C# How can I return my base class in a webservice

Question

I have a class Car and a derived SportsCar: Car
Something like this:

public class Car
{
    public int TopSpeed{ get; set; }
}


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

I have a webservice with methods returning Cars i.e:

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

It returns:

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

I have another method that also returns cars like this:

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

It compiles fine and everything, but when invoking it I get:
System.InvalidOperationException: There was an error generating the XML document. ---> System.InvalidOperationException: The type wsBaseDerived.SportsCar was not expected. Use the XmlInclude or SoapInclude attribute to specify types that are not known statically.

I find it strange that it can't serialize this as a straight car, as mycar is a car.

Adding XmlInclude on the WebMethod of ourse removes the error:

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

and it now returns:

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

But I really want the base class returned, without the extra properties etc from the derived class.

Is that at all possible without creating mappers etc?

Please say yes ;)

Answer 1

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

Good luck.

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

Answer 2

Just a stab, but have you tried this? Cast on the return.

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

Answer 3

Do this:

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

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

I know it's a pain but i don't know of an alternative.

Answer 4

Try WCF, there is [KnownType].

Don't know, however, if thats possible with oldskool SOAP webservices, though.

Answer 5

The other comments and answers here got me thinking, and if I need to make a Mapper-method so be it:

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

}

and the webmethod:

[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 purpose of having a derived object in my view, but until someone comes up with another solution thats the way I'll fly...

Answer 6

If you want it to always serialize as "Car" instead of "SportsCar" add an [XmlRoot("Car")] to it.

public class Car {
  //stuff
}

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

Edit: Using the XmlSerializer like this:

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>

Answer 7

I would implement a 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; }
    }

Then you can return a new Car based on the SportsCar in the GetMyCar-method. I think this way clearly express the intent of the method.

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