NullPointerException help?

[fab5freddy]

I am fairly new to java. I have only been working with it for about two weeks. I'm trying to teach myself as I go to see what it's like before I try to enroll in any classes. The code I am writing is meant to tell a person how much to pay if they were to go to the movies and also if they have a coupon or not. It does this by asking your age and if you have a coupon or not. However each time I run the code I'm able to type in the age, but when it ask if you have a coupon or not I get the error message. "Exception in thread "main" java.lang.NullPointerException at YouCanDoIt.main(YouCanDoIt.java:16)" Here is a listing of my code:

import java.util.Scanner;

class YouCanDoIt {

public static void main(String args[]) {
Scanner myScanner = new Scanner(System.in);
int age;
double price = 0.00;
char reply;
boolean isKid, isSenior, hasCoupon, hasNoCoupon;

System.out.print("How old are you? ");
age = myScanner.nextInt();

System.out.print("Have a coupon? (Y/N) ");
reply = myScanner.findInLine(".").charAt(0);

isKid = age < 12;
isSenior = age >= 65;
hasCoupon = reply == 'Y' || reply == 'y';
hasNoCoupon = reply == 'N' || reply == 'n';

if (!isKid && !isSenior) {
price = 9.25;
}
if (isKid || isSenior) {
price = 5.25;
}
if (hasCoupon) {
price -= 2.00;
}
if (!hasCoupon && !hasNoCoupon) {
System.out.println("Huh?");
}

System.out.print("Please pay $");
System.out.print(price);
System.out.print(". ");
System.out.println("Enjoy the show!");
}
}
The more detail you can provide would help me greatly as I said I am quite new to this.

[jojo]

This will probally return null:
myScanner.findInLine(".")

make sure ur Scanner object is also not null

[[RaIdEn]]

System.out.print("Have a coupon? (Y/N) ");

-------------------------------------
reply = myScanner.findInLine(".").charAt(0);
-------------------------------------

do you know whats findinline doing?Its purpose is not meant to read the users input,its purpose is to read the already specified Strings.

Attempts to find the next occurrence of a pattern constructed from the specified string, ignoring delimiters.

Instead try using Scanner.nextLine() or Scanner.next.

String whatever = Scanner.nextLine();
//and check whether that String is Y or N

This does the job but got some minor issues.:)