How to check the input is an integer or not in Java?

Question

In my program I want an integer input by the user. I want an error message to be show when user inputs a value which is not an integer. How can I do this. My program is to find area of circle. In which user will input the value of radius. But if user inputs a character I want a message to be shown saying Invalid Input.

This is my code:

int radius, area;
Scanner input=new Scanner(System.in);
System.out.println("Enter the radius:\t");
radius=input.nextInt();
area=3.14*radius*radius;
System.out.println("Area of circle:\t"+area);

score 28 · Accepted Answer · answered Nov 12 '13 at 09:14

If you are getting the user input with Scanner, you can do:

if(yourScanner.hasNextInt()) {
    yourNumber = yourScanner.nextInt();
}

If you are not, you'll have to convert it to int and catch a NumberFormatException:

try{
    yourNumber = Integer.parseInt(yourInput);
}catch (NumberFormatException ex) {
    //handle exception here
}

score 7 · Answer 2 · answered Nov 12 '13 at 09:18

You can try this way

 String input = "";
 try {
   int x = Integer.parseInt(input); 
   // You can use this method to convert String to int, But if input 
   //is not an int  value then this will throws NumberFormatException. 
   System.out.println("Valid input");
 }catch(NumberFormatException e) {
   System.out.println("input is not an int value"); 
   // Here catch NumberFormatException
   // So input is not a int.
 }

score 2 · Answer 3 · edited Nov 12 '13 at 10:24

2

Using Integer.parseIn(String), you can parse string value into integer. Also you need to catch exception in case if input string is not a proper number.

int x = 0;

try {       
    x = Integer.parseInt("100"); // Parse string into number
} catch (NumberFormatException e) {
    e.printStackTrace();
}

edited Nov 12 '13 at 10:24

Steve

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answered Nov 12 '13 at 09:12

Dark Knight

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I am very beginner and cannot understand your code. Can you please explain it? – Belal Khan Nov 12 '13 at 09:13
Also, you may want to change area from int to double. As with int you'll loose precision value. – Dark Knight Nov 12 '13 at 09:22
1

Yeah that seems a bit complicated. What you can read with the Scanner is a String. What @DarkKnight does right here, it's trying to convert the string into an integer (the parseInt line). The try/catch semantics is just there to notice if the program crashes. It means that the program will try something and if it *crashes*, it will do something. – Fabinout Nov 12 '13 at 09:22
@Fabinout Thanks a lot for putting it clearly. – Dark Knight Nov 12 '13 at 09:23

score 1 · Answer 4 · answered Nov 12 '13 at 09:13

If the user input is a String then you can try to parse it as an integer using parseInt method, which throws NumberFormatException when the input is not a valid number string:

try {

    int intValue = Integer.parseInt(stringUserInput));
}(NumberFormatException e) {
    System.out.println("Input is not a valid integer");
}

score 1 · Answer 5 · answered Nov 12 '13 at 09:18

You can use try-catch block to check for integer value

for eg:

User inputs in form of string

try
{
   int num=Integer.parseInt("Some String Input");
}
catch(NumberFormatException e)
{
  //If number is not integer,you wil get exception and exception message will be printed
  System.out.println(e.getMessage());
}

score 1 · Answer 6 · answered Nov 12 '13 at 09:29

        String input = "";
        int inputInteger = 0;
        BufferedReader br    = new BufferedReader(new InputStreamReader (System.in));

        System.out.println("Enter the radious: ");
        try {
            input = br.readLine();
            inputInteger = Integer.parseInt(input);
        } catch (NumberFormatException e) {
            System.out.println("Please Enter An Integer");
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        float area = (float) (3.14*inputInteger*inputInteger);
        System.out.println("Area = "+area);

How to check the input is an integer or not in Java?

6 Answers6

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