将XML文档转换为对象的最简单方法是什么?

I have an XMLDocument that i need to read in and convert into a set of objects. I have the following objects

我有一个XMLDocument，需要读取它并将其转换为一组对象。我有以下的对象

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

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

and i have the following XML file:

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>

What is the best way of doing this? Should I be serializing the xmldocument to the object automatically or do i need to parse out each element and convert into my object manually? In particular, I am trying to figure out how to convert the collections (locations, buildings, etc).

最好的方法是什么?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工转换为对象?特别是，我正在设法弄清楚如何转换集合(位置、建筑物等)。

What is the best suggestion to convert this XML file into basically a

将这个XML文件转换为a的最佳建议是什么

List<Location>

objects?

对象?

2 个解决方案

#1

You could start by fixing your XML because in the example you have shown you have unclosed tags. You might also wrap the <Building> tags into a <Buildings> collection in order to be able to have other properties in this Location class other than buildings.

您可以从修改XML开始，因为在示例中显示了未关闭的标记。您还可以将 <构建> 标记放入 <建筑物> 集合中，以便能够在此位置类中拥有其他属性，而不是建筑物。

<?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>

Once you have fixed your XML you could adapt your models. I would recommend you using properties instead of fields in your classes:

一旦修复了XML，就可以调整模型。我建议您在您的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; }
}

and now all that's left is deserialize the XML:

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
}

#2

Just use the XML serialization attributes- for example:

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;
}

Just remember - everything will be serialized as XML Elements by default - with the sames the same as the names of the objects :)

只要记住——一切都将被序列化为XML元素——sames的名称与对象的名称相同:)

Do this to load:

做这个加载:

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