C# 如何从一组类中创建一个包含一个类的实例的可序列化类
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原文地址: http://stackoverflow.com/questions/14326832/
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How to make a serializable class that contains an instance of one class from a set of classes
提问by John
In .Net 4 or 4.5, how would you design a serializable class that contains an instance of one class from a set of classes? For instance, suppose I have a Garage class, which can hold an instance of any "vehicle" type classes, say Car, Boat, Motorcycle, Motorhome. But the Garage can only hold an instance of one of those classes. I have tried a few different ways of doing this, but my problem is to make it serializable.
在 .Net 4 或 4.5 中，您将如何设计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，它可以包含任何“车辆”类型类的实例，比如汽车、船、摩托车、房车。但是 Garage 只能保存这些类之一的实例。我尝试了几种不同的方法来做到这一点，但我的问题是让它可序列化。
Here is a starting example where there is only one option for the instance in the Garage class. You should be able to plug it right into a new console app and try it.
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尝试使用。
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;
}
}
}
}
回答by Eric Herlitz
To serialize sequences of serializable classes you can use Generic list instances.
要序列化可序列化类的序列，您可以使用通用列表实例。
I generated this
我生成了这个
<?xml version="1.0"?>
<Garage xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<GarageOwner>John</GarageOwner>
<MyVehicles>
<Vehicle>
<VehicleType>Car</VehicleType>
<VehicleNumber>1234</VehicleNumber>
</Vehicle>
<Vehicle>
<VehicleType>Boat</VehicleType>
<VehicleNumber>56234</VehicleNumber>
</Vehicle>
</MyVehicles>
</Garage>
By simply converting the MyVehicle to a generic list
通过简单地将 MyVehicle 转换为通用列表
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<Vehicle> 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<Vehicle>()
{
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;
}
}
}
}
回答by John
Based on another SO article, this is what finally worked for me. It can serialize and deserialize cleanly. Using this example, I can design a "tree" of objects which have options for what is used. So this could be extended that a Car can have one Engine of several different Engine type classes and one Interior of several different Interior types... and so on.
基于另一篇 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”，其中有使用的选项。因此，这可以扩展为一辆汽车可以拥有一个具有多种不同发动机类型类别的发动机和一种具有多种不同内饰类型的内饰......等等。
The code began to work by adding the statements below like: [XmlInclude(typeof(Car))]
通过添加以下语句，代码开始工作：[XmlInclude(typeof(Car))]
But please let me know if there are better ways!
但是请让我知道是否有更好的方法！
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;
}
}
}
}
