[英]Serialize/Deserialize a dynamic object
我有以下課程:
public abstract class Animal
{
public Animal() { _myType = getAnimal(this.GetType().Name); }
private dynamic _myType;
public dynamic myType { get { return _myType; } }
}
public class Cat : Animal
{
public Cat() : base() { }
}
及其輔助功能:
public static T CreateAnimal<T>(string animal)
{
Type type = Type.GetType(typeof(Form1).FullName + "+" + animal);
return (T)Activator.CreateInstance(type);
}
public static dynamic getAnimal(string name)
{
dynamic theAnimal = Activator.CreateInstance(MyAnimals); // Will default to 'Cat'
FieldInfo fi = MyAnimals.GetField(name);
int iEnum = (int)fi.GetValue(MyAnimals);
return Enum.ToObject(MyAnimals, iEnum);
}
它從動態創建的枚舉“ MyAnimals”獲取其“ myType”:
public static Type MyAnimals;
public static void CreateAnimalEnum()
{
// Get the current application domain for the current thread.
AppDomain currentDomain = AppDomain.CurrentDomain;
// Create a dynamic assembly in the current application domain,
// and allow it to be executed and saved to disk.
AssemblyName aName = new AssemblyName("TempAssembly");
AssemblyBuilder ab = currentDomain.DefineDynamicAssembly(
aName, AssemblyBuilderAccess.Run);
// Define a dynamic module in "TempAssembly" assembly. For a single-
// module assembly, the module has the same name as the assembly.
ModuleBuilder mb = ab.DefineDynamicModule(aName.Name);
// Define a public enumeration with an underlying type of Integer.
EnumBuilder eb = mb.DefineEnum("MyAnimalType", TypeAttributes.Public, typeof(int));
var types = new List<Type>();
int Count = 0;
foreach (var assembly in AppDomain.CurrentDomain.GetAssemblies())
try
{
types.AddRange(assembly.GetTypes().Where(x => x.IsSubclassOf(typeof(Animal))));
}
catch { }
foreach (var type in types)
eb.DefineLiteral(type.Name, Count++);
// Create the type and save the assembly.
MyAnimals = eb.CreateType();
}
所以現在當我創建一只貓時,我無法對其進行序列化。 “ InvalidOperationException:生成XML文檔時出錯。” 我已經嘗試過使用DynamicObject,但發現了一個動態幫助程序類( https://gist.github.com/martinnormark/2574972 ),但是當我想要序列化一個封裝在另一個類中的Cat對象時,這沒有幫助。
public static bool Save(Animal animal)
{
System.Xml.Serialization.XmlSerializer ListSer = new System.Xml.Serialization.XmlSerializer(typeof(Animal));
System.IO.StreamWriter mywriter = new System.IO.StreamWriter(@"test.txt", false);
ListSer.Serialize(mywriter, animal);
mywriter.Flush();
mywriter.Close();
return true;
}
public Form1()
{
InitializeComponent();
GetEDIDeviceTypesEnums();
Animal c = new Cat();
Save(c);
// This way fails too
dynamic cat = CreateAnimal<Animal>("Cat");
Save(cat);
}
為了序列化Cat,我缺少什么?
XMLSerializer需要提前知道它可以序列化什么類型。 如果在抽象類上初始化XMLSerializer,則它將只知道如何序列化該類,而不會繼承任何類。
XMLSerializer有另一個構造函數,允許您輸入一個額外類型的數組供嘗試序列化時使用。 您可以從GetAssemblies動態構建該類型的數組(類似於構建自定義MyAnimals枚舉的過程):
public static bool Save(Animal animal)
{
var lListOfAnimals = (from lAssembly in AppDomain.CurrentDomain.GetAssemblies()
from lType in lAssembly.GetTypes()
where typeof(Animal).IsAssignableFrom(lType)
select lType).ToArray();
System.Xml.Serialization.XmlSerializer ListSer = new System.Xml.Serialization.XmlSerializer(typeof(Animal), lListOfAnimals);
在此線程中,代碼直接從Yahoo認真的回答中提出。 正如Yahoo認真提到的那樣,如果您經常調用Save,則以這種方式使用Reflection可能會導致性能下降,因此您可以緩存Animal類型數組,而不是每次序列化時都對其進行重建。
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