在 C# 中使用自定义序列化时如何控制 XML 元素的名称？

Question

我正在使用需要遵守的外部开发 XSD，其中字段 Engine 可以是以下两种类型之一：Electric 或 Combustion。两者都派生自超类Engine。

有多种方法可以让 C# 中的 XmlSerializer 处理此问题，但似乎都与相同的副作用相关：使用自定义序列化程序时，默认的序列化命名约定被抛弃了。我想模仿 Vehicle 类的其余字段的默认行为。

考虑以下几点：

using System;
using System.Xml;
using System.Xml.Schema;
using System.Xml.Serialization;

namespace DMV
{
    public class Vehicle : IXmlSerializable
    {
        public Engine Engine { get; set; }
        [XmlElement("Manufacturer")]
        public Company Manufacturer { get; set; }

        public void WriteXml(XmlWriter writer)
        {
            XmlSerializer engineSerializer = new XmlSerializer(Engine.GetType());
            engineSerializer.Serialize(writer, Engine);
            
            XmlSerializer manufacturerSerializer = new XmlSerializer(Manufacturer.GetType());
            manufacturerSerializer.Serialize(writer, Manufacturer);
        }
    }

    public abstract class Engine { }

    public class Electric : Engine
    {
        public int Poles { get; set; }
    }

    public class Combustion : Engine
    {
        public int CC { get; set; }
    }

    public class Company 
    {
        public string Name { get; set; }
    }

}

我尝试使用以下序列化代码运行它：

Vehicle vehicle = new Vehicle();
vehicle.Engine = new Electric();
vehicle.Manufacturer = new Company();

XmlSerializer serializer = new XmlSerializer(typeof(Vehicle));
TextWriter writer = new StringWriter();
serializer.Serialize(writer, vehicle);

Console.WriteLine(writer.ToString());
writer.Close();

结果是下面一堆令人失望的 caballus 粪便：

<Vehicle>
  <Electric>
    <Poles>0</Poles>
  </Electric>
  <Company/>
</Vehicle>

（是的，Engine 字段已正确序列化，但Company 没有那么多。）

如果我使用默认（内置）序列化，我会得到以下（期望的！）输出：

<Vehicle>
  <Manufacturer />
</Vehicle>

如何既序列化派生类Electric 或Combustion，同时保留字段Manufacturer（Company 类型）的默认命名约定？

Answer 1

您可以通过这样做摆脱自定义序列化程序：

public class Vehicle
{
    [XmlElement(typeof(Electric), ElementName = "Electric")]
    [XmlElement(typeof(Combustion), ElementName = "Combustion")]
    public Engine Engine { get; set; }
    [XmlElement("Manufacturer")]
    public Company Manufacturer { get; set; }
}

Answer 2

尝试将specific constructor 用于XmlSerializer，它接受“其他类型”的数组来处理您的派生类，而不是实现IXmlSerializable 并手动进行序列化。序列化程序现在将处理您的派生类，但也会以您想要的方式获取您的其他属性（例如XmlElement）：

using System;
using System.IO;
using System.Xml.Serialization;

namespace SomeNamespace
{
    public class Program
    {
        static void Main()
        {
            Vehicle vehicle = new Vehicle();
            vehicle.Engine = new Electric();
            vehicle.Manufacturer = new Company();

            XmlSerializer serializer = new XmlSerializer(typeof(Vehicle), new Type[] { typeof(Combustion), typeof(Electric)});
            TextWriter writer = new StringWriter();
            serializer.Serialize(writer, vehicle);

            Console.WriteLine(writer.ToString());
            writer.Close();
        }
    }

    public class Vehicle
    {
        public Engine Engine { get; set; }
        [XmlElement("Manufacturer")]
        public Company Manufacturer { get; set; }
    }

    public abstract class Engine { }

    public class Electric : Engine
    {
        public int Poles { get; set; }
    }

    public class Combustion : Engine
    {
        public int CC { get; set; }
    }

    public class Company
    {
        public string Name { get; set; }
    }

}

输出：

<?xml version="1.0" encoding="utf-16"?>
<Vehicle xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Engine xsi:type="Electric">
    <Poles>0</Poles>
  </Engine>
  <Manufacturer />
</Vehicle>

与公认的答案不同，这种方法的优点是不会将基类链接到所有可能的派生类。 （任何时候您在该方法中添加一个新的派生类型，您都必须返回并编辑基本类型。）虽然这个例子有Program.Main() 硬编码派生类型的数组，但您可以很容易地使用反射来构建该数组，使得添加新的派生类型变得更加容易。

Answer 3

有两种方法可以做到这一点。

1. 将XmlRoot 属性添加到Company 类定义中：

    [XmlRoot("Manufacturer")]
    public class Company
   

2. 将XmlRootAttribute 添加到XmlSerializer 构造函数。

    manufacturerSerializer = new XmlSerializer(Manufacturer.GetType(),
        new XmlRootAttribute("Manufacturer"));
   

   注意这个：Dynamically Generated Assemblies。