如何创建一个包含一组类中一个类的实例的可序列化类
在.Net 4或4.5中，您如何设计一个包含一组类中一个类的实例的可序列化类？ 例如，假设我有一个Garage类，它可以容纳任何“车辆”类型的实例，比如Car，Boat，Motorcycle，Motorhome。 但是Garage只能拥有其中一个类的实例。 我尝试了几种不同的方法，但我的问题是使其可序列化。
这是一个起始示例，其中Garage类中的实例只有一个选项。 您应该能够将其直接插入新的控制台应用程序并进行尝试。
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.IO; using System.Xml; using System.Xml.Serialization; namespace Patterns { [Serializable()] public class Garage { private Vehicle _MyVehicle; public Garage() { } public string GarageOwner { get; set; } public Vehicle MyVehicle { get { return _MyVehicle; } set { _MyVehicle = value; } } } [Serializable()] public class Vehicle { public string VehicleType { get; set; } public int VehicleNumber { get; set; } } class Serializer { static string _StartupPath = @"C:\Projects\Patterns\Data\"; static string _StartupFile = "SerializerTest.xml"; static string _StartupXML = _StartupPath + _StartupFile; static void Main(string[] args) { Console.Write("Press w for write. Press r for read:"); ConsoleKeyInfo cki = Console.ReadKey(true); Console.WriteLine("Pressed: " + cki.KeyChar.ToString()); if (cki.KeyChar.ToString() == "w") { Garage MyGarage = new Garage(); MyGarage.GarageOwner = "John"; MyGarage.MyVehicle = new Vehicle(); MyGarage.MyVehicle.VehicleType = "Car"; MyGarage.MyVehicle.VehicleNumber = 1234; WriteGarageXML(MyGarage); Console.WriteLine("Serialized"); } else if (cki.KeyChar.ToString() == "r") { Garage MyGarage = ReadGarageXML(); Console.WriteLine("Deserialized Garage owned by " + MyGarage.GarageOwner); } Console.ReadKey(); } public static void WriteGarageXML(Garage pInstance) { XmlSerializer writer = new XmlSerializer(typeof(Garage)); using (FileStream file = File.OpenWrite(_StartupXML)) { writer.Serialize(file, pInstance); } } public static Garage ReadGarageXML() { XmlSerializer reader = new XmlSerializer(typeof(Garage)); using (FileStream input = File.OpenRead(_StartupXML)) { return reader.Deserialize(input) as Garage; } } } } 根据另一篇SO文章 ，这最终对我有用。 它可以干净地序列化和反序列化。 使用这个例子，我可以设计一个对象的“树”，它可以选择使用的东西。 所以这可以扩展到Car可以有一个引擎有几个不同的引擎类型类和一个内部有几种不同的内部类型……等等。
代码开始工作，添加如下语句：[XmlInclude（typeof（Car））]
但如果有更好的方法，请告诉我！
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.IO; using System.Xml; using System.Xml.Serialization; namespace Patterns { public class Garage { private Vehicle _MyVehicle; public Garage() { } public string GarageOwner { get; set; } public Vehicle MyVehicle { get { return _MyVehicle; } set { _MyVehicle = value; } } } [XmlInclude(typeof(Car))] [XmlInclude(typeof(Boat))] [XmlInclude(typeof(Motorcycle))] [XmlInclude(typeof(Motorhome))] public abstract class Vehicle { public string VehicleType { get; set; } public int VehicleNumber { get; set; } } public class Car : Vehicle { public int Doors { get; set; } } public class Boat : Vehicle { public int Engines { get; set; } } public class Motorcycle : Vehicle { public int Wheels { get; set; } } public class Motorhome : Vehicle { public int Length { get; set; } } class Serializer { static string _StartupPath = @"C:\Projects\Patterns\Data\"; static string _StartupFile = "SerializerTest.xml"; static string _StartupXML = _StartupPath + _StartupFile; static void Main(string[] args) { Console.Write("Press w for write. Press r for read:"); ConsoleKeyInfo cki = Console.ReadKey(true); Console.WriteLine("Pressed: " + cki.KeyChar.ToString()); if (cki.KeyChar.ToString() == "w") { Garage MyGarage = new Garage(); MyGarage.GarageOwner = "John"; Car c = new Car(); c.VehicleType = "Lexus"; c.VehicleNumber = 1234; c.Doors = 4; MyGarage.MyVehicle = c; WriteGarageXML(MyGarage); Console.WriteLine("Serialized"); } else if (cki.KeyChar.ToString() == "r") { Garage MyGarage = ReadGarageXML(); Console.WriteLine("Deserialized Garage owned by " + MyGarage.GarageOwner); } Console.ReadKey(); } public static void WriteGarageXML(Garage pInstance) { XmlSerializer writer = new XmlSerializer(typeof(Garage)); using (FileStream file = File.OpenWrite(_StartupXML)) { writer.Serialize(file, pInstance); } } public static Garage ReadGarageXML() { XmlSerializer reader = new XmlSerializer(typeof(Garage)); using (FileStream input = File.OpenRead(_StartupXML)) { return reader.Deserialize(input) as Garage; } } } }
要序列化可序列化类的序列，可以使用通用列表实例。
我生成了这个
John Car 1234 Boat 56234
只需将MyVehicle转换为通用列表即可
using System; using System.Collections.Generic; using System.IO; using System.Xml.Serialization; namespace Patterns { [Serializable()] public class Garage { public string GarageOwner { get; set; } public List MyVehicles { get; set; } } [Serializable()] public class Vehicle { public string VehicleType { get; set; } public int VehicleNumber { get; set; } } class Serializer { static string _StartupPath = @"C:\temp\"; static string _StartupFile = "SerializerTest.xml"; static string _StartupXML = _StartupPath + _StartupFile; static void Main(string[] args) { Console.Write("Press w for write. Press r for read:"); ConsoleKeyInfo cki = Console.ReadKey(true); Console.WriteLine("Pressed: " + cki.KeyChar.ToString()); if (cki.KeyChar.ToString() == "w") { Garage MyGarage = new Garage(); MyGarage.GarageOwner = "John"; // Create some vehicles var myVehicle1 = new Vehicle(); myVehicle1.VehicleType = "Car"; myVehicle1.VehicleNumber = 1234; var myVehicle2 = new Vehicle(); myVehicle2.VehicleType = "Boat"; myVehicle2.VehicleNumber = 56234; // Create a new instance and add the vehicles MyGarage.MyVehicles = new List () { myVehicle1, myVehicle2 }; WriteGarageXML(MyGarage); Console.WriteLine("Serialized"); } else if (cki.KeyChar.ToString() == "r") { Garage MyGarage = ReadGarageXML(); Console.WriteLine("Deserialized Garage owned by " + MyGarage.GarageOwner); } Console.ReadKey(); } public static void WriteGarageXML(Garage pInstance) { XmlSerializer writer = new XmlSerializer(typeof(Garage)); using (FileStream file = File.OpenWrite(_StartupXML)) { writer.Serialize(file, pInstance); } } public static Garage ReadGarageXML() { XmlSerializer reader = new XmlSerializer(typeof(Garage)); using (FileStream input = File.OpenRead(_StartupXML)) { return reader.Deserialize(input) as Garage; } } } }