CODEWITHSUNDEEP
c#classserializationhierarchy
JaeRae
JaeRae

Reputation: 31

How Do I use C# deserialization with a hierarchichal class structure

I'm trying to deserialize some xml files into some classes which have been simplified to the following:

[XmlRoot("person")]
[Serializable]
public class Person
{
    [XmlElement]
    public Toy Toy { get; set; }
}

[Serializable]
public class ActionMan : Toy
{
    [XmlElement("guns")]
    public string Guns;
}

[Serializable]

public class Doll : Toy
{
    [XmlElement("name")]
    public  String Name;
}

[XmlInclude(typeof(Doll))]
[XmlInclude(typeof(ActionMan))]
public class Toy
{
}

[TestFixture]
public class ToyTest
{
    [Test]
    public void testHierarchy()
    {
        String filePath = @"test\brother.xml";
        String sisfilePath = @"test\sister.xml";
        var serializer = new XmlSerializer(typeof(Person));
        Person brother = (Person)serializer.Deserialize(new FileStream(filePath, FileMode.Open));
        Person sister = (Person)serializer.Deserialize(new FileStream(sisfilePath, FileMode.Open));

        Assert.IsNotNull(brother);
        Assert.IsNotNull(sister);
        Assert.IsAssignableFrom(typeof(ActionMan),brother.Toy);
        Assert.IsAssignableFrom(typeof(Doll),sister.Toy);
    }
}

I want to use the c# Serialisation (I know I can use my own deserialiser) and I think I'm perhaps simply missing a particular tag that I don't know about (and I'm sure I've got superfluous tags).

here is one fo the xml files:

<person>
  <doll>
    <name>Jill</name>
  </doll>
</person>

the error I get is "Expected: assignable from " on the third assert

Upvotes: 3

Views: 1314

Answers (4)

Carlos Blanco
Carlos Blanco

Reputation: 8772

Try the following

public class Person 
{

    public Toy toy
    {
        get
        {
            return (doll == null) ? (Toy)actionMan : (Toy)doll;
        }

    }


    public Doll doll;
    public ActionMan actionMan;
}


public class Toy 
{

}


public class Doll : Toy
{
    public String name;

}


public class ActionMan : Toy
{
    public String guns;
}

class Program
{
    static void Main(string[] args)
    {

        Person brother = new Person();
        ActionMan am = new ActionMan();
        am.guns = "Laser Beam";
        brother.actionMan = am;

        Person sister = new Person();
        Doll d = new Doll();
        d.name = "Jill";
        sister.doll = d;

        Serialize(brother, "brother.xml");
        Serialize(sister, "sister.xml");

        Person b = Deserialize("brother.xml");
        Person s = Deserialize("sister.xml");

        Console.WriteLine(((ActionMan)b.toy).guns);
        Console.WriteLine(((Doll)s.toy).name);
        Console.Read();
    }

    public static Person Deserialize(String filename)
    {
        var serializer = new XmlSerializer(typeof(Person));
        return (Person)serializer.Deserialize(new FileStream(filename, FileMode.Open));

    }

    public static void Serialize(Person p, String filename){
        Stream stream = File.Open(filename, FileMode.Create);
        XmlSerializer s = new XmlSerializer(typeof(Person));
        s.Serialize(stream, p);
        stream.Close();

    }

You can expand from here. Remember that, in the attribute names, casing matters. The serialization output I got is

brother.xml

<?xml version="1.0"?>
<Person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <actionMan>
    <guns>Laser Beam</guns>
  </actionMan>
</Person>

sister.xml

<?xml version="1.0"?>
<Person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <doll>
    <name>Jill</name>
  </doll>
</Person>

The output is

Laser Beam
Jill

Upvotes: 0

John Alexiou
John Alexiou

Reputation: 29264

What I have done, is designed the class structure the way I want, fill in some basic data and then serialize it. Then examine how it serializes and adjust the Xml attributes. If your Toy class has only a few derivatives, then you can de-serialize them seperately with a action-man field and a doll field that might be null or not.

Alternatively to go from an Xml file to a c# class I use the xsd.exe tool to generate a .xsd file with xsd mydata.xml and then from that a c# class file with xsd /c /l:cs mydata.xsd. Then I examine the class data to get clues as to how to define my class and what attributes to use.

link to xsd tool from Microsoft.

Upvotes: 1

Euphoric
Euphoric

Reputation: 12849

When I try to serialize your structure (person, who has ActionMan as Toy) i get

<person>
  <Toy xsi:type="ActionMan" />
</person>

I guess this is how you can handle inheritance of your types. But I guess you cant change your already serialised XML.

Upvotes: 1

Carlos Blanco
Carlos Blanco

Reputation: 8772

Class person should contain a "doll" attribute instead of a "Toy" attribute, I mean the name. The XML node must have the same name as the attribute name -casing matters.

Upvotes: 1

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