C # How can I return the base class in webservice

Question

I have a Car class and a derivative of SportsCar: Car
Something like that:

public class Car { public int TopSpeed{ get; set; } } public class SportsCar : Car { public string GirlFriend { get; set; } } 

I have a web service with methods that return cars ie:

 [WebMethod] public Car GetCar() { return new Car() { TopSpeed = 100 }; } 

It returns:

 <Car> <TopSpeed>100</TopSpeed> </Car> 

I have another method that also returns such cars:

 [WebMethod] public Car GetMyCar() { Car mycar = new SportsCar() { GirlFriend = "JLo", TopSpeed = 300 }; return mycar; } 

It compiles everything and everything, but when I call, I get:
System.InvalidOperationException: An error occurred while generating an XML document. ---> System.InvalidOperationException: Type wsBaseDerived.SportsCar was not expected. Use the XmlInclude or SoapInclude attribute to indicate types that are not known statically.

It seems strange to me that he cannot serialize it as a direct car, since my car is a car.

Adding XmlInclude to WebMethod of our method fixes the error:

 [WebMethod] [XmlInclude(typeof(SportsCar))] public Car GetMyCar() { Car mycar = new SportsCar() { GirlFriend = "JLo", TopSpeed = 300 }; return mycar; } 

and now it returns:

 <Car xsi:type="SportsCar"> <TopSpeed>300</TopSpeed> <GirlFriend>JLo</GirlFriend> </Car> 

But I really want to return the base class, without additional properties, etc. from a derived class.

Is this possible without creating maps, etc.?

Say yes;)

Answer 1

I would execute the copy constructor in the base class.

  public class Car { public int TopSpeed { get; set; } public Car(Car car) { TopSpeed = car.TopSpeed; } public Car() { TopSpeed = 100; } } public class SportsCar : Car { public string GirlFriend { get; set; } } 

You can then return the new SportsCar-based car in the GetMyCar method. I think this method clearly expresses the intention of the method.

  public Car GetMyCar() { var sportsCar = new SportsCar { GirlFriend = "JLo", TopSpeed = 300 }; return new Car(sportsCar); } 

Answer 2

Do it:

 [WebMethod] public Car GetMyCar() { Car mycar = new SportsCar() { GirlFriend = "JLo", TopSpeed = 300 }; return new Car() {TopSpeed = mycar.TopSpeed}; } 

The reason is because the XMLSerializer checks the type of the GetType () object and expects it to be the same as the declared one.

I know this is a pain, but I do not know an alternative.

Answer 3

Or use the XmlIgnoreAttribute attributes that you want to ignore. Or an even more painful way would be to implement custom serialization for your attributes.

Good luck.

 public class SportsCar : Car { [XmlIgnoreAttribute] public string GirlFriend { get; set; } } 

Answer 4

Just a hit, but have you tried it? Throw on a refund.

 [WebMethod] public Car GetMyCar() { Car mycar = new SportsCar() { GirlFriend = "JLo", TopSpeed = 300 }; return (Car)mycar; } 

Answer 5

Try WCF, [KnownType] exists.

I do not know, however, if this is possible using the oldskool SOAP web services.

Answer 6

Other comments and answers here made me think, and if I need to create a Mapper method, let it be like this:

 public class Car: ICloneable { public int TopSpeed{ get; set; } public object Clone() { return new Car() { TopSpeed = this.TopSpeed }; } } 

and web method:

 [WebMethod] public Car GetMyNewCar() { Car mycar = new SportsCar() { GirlFriend = "HiLo", TopSpeed = 300 }; return (Car)mycar.Clone(); } 

This works as expected:

 <Car> <TopSpeed>300</TopSpeed> </Car> 

This defeats the goal of having a derived object in my view, but until someone comes up with another solution that I will fly ...

Answer 7

If you want to always be serialized as "Car" instead of "SportsCar", add [XmlRoot("Car")] .

 public class Car { //stuff } [XmlRoot("Car")] public class SportsCar { //Sporty stuff } 

Edit: using XmlSerializer as follows:

 XmlSerializer ser = new XmlSerializer(typeof(SportsCar)); SportsCar car = new SportsCar() //stuff ser.Serialize(Console.out, car) 

you should get

 <Car> <TopSpeed>300</TopSpeed> <GirlFriend>JLo</GirlFriend> </Car>