Figuring out if a number is an input in Java? [duplicate]

Question

I made a program where I ask the user to input a number out of 0-8, and if it isn't a number a message will appear asking to input a number. If a letter is typed, on the next line it will say "please enter a number."

How do I do this? If it's more than 8 it stays 8 while if it is less than 0 it stays 0.

        if (number >= 8) {
            number = 8;
        } else if (number <= 0) {
            number = 0;
        }

UPDATE:

    System.out.println("Enter The Number you want:<0 - 8> "); 
    number = in.nextInt(); \\ in is the name of scanner 

    try {
      if number = in.nextInt(); 
      if(number >= 8) {
        number = 8;
      } else if (number <= 0) { 
        number = 0;
      }
    } catch (NumberFormatException e) {
      System.out.println("Enter a number");
    }

This didn't work, what am I doing wrong?

Look into Integer.parseInt(String), and see how it handles receiving something that is not a number. — Henrik Aasted Sørensen
– Henrik Aasted Sørensen, Commented Jan 27, 2014 at 10:53
Research.. try something.. if it doesn't work, come back for help with the code you attempted. — Paul Samsotha
– Paul Samsotha, Commented Jan 27, 2014 at 10:53
*** I did try the integer.parseint(String) method and the try catch method. I'll look into those links. — user3240125
– user3240125, Commented Jan 27, 2014 at 10:55
Perhaps using a third party library? commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/… — Ben Dale
– Ben Dale, Commented Jan 27, 2014 at 11:05

Nigel Tufnel · Accepted Answer · 2014-01-27 10:53:53Z

2

You can use the Scanner class to read from the System.in and check if the string is a number:

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

Scanner sc = new Scanner(System.in);
try {
    int i = sc.nextInt();
} catch (InputMismatchException exc) {
    System.out.println("not a number");
}

answered Jan 27, 2014 at 10:53

Nigel Tufnel

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6 Comments

user3240125 Over a year ago

' System.out.println("Enter The Number you want:<0 - 8> "); number = in.nextInt(); \\ in is the name of scanner try { if number = in.nextInt(); if(number >= 8) { number = 8; } else if (number <= 0) { number = 0; } } catch (NumberFormatException e) { System.out.println("Enter a number"); } ' When I tried running it the catch didn't work.

Nigel Tufnel Over a year ago

You should catch an InputMismatchException, not a NumberFormatException.

user3240125 Over a year ago

` System.out.println("Enter The Number you want:<0 - 8> "); number = in.nextInt(); try { number = in.nextInt(); if(number >= 8) { number = 8; } else if (number <= 0) { number = 0; } } catch (InputMismatchException e) { System.out.println("Enter a number please"); }` Ran it again and the command line said "Exception in thread "main" java.util.InputMismatchException" etc.

Nigel Tufnel Over a year ago

Well, you're calling in.nextInt outside of the try-catch block. I don't even know why do you need two in.nextInt calls.

user3240125 Over a year ago

So I did the code that you provided from scratch again, and it worked thank you! But it didn't stop the program, the program should end right when the user inputs a letter when it should be a number. How do I do that?

|

Ben Barkay · Accepted Answer · 2014-01-27 10:56:26Z

0

In addition to Nigel Tufnel's answer, if you are not using Scanner (but BufferedReader instead, for instance), you can use the following:

System.out.print("Please enter a number: ");
String numberString = in.readLine();
try {
    int number = Integer.parseInt(numberString);
    System.out.println("You chose number " + number + "!");
} catch (NumberFormatException nfe) {
    System.out.println("Please enter a number");
}

answered Jan 27, 2014 at 10:56

Ben Barkay

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1 Comment

user3240125 Over a year ago

I am using Scanner but thank you for answering.

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