下面的代码段应该可以做到这一点(你可以忽略大部分序列化属性):

public class Car
{
  public string StockNumber { get; set; }
  public string Make { get; set; }
  public string Model { get; set; }
}

[XmlRootAttribute("Cars")]
public class CarCollection
{
  [XmlElement("Car")]
  public Car[] Cars { get; set; }
}

...

using (TextReader reader = new StreamReader(path))
{
  XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
  return (CarCollection) serializer.Deserialize(reader);
}

看看这是否有帮助:

[Serializable()]
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
    [XmlArrayItem(typeof(Car))]
    public Car[] Car { get; set; }
}

.

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElement()]
    public string StockNumber{ get; set; }

    [System.Xml.Serialization.XmlElement()]
    public string Make{ get; set; }

    [System.Xml.Serialization.XmlElement()]
    public string Model{ get; set; }
}

如果做不到这一点，请使用visual studio附带的xsd.exe程序来基于该xml文件创建一个模式文档，然后再次使用它来基于该模式文档创建一个类。

我不认为.net对反序列化数组有什么挑剔。第一个xml文档格式不正确。 没有根元素，尽管看起来有。规范xml文档有一个根和至少一个元素(如果有的话)。在你的例子中:

<Root> <-- well, the root
  <Cars> <-- an element (not a root), it being an array
    <Car> <-- an element, it being an array item
    ...
    </Car>
  </Cars>
</Root>

使用泛型类来反序列化XML文档如何

//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// Generic class to load any xml into a class
// used like this ...
// YourClassTypeHere InfoList = LoadXMLFileIntoClass<YourClassTypeHere>(xmlFile);

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

public static T LoadXMLFileIntoClass<T>(string xmlFile)
{
    T returnThis;
    XmlSerializer serializer = new XmlSerializer(typeof(T));
    if (!FileAndIO.FileExists(xmlFile))
    {
        Console.WriteLine("FileDoesNotExistError {0}", xmlFile);
    }
    returnThis = (T)serializer.Deserialize(new StreamReader(xmlFile));
    return (T)returnThis;
}

这一部分可能是必要的，也可能不是。在Visual Studio中打开XML文档，右键单击XML，选择属性。然后选择您的模式文件。

如果您的.xml文件已经在磁盘的某个地方生成，并且您使用了List<T>，请尝试此代码块:

//deserialization

XmlSerializer xmlser = new XmlSerializer(typeof(List<Item>));
StreamReader srdr = new StreamReader(@"C:\serialize.xml");
List<Item> p = (List<Item>)xmlser.Deserialize(srdr);
srdr.Close();`

注意:C:\serialize.xml是我的.xml文件的路径。你可以根据自己的需要进行更改。

下面的代码段应该可以做到这一点(你可以忽略大部分序列化属性):

public class Car
{
  public string StockNumber { get; set; }
  public string Make { get; set; }
  public string Model { get; set; }
}

[XmlRootAttribute("Cars")]
public class CarCollection
{
  [XmlElement("Car")]
  public Car[] Cars { get; set; }
}

...

using (TextReader reader = new StreamReader(path))
{
  XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
  return (CarCollection) serializer.Deserialize(reader);
}