将此 XML 文档转换为我的对象的最简单方法是什么？答案

【问题标题】：What is the easiest way to convert this XML document to my object?将此 XML 文档转换为我的对象的最简单方法是什么？
【发布时间】：2013-02-01 11:27:51
【问题描述】：

我有一个 XMLDocument 需要读取并转换为一组对象。我有以下对象

public class Location
{
      public string Name;
      public List<Building> Buildings;
}

public class Building
{
     public string Name;
     public List<Room> Rooms;
}

我有以下 XML 文件：

 <?xml version="1.0" encoding="utf-8" ?>
 <info>
 <locations>
  <location name="New York">
  <Building name="Building1">
    <Rooms>
      <Room name="Room1">
        <Capacity>18</Capacity>
      </Room>
      <Room name="Room2">
        <Capacity>6</Capacity>
      </Room>
    </Rooms>
  </Building>

  <Building name="Building2">
    <Rooms>
      <Room name="RoomA">
        <Capacity>18</Capacity>
      </Room>
    </Rooms>
  </Building>
</location>
<location name ="London">
  <Building name="Building45">
    <Rooms>
      <Room name="Room5">
        <Capacity>6</Capacity>
      </Room>
  </Building>
</location>
</locations>
</info>

这样做的最佳方法是什么？我应该自动将 xmldocument 序列化为对象，还是需要解析每个元素并手动转换为我的对象？特别是，我试图弄清楚如何转换集合（位置、建筑物等）。

将这个 XML 文件基本上转换成一个文件的最佳建议是什么

List<Location>

对象？

【问题讨论】：

标签： c# xml converter xmldocument

【解决方案1】：

您可以从修复 XML 开始，因为在示例中您已经展示了未闭合的标签。您还可以将 <Building> 标记包装到 <Buildings> 集合中，以便能够在此 Location 类中拥有除建筑物之外的其他属性。

<?xml version="1.0" encoding="utf-8" ?>
<info>
  <locations>
    <location name="New York">
      <Buildings>
        <Building name="Building1">
          <Rooms>
            <Room name="Room1">
              <Capacity>18</Capacity>
            </Room>
            <Room name="Room2">
              <Capacity>6</Capacity>
            </Room>
          </Rooms>
        </Building>

        <Building name="Building2">
          <Rooms>
            <Room name="RoomA">
              <Capacity>18</Capacity>
            </Room>
          </Rooms>
        </Building>
      </Buildings>
    </location>
    <location name="London">
      <Buildings>
        <Building name="Building45">
          <Rooms>
            <Room name="Room5">
              <Capacity>6</Capacity>
            </Room>
          </Rooms>
        </Building>
      </Buildings>
    </location>
  </locations>
</info>

一旦您修复了您的 XML，您就可以调整您的模型。我建议您在类中使用属性而不是字段：

public class Location
{
    [XmlAttribute("name")]
    public string Name { get; set; }

    public List<Building> Buildings { get; set; }
}

public class Building
{
    [XmlAttribute("name")]
    public string Name { get; set; }
    public List<Room> Rooms { get; set; }
}

public class Room
{
    [XmlAttribute("name")]
    public string Name { get; set; }
    public int Capacity { get; set; }
}

[XmlRoot("info")]
public class Info
{
    [XmlArray("locations")]
    [XmlArrayItem("location")]
    public List<Location> Locations { get; set; }
}

现在剩下的就是反序列化 XML：

var serializer = new XmlSerializer(typeof(Info));
using (var reader = XmlReader.Create("locations.xml"))
{
    Info info = (Info)serializer.Deserialize(reader);
    List<Location> locations = info.Locations;
    // do whatever you wanted to do with those locations
}

【讨论】：

【解决方案2】：

只需使用 XML 序列化属性——例如：

public class Location
{
      [XmlAttribute("name");
      public string Name;
      public List<Building> Buildings;
}

public class Building
{
     [XmlAttribute("name");
     public string Name;
     public List<Room> Rooms;
}

请记住 - 默认情况下，所有内容都将被序列化为 XML 元素 - 与对象名称相同:)

执行此操作以加载：

using(var stream = File.OpenRead("somefile.xml"))
{
   var serializer = new XmlSerializer(typeof(List<Location>));
   var locations = (List<Location>)serializer.Deserialize(stream );
}

【讨论】：