get.

Upvotes: 0

christopher
christopher

Reputation: 27356

No effort? Then tutorials for you!

Firstly, you need to learn about basic validation in Java. In short, you're trying to put some letters into a denary number, which it isn't going to like. You need to read a bit about Regex matching. You can use regular expressions to test the content of String variables.

Secondly, look into Exception Handling. This allows you to catch an error that is thrown, and output a meaningful message to the user.

Your Problem...

a = Integer.parseInt(num1);

Will throw an exception because you're not checking the value.

Upvotes: 0

Suresh Atta
Suresh Atta

Reputation: 122008

cause:

a = Integer.parseInt(num1);

The above line executes successfully then and then only when user enters a correct integer string.

That obvious that the string "s" is not a valid integer.

solution:

One thing you can do wrap your parseInt code in try catch block and prompt user in catch block with message "invalid input"

Upvotes: 0

Pracede
Pracede

Reputation: 4361

a = Integer.parseInt(num1);

Your are parsing num1 which is String to Integer. This causes exception

Upvotes: 1

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