Re: quick java question

Here is an example using an integer. (Sorry, but I had not remembered that you were using a double when I began quickly working up this example, and I'm off to bed for now.) Remember that if you enter a char type, it will still work as it will output the ascii value! You can swap the int for a double and it should work. Hope this helps! 

import java.util.Scanner;


public class TestInt {


public static void main(String[] args) {

System.out.print("Please enter an integer: ");

Scanner sc = new Scanner(System.in);

if (sc.hasNextInt()) {

System.out.println("Your integer is " + sc.nextInt());

} else {

System.out.println("You did not enter an integer!");

}

}


}



Kind regards,

Jeff
----- Original Message ----- From: "Sina Bahram" <sbahram@xxxxxxxxx> 
To: <programmingblind@xxxxxxxxxxxxx>
Sent: Thursday, November 29, 2007 10:19 PM
Subject: RE: quick java question
If you'd like some help on this tonight, shoot me the file off list ... I'm busy tomorrow, but hopefully Suzanne can help you then, if we don't get it 
tonight?

sbahram@xxxxxxxxx

Take care,
Sina


-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Alex Parks
Sent: Thursday, November 29, 2007 9:55 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: RE: quick java question

Okay, thanks! I will attach the .java file to an email and send it to you
privately.  It will save my poor cs professor from answering all my
questions and give me a new perspective.  Thanks again.

Have a great day,
Alex


----- Original Message -----
From: "Suzanne Balik" <spbalik@xxxxxxxxxxxxxx
To: programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 21:30:35 -0500 (EST)
Subject: RE: quick java question



You typically use double's rather than float's in Java.  How

about this --

would this work for you?



Scanner sc = new Scanner(System.in);
if (sc.hasNextDouble())
 double value = sc.nextDouble();



I'll be in my office some time after 3:00 PM tomorrow (Friday).

If you

want to send me your code, I can help you with it then.



Suzanne Balik







I tried it, but have changed the var to a float after I sent the
message.  I tried var.hasNextFloat() but I get an error on that line
that says something like "float cannot be dereferenced".  I am
checking to be sure a number (the var in question) is greater than 1
but less than another var and so thought I could just put this simple
statement as another condition in the if statement, but I got the
above error.  I am using the latest Java.



Have a great day,
Alex



----- Original Message -----
From: "Suzanne Balik" <spbalik@xxxxxxxxxxxxxx
To: programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 21:00:11 -0500 (EST)
Subject: RE: quick java question



You can use the Scanner class to do what you want to do easily,

if you're

using Java 5.0.  For example,



Scanner sc = new Scanner(System.in);
if (sc.hasNextInt())
 int value = sc.nextInt();





Suzanne Balik



NC State University      EB II 2318    (919)515-5617



We'll take a cup o' kindness yet, for auld lang syne.
                                      --Robert Burns




Do a search for
Try catch tutorial
Try statements and catch clauses are the way to go if you are

going to

seriously work with java.




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

Parks

Sent: Thursday, November 29, 2007 7:55 AM
To: programmingblind@xxxxxxxxxxxxx
Subject: RE: quick java question




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
http://www.freelists.org/list/programmingblind



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




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




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





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





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





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




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

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





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

Other related posts: