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 ittonight? 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 questionYou typically use double's rather than float's in Java. Howabout 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 youwant to send me your code, I can help you with it then.Suzanne BalikI 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 questionYou can use the Scanner class to do what you want to do easily,if you'reusing Java 5.0. For example,Scanner sc = new Scanner(System.in); if (sc.hasNextInt()) int value = sc.nextInt();Suzanne BalikNC State University EB II 2318 (919)515-5617We'll take a cup o' kindness yet, for auld lang syne. --Robert BurnsDo a search for Try catch tutorial Try statements and catch clauses are the way to go if you aregoing toseriously work with java.-----Original Message----- From: programmingblind-bounce@xxxxxxxxxxxxx [mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of AlexParksSent: Thursday, November 29, 2007 7:55 AM To: programmingblind@xxxxxxxxxxxxx Subject: RE: quick java questionI 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 questionIf you're expecting an integer, then you can doInteger.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 AlexParksSent: Thursday, November 29, 2007 12:51 AM To: programmingblind@xxxxxxxxxxxxx Subject: quick java questionHi all, Is there a quick way of determining if the user input is anumber? I need itto be a number; it throws an exception and exits the program ifit isn't.Maybe something like: if (input!=\int I have no idea about the syntax, but something along those linesis what Iam 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__________ 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__________ 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
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