RE: quick java question

  • From: Alex Parks <mehgcap@xxxxxxx>
  • To: programmingblind@xxxxxxxxxxxxx
  • Date: Thu, 29 Nov 2007 08:54:38 -0500

I am new to this language. What do I put in place of "aString" and how do I use the exception in my if statement? Add a throwsException to the end? Thanks. 


Have a great day,
Alex


----- Original Message -----
From: "Sina Bahram" <sbahram@xxxxxxxxx
To: <programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 07:07:27 -0500
Subject: RE: quick java question



If you're expecting an integer, then you can do



Integer.parseInt(aString)



That will throw a NumberFormatException if it isn't a number.



Take care,
Sina




-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Alex 
Parks

Sent: Thursday, November 29, 2007 12:51 AM
To: programmingblind@xxxxxxxxxxxxx
Subject: quick java question



Hi all,
Is there a quick way of determining if the user input is a 
number? I need it
to be a number; it throws an exception and exits the program if 
it isn't.

Maybe something like:
if (input!=\int
I have no idea about the syntax, but something along those lines 
is what I

am looking for.  Thanks for any help.



Have a great day,
Alex
__________
View the list's information and change your settings at
//www.freelists.org/list/programmingblind



__________
View the list's information and change your settings at
//www.freelists.org/list/programmingblind



__________
View the list's information and change your settings at //www.freelists.org/list/programmingblind

Other related posts:

  1. Home
  2. programmingblind
  3. Archive
  4. 11-2007