Question

我正在尝试使用System.Dynamic.ExpandoObject，因此我可以在运行时动态创建属性。稍后，我需要传递此对象的实例，并且使用的机制需要序列化。

当然，当我尝试序列化我的动态对象时，我得到了异常：

System.Runtime.Serialization.SerializationException未处理。

在程序集'System.Core中输入'System.Dynamic.ExpandoObject'，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089'未标记为可序列化。

我可以序列化ExpandoObject吗？是否有另一种方法来创建可序列化的动态对象？也许使用DynamicObject包装器？

我创建了一个非常简单的Windows窗体示例来复制错误：

using System;
using System.Windows.Forms;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Formatters.Binary;
using System.Dynamic;

namespace DynamicTest
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {            
            dynamic dynamicContext = new ExpandoObject();
            dynamicContext.Greeting = "Hello";

            IFormatter formatter = new BinaryFormatter();
            Stream stream = new FileStream("MyFile.bin", FileMode.Create,
                                           FileAccess.Write, FileShare.None);
            formatter.Serialize(stream, dynamicContext);
            stream.Close();
        }
    }
}

Answer 1

我无法序列化ExpandoObject，但我可以手动序列化DynamicObject。因此，使用DynamicObject的TryGetMember / TrySetMember方法并实现ISerializable，我可以解决我真正要序列化动态对象的问题。

我在我的简单测试应用程序中实现了以下内容：

using System;
using System.Windows.Forms;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Formatters.Binary;
using System.Collections.Generic;
using System.Dynamic;
using System.Security.Permissions;

namespace DynamicTest
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {            
            dynamic dynamicContext = new DynamicContext();
            dynamicContext.Greeting = "Hello";
            this.Text = dynamicContext.Greeting;

            IFormatter formatter = new BinaryFormatter();
            Stream stream = new FileStream("MyFile.bin", FileMode.Create, FileAccess.Write, FileShare.None);
            formatter.Serialize(stream, dynamicContext);
            stream.Close();
        }
    }

    [Serializable]
    public class DynamicContext : DynamicObject, ISerializable
    {
        private Dictionary<string, object> dynamicContext = new Dictionary<string, object>();

        public override bool TryGetMember(GetMemberBinder binder, out object result)
        {
            return (dynamicContext.TryGetValue(binder.Name, out result));
        }

        public override bool TrySetMember(SetMemberBinder binder, object value)
        {
            dynamicContext.Add(binder.Name, value);
            return true;
        }

        [SecurityPermissionAttribute(SecurityAction.Demand, SerializationFormatter = true)]
        public virtual void GetObjectData(SerializationInfo info, StreamingContext context)
        {
            foreach (KeyValuePair<string, object> kvp in dynamicContext)
            {
                info.AddValue(kvp.Key, kvp.Value);
            }
        }

        public DynamicContext()
        {
        }

        protected DynamicContext(SerializationInfo info, StreamingContext context)
        {
            // TODO: validate inputs before deserializing. See http://msdn.microsoft.com/en-us/library/ty01x675(VS.80).aspx
            foreach (SerializationEntry entry in info)
            {
                dynamicContext.Add(entry.Name, entry.Value);
            }
        }

    }
}

和Why does SerializationInfo not have TryGetValue methods?有一个缺失的拼图，以保持简单。

Answer 2

ExpandoObject实现IDictionary<string, object>，例如：

class Test
{
    static void Main()
    {
        dynamic e = new ExpandoObject();
        e.Name = "Hello";

        IDictionary<string, object> dict = (IDictionary<string, object>)e;

        foreach (var key in dict.Keys)
        {
            Console.WriteLine(key);
        }

        dict.Add("Test", "Something");

        Console.WriteLine(e.Test);

        Console.ReadKey();
    }
}

您可以将字典的内容写入文件，然后通过反序列化创建一个新的ExpandoObject，将其强制转换回字典并将属性写回？

Answer 3

回答可能有点迟，但是我使用jsonFx来序列化和反序列化expandoObjects并且效果非常好：

序列：

dim XMLwriter As New JsonFx.Xml.XmlWriter
dim serializedExpando as string =XMLwriter.Write(obj)

反序列化

dim XMLreader As New JsonFx.Xml.XmlReader
Dim obj As ExpandoObject = XMLreader.Read(Str)

我可以在.NET 4中序列化ExpandoObject吗？

3 个答案: