使用IXmlSerializable对派生类进行C＃序列化

Question

我会尽量让问题变得简单。以下是2个示例程序来演示它。如果你需要分析它，你可以将它复制并粘贴到Visual Studio中。

随意评论并更正我的英语。希望你能理解我。

我的目标是： 我们有2个类 - BaseClass和DerivedClass。

我们需要：

1. 创建DerivedClass的实例
2. 将其分配给BaseClass类型的变量
3. 将该变量序列化为XML文件
4. 将该文件反序列化为BaseClass类型的变量
5. 该变量应包含DerivedClass类型的对象

第一个示例程序使用“内置”序列化 - 并且工作正常。

但是我希望对序列化过程有更多的控制（使用IXmlSerializable接口） AND 达到同样的效果。

第一个例子：

[Serializable]
public class BaseClass
{
    public string property1;
    public string property2;

    public BaseClass()
    {
    }
}

[Serializable]
public class DerivedClass : BaseClass
{
    public string property3;

    public DerivedClass()
    {
    }
}

class Program
{
    static void Main(string[] args)
    {
        XmlSerializer mySer = new XmlSerializer(typeof(BaseClass), new Type [] {typeof(DerivedClass)});

        BaseClass exampleBaseClass;
        DerivedClass exampleDerivedClass = new DerivedClass();
        exampleDerivedClass.property1 = "Foo";
        exampleDerivedClass.property2 = "Bar";
        exampleDerivedClass.property3 = "Sth";

        exampleBaseClass = exampleDerivedClass;

        Console.WriteLine(exampleBaseClass.GetType());   // <----- 1
        StreamWriter sw = new StreamWriter("test.xml");
        mySer.Serialize(sw, exampleBaseClass);           // <----- 2
        sw.Close();
        StreamReader sr = new StreamReader("test.xml");
        exampleBaseClass = (BaseClass)mySer.Deserialize(sr);
        Console.WriteLine(exampleBaseClass.GetType());   // <----- 3
        Console.ReadKey();
    }
}

1. 打印“ConsoleApplication1.DerivedClass - 它没关系。
2. 这会创建包含所有3个属性的文件 - 而且没关系。
3. 打印“ConsoleApplication1.DerivedClass - 它没关系。

第二个例子：

[Serializable]
public class BaseClass : IXmlSerializable
{
    public string property1;
    public string property2;

    public BaseClass()
    {
    }

    public System.Xml.Schema.XmlSchema GetSchema()
    {
        return null;
    }
    public virtual void WriteXml(XmlWriter writer)
    {
        writer.WriteElementString("property1", property1);
        writer.WriteElementString("property2", property2);
    }
    public virtual void ReadXml(XmlReader reader)
    {
        while (reader.Read())
            switch (reader.Name)
            {
                case "property1":
                    property1 = reader.Value;
                    break;
                case "property2":
                    property2 = reader.Value;
                    break;
            }
    }        
}

[Serializable]
public class DerivedClass : BaseClass, IXmlSerializable
{
    public string property3;

    public DerivedClass()
    {
    }


    public override void WriteXml(XmlWriter writer)
    {
        writer.WriteElementString("property1", property1);
        writer.WriteElementString("property2", property2);
        writer.WriteElementString("property3", property3);
    }

    public override void ReadXml(XmlReader reader)
    {
        while (reader.Read())
            switch (reader.Name)
            {
                case "property1":
                    property1 = reader.Value;
                    break;
                case "property2":
                    property2 = reader.Value;
                    break;
                case "property3":
                    property3 = reader.Value;
                    break;
            }
    }        
}

class Program
{
    static void Main(string[] args)
    {
        XmlSerializer mySer = new XmlSerializer(typeof(BaseClass), new Type [] {typeof(DerivedClass)});

        BaseClass exampleBaseClass;
        DerivedClass exampleDerivedClass = new DerivedClass();
        exampleDerivedClass.property1 = "Foo";
        exampleDerivedClass.property2 = "Bar";
        exampleDerivedClass.property3 = "Sth";

        exampleBaseClass = exampleDerivedClass;

        Console.WriteLine(exampleBaseClass.GetType());   // This prints "ConsoleApplication1.DerivedClass - and it's OK.
        StreamWriter sw = new StreamWriter("test.xml");
        mySer.Serialize(sw, exampleBaseClass);           // This creates file, that starts with xsi:type="DerivedClass" and contains all 3 properties - and it's OK.
        sw.Close();
        StreamReader sr = new StreamReader("test.xml");
        exampleBaseClass = (BaseClass)mySer.Deserialize(sr);
        Console.WriteLine(exampleBaseClass.GetType());   // This prints "ConsoleApplication1.DerivedClass - and it's OK. Everything works just great.
        Console.ReadKey();
    }
}

1. 打印“ConsoleApplication1.DerivedClass - 它没关系。
2. 这会创建包含所有3个属性的文件 - 而且没关系。
3. 这会打印“ConsoleApplication1.BaseClass-并且它不合适。结果对象只包含2个属性。

主要问题是： 如何更改简单类的定义？

感谢您的帮助。