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Start your free trialmichael lee
5,179 Pointsrename a new char but doesn't work
is it better to use the same name the use this.name or simply give it a new one? in my code there's something wrong, but I don't get it. Also how do I use my code normalizeDiscountCode in the applyDiscountCode method? thank you
public class Order {
private String itemName;
private int priceInCents;
private String discountCode;
public Order(String itemName, int priceInCents) {
this.itemName = itemName;
this.priceInCents = priceInCents;
}
public String getItemName() {
return itemName;
}
public int getPriceInCents() {
return priceInCents;
}
public String getDiscountCode() {
return discountCode;
}
private char normalizeDiscountCode(char normalDiscountCode){
normalDiscountCode = discountCode.toUpperCase(normalDiscountCode);
return normalDiscountCode;
}
public void applyDiscountCode(String discountCode) {
this.discountCode = discountCode;
}
}
public class Example {
public static void main(String[] args) {
// This is here just for example use cases.
Order order = new Order(
"Yoda PEZ Dispenser",
600);
// These are valid. They are letters and the $ character only
order.applyDiscountCode("abc");
order.getDiscountCode(); // ABC
order.applyDiscountCode("$ale");
order.getDiscountCode(); // $ALE
try {
// This will throw an exception because it contains numbers
order.applyDiscountCode("ABC123");
} catch (IllegalArgumentException iae) {
System.out.println(iae.getMessage()); // Prints "Invalid discount code"
}
try {
// This will throw as well, because it contains a symbol.
order.applyDiscountCode("w@w");
}catch (IllegalArgumentException iae) {
System.out.println(iae.getMessage()); // Prints "Invalid discount code"
}
}
}
1 Answer
Binyamin Friedman
14,615 PointsThe challenge says that normalizeDiscountCode should accept and return a String. The method should look like:
private String normalizeDiscountCode(String discountCode){
return discountCode.toUpperCase();
}
Remember that you can call a method with the method name and all the arguments separated by commas.
method(argument1, argument2)
In your case, the name of the method is normalizeDiscountCode and it requires one string as an argument (the discount code or discountCode). The method will return a String which is what you will set this.discountCode to. Try to figure the rest out by yourself ;)
michael lee
5,179 Pointsmichael lee
5,179 Pointsthis is what I did but it still didn't work, am I missing something?
``` public void applyDiscountCode(String discountCode) {
}
private String normalizeDiscountCode(String discountCode){ return discountCode.toUpperCase();
} }```
Binyamin Friedman
14,615 PointsBinyamin Friedman
14,615 PointsAlmost! When you call a method, (unless the return type is void) it will return a value, but generally won't modify the arguments.
normalizeDiscountCode(discountCode);
The method doesn't actually change discountCode, but returns a value which you can then set this.discountCode to:
this.discountCode = normalizeDiscountCode(discountCode)