How to find if a String is numeric?

[Cindy Glass]

How can I analyze the args[0] input on main and find if it is numeric before using it?

[Paul Caudle]

Hi Cindy,
Java 2 has a static method of the Double class called parseDouble that takes a String argument and returns a double.
If the String argument cannot be converted into a double, it throws a NumberFormatException.
If you put the Double.parseDouble("5.6") in a try block then an Exception will be thrown to your catch block. If the parse was successful, no exception will be thrown.
Let me know if that was too complicated or if you have found an easier way to do it (That would be great).
Hope that helps.
Paul

[Grant Crofton]

You could try
whatever = Double.parseDouble( args[0] );
which would throw a NumberFormatException if it wasn't a number.

[Grant Crofton]

Great minds...

[Paul Caudle]

Here's an example:
in a method somewhere:
String numWord = JTextField1.getText();
double num;
try{
//Attempt to convert to a number
num = Double.parseDouble(numWord);
}
catch(Exception e){
//tell the user if it wasn't a number
JOptionPane.showMessageDialog(this, "Not a valid Number");
//exit the method
return;
}
//here you can do whatever it is you were going to do with that number

That might shed some light on the convoluted explanation from before
Paul

[Anonymous]

Try this:

boolean isNumeric = true;
for(i=0;i<args[0].getLength();i++){>
    if(!Character.isDigit(args[0].charAt(i))){
        isNumeric = false;
    }
}
if(isNumeric){
    // your argument is numeric, so go ahead and use it...
}

hope that helps.

[Paul Caudle]

that's funny, Grant...agreed about the minds, too

[Cindy Glass]

Both these methods are great! Thanks - it just seems such a common place thing to do that I am amazed that the String class doesn't have an isNumeric method.