这个问题的意思是什么？

Question

我有一个XML文档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;
}

我将要翻译以下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>

什么是最好的方法？我应该自动将xml文档序列化为对象，还是需要手动解析每个元素并转换为我的对象？特别是，我正在努力弄清如何转换集合（位置、建筑物等）。

如何将这个XML文件转换成一个基本的。请提出最佳建议。

List<Location>

对象？

- leora

Answer 1

你可以先修复XML，因为在你展示的示例中有未关闭的标签。此外，你可能还需要将<Building>标签包装到一个<Buildings>集合中，以便能够在此位置类中拥有其他属性而不只是建筑物。

<?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XML，您可以调整模型。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; }
}

现在只剩下反序列化XML了：

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
}

- Darin Dimitrov

Answer 2

只需使用XML序列化属性即可，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，并使用对象名称相同的名称 :)

要加载，请执行以下操作：

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

- Dave Kerr