How to avoid Number Format Exception in java? [dup

Question

This question already has an answer here:

• What is a NumberFormatException and how can I fix it? [duplicate] 9 answers

In my day to day web application development there are many instances where we need to take some number inputs from the user.

Then pass on this number input to may be service or DAO layer of the application.

At some stage since its a number (integer or float), we need to convert it into Integer as shown in the following code snippet.

String cost = request.getParameter("cost");

if (cost !=null && !"".equals(cost) ){
    Integer intCost = Integer.parseInt(cost);
    List<Book> books = bookService . findBooksCheaperThan(intCost);  
}

Here in the above case I have to check if the input is not null or if there is no input (blank) or sometimes there is a possibility of a non number inputs e.g. blah, test etc.

What is the best possible way of handling such situations?

Answer 1

public class Main {
    public static void main(String[] args) {

        String number;

        while(true){

            try{
                number = JOptionPane.showInputDialog(null);

                if( Main.isNumber(number) )
                    break;

            }catch(NumberFormatException e){
                System.out.println(e.getMessage());
            }

        }

        System.out.println("Your number is " + number);

    }

    public static boolean isNumber(Object o){
        boolean isNumber = true;

        for( byte b : o.toString().getBytes() ){
            char c = (char)b;
            if(!Character.isDigit(c))
                isNumber = false;
        }

        return isNumber;
    }

}

Answer 2

I suggest to do 2 things:

• validate the input on client side before passing it to the Servlet
• catch the exception and show an error message within the user frontend as Tobiask mentioned. This case should normally not happen, but never trust your clients. ;-)

Answer 3

Exceptions in recent versions of Java aren't expensive enough to make their avoidance important. Use the try/catch block people have suggested; if you catch the exception early in the process (i.e., right after the user has entered it) then you're not going to have the problem later in the process (because it'll be the right type anyway).

Exceptions used to be a lot more expensive than they are now; don't optimize for performance until you know the exceptions are actually causing a problem (and they won't, here.)

Answer 4

Just catch your exception and do proper exception handling:

if (cost !=null && !"".equals(cost) ){
        try {
           Integer intCost = Integer.parseInt(cost);
           List<Book> books = bookService . findBooksCheaperThan(intCost);  
        } catch (NumberFormatException e) {
           System.out.println("This is not a number");
           System.out.println(e.getMessage());
        }
    }

Answer 5

To Determine if a string is Int or Float and to represent in longer format.

Integer

 String  cost=Long.MAX_VALUE+"";
  if (isNumeric (cost))    // returns false for non numeric
  {  
      BigInteger bi  = new BigInteger(cost);

  }

public static boolean isNumeric(String str) 
{ 
  NumberFormat formatter = NumberFormat.getInstance(); 
  ParsePosition pos = new ParsePosition(0); 
  formatter.parse(str, pos); 
  return str.length() == pos.getIndex(); 
} 

Answer 6

In Java there's sadly no way you can avoid using the parseInt function and just catching the exception. Well you could theoretically write your own parser that checks if it's a number, but then you don't need parseInt at all anymore.

The regex method is problematic because nothing stops somebody from including a number > INTEGER.MAX_VALUE which will pass the regex test but still fail.