How do I deserialize XML without knowing the type beforehand?

Question

Say I have a couple of basic objects like so:

[Serializable]
public class Base
{

    public string Property1 { get; set; }
    public int Property2 { get; set; }
}

[Serializable]
public class Sub: Base
{
    public List<string> Property3 { get; set; }

    public Sub():base()
    {
        Property3 = new List<string>();
    }        
}

And I serialize them like so:

Sub s = new Sub {Property1 = "subtest", Property2 = 1000};
s.Property3.Add("item 1");
s.Property3.Add("item 2");

XmlSerializer sFormater = new XmlSerializer(typeof(Sub));
using (FileStream fStream = new FileStream("SubData.xml", 
    FileMode.Create, FileAccess.Write, FileShare.None))
{
    sFormater.Serialize(fStream, s);
}

How can I deserialize them, so that I get back the correct class?

As in, I'd want something like this

XmlSerializer bFormater = new XmlSerializer(typeof (Base));
Base newBase;
using (FileStream fStream = new FileStream("BaseData.xml", 
    FileMode.Open, FileAccess.Read, FileShare.Read))
{
    newBase = (Base) bFormater.Deserialize(fStream);
}

Except I'd be able to pass it an XML file for any class that descends from Base and the correct class would be created.

I'm thinking I could read the name of the root node of the XML and use a switch statement to create the correct XmlSerializer, but I was wondering if there was a simpler way.

Answer 1

You can read the XML file's root node and instead of using a switch statement, you can write your code like this -

Type yourType = Type.GetType("Your Type");
XmlSerializer xs = new XmlSerializer(yourType);

I don't think there's any way other than reading the XML because if you don't know the type, you can't do anything.

Answer 2

Use the [XmlInclude] attribute on your base class to tell the XML serializer about derived classes, so it can figure out what to create. Your last code snippet should then work correctly.