c# - 通用类型层次结构的 DataContractSerialization

Question

以下程序尝试从层次结构中序列化然后反序列化通用类型的对象，失败并出现代码下方列出的错误。

如何让它发挥作用？

代码:

[DataContract]
[KnownType(nameof(GetKnownTypes))]
public abstract class Base<T>
{
    private static Type[] GetKnownTypes()
    {
        return new[] {typeof(DerivedA<T>)};
    }
}


[DataContract]
public class DerivedA<T> : Base<T>
{
    [DataMember]
    public T Foo { get; set; }
}

class Program
{
    private static void Main()
    {
        // This works fine
        var foo = new DerivedA<string> { Foo = "foo" };
        var xml = Serialize(foo);
        var foo2 = Deserialize<DerivedA<string>>(xml);
        Console.WriteLine(foo2.Foo); // foo

        // This throws the exception below
        // (from serializer.ReadObject() in the Deserialize method)
        var foo3 = Deserialize<Base<string>>(xml);
    }

    private static string Serialize<T>(T o)
    {
        string result = null;
        var serializer = new DataContractSerializer(o.GetType());
        using (var stream = new MemoryStream())
        {
            serializer.WriteObject(stream, o);
            stream.Position = 0;

            var reader = new StreamReader(stream);
            result = reader.ReadToEnd();
        }
        return result;
    }

    private static T Deserialize<T>(string xml)
    {
        var result = default(T);
        var serializer = new DataContractSerializer(typeof(T));
        using (var stream = new MemoryStream())
        {
            var writer = new StreamWriter(stream);
            writer.Write(xml);
            writer.Flush();
            stream.Position = 0;

            result = (T)serializer.ReadObject(stream);
        }
        return result;
    }
}

异常:

Unhandled Exception: System.Runtime.Serialization.SerializationException: Error in line 1 position 139. Expecting element 'BaseOfstring' from namespace 'http://schemas.datacontract.org/2004/07/ConsoleApplication7'.. Encountered 'Element'  with name 'DerivedAOfstring', namespace 'http://schemas.datacontract.org/2004/07/ConsoleApplication7'.
  at System.Runtime.Serialization.DataContractSerializer.InternalReadObject(XmlReaderDelegator xmlReader, Boolean verifyObjectName, DataContractResolver dataContractResolver)
  at System.Runtime.Serialization.XmlObjectSerializer.ReadObjectHandleExceptions(XmlReaderDelegator reader, Boolean verifyObjectName, DataContractResolver dataContractResolver)
  at System.Runtime.Serialization.XmlObjectSerializer.ReadObject(XmlDictionaryReader reader)
  at System.Runtime.Serialization.XmlObjectSerializer.ReadObject(Stream stream)
  at ConsoleApplication7.Program.Deserialize[T](String xml) in c:\users\tly01\documents\visual studio 2015\Projects\ConsoleApplication7\ConsoleApplication7\Program.cs:line 65

消息中最有趣的部分非常清楚出了什么问题:

Expecting element 'BaseOfstring' from namespace . Encountered 'Element' with name 'DerivedAOfstring', namespace .

果然，上面main方法中的xml是这样的(加了空格):

<DerivedAOfstring
    xmlns="http://schemas.datacontract.org/2004/07/ConsoleApplication7"
    xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
  <Foo>foo</Foo>
</DerivedAOfstring>

我意识到我必须以某种方式让序列化程序知道我的类型层次结构，但我已经尝试了上面列表中 KnownType 属性的许多不同变体，但没有一个有卓有成效。使这项工作正常进行的正确方法是什么？

Answer 1

您需要做的是始终使用 DataContractSerializer 序列化和反序列化您的派生类型。 constructed using identical Type arguments .您正在尝试使用 DerivedA<T>用于序列化和 Base<T>用于反序列化。这种不一致会导致您看到的问题。

发生这种情况的原因如下。通常，有两种常规方法可以将多态类型序列化为 XML:

• 可以更改元素名称以表示派生类型。与 XmlSerializer , 这是通过 [XmlElement(elementName, subType)] 支持的属性等等。

• 元素可以使用标准 {http://www.w3.org/2001/XMLSchema}type 显式断言其类型属性，通常缩写为 xsi:type . XmlSerializer通过 [XmlInclude(subType)] 支持此机制属性。 DataContractSerializer通过 KnownTypeAttribute 支持这一点.

事实DataContractSerializer使用第二种机制是导致您的问题的原因。当您序列化派生类的实例时，派生类协定名称将作为根元素名称发出。但是，当您将派生类的实例作为其基类的实例 序列化时，基类协定名称将与附加的 xsi:type 一起使用。 .要看到这一点，请按如下方式重构您的代码:

    private static string Serialize<T>(T o)
    {
        return Serialize(o, null);
    }

    private static string Serialize<T>(T o, DataContractSerializer serializer)
    {
        string result = null;
        serializer = serializer ?? new DataContractSerializer(o.GetType());
        using (var stream = new MemoryStream())
        {
            serializer.WriteObject(stream, o);
            stream.Position = 0;

            var reader = new StreamReader(stream);
            result = reader.ReadToEnd();
        }
        return result;
    }

现在如果你这样做:

var xml = Serialize(foo);
Debug.WriteLine(xml);

你会看到

<DerivedAOfstring 
 xmlns="http://schemas.datacontract.org/2004/07/Question37284138" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
     <Foo>foo</Foo>
</DerivedAOfstring>

但如果你这样做

var baseXml = Serialize(foo, new DataContractSerializer(typeof(Base<string>)));
Debug.WriteLine(baseXml);

你会看到:

<BaseOfstring i:type="DerivedAOfstring"
 xmlns="http://schemas.datacontract.org/2004/07/Question37284138" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
    <Foo>foo</Foo>
</BaseOfstring>

注意到区别了吗？这意味着前者不能使用 DataContractSerializer(typeof(Base<string>)) 反序列化。 .但是如果你序列化使用 DataContractSerializer(typeof(Base<string>))然后反序列化将起作用 - 并正确构造 DerivedA<string> 的实例感谢 xsi:type 的存在属性。