您只需将Cars汽车属性的一个属性从XmlArrayItem更改为xmlelement。也就是说，从

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

to

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

一个衬套:

var object = (Cars)new XmlSerializer(typeof(Cars)).Deserialize(new StringReader(xmlString));

如果使用xsd.exe创建xsd文件时出现错误，请使用msdn上提到的XmlSchemaInference类。下面是一个要演示的单元测试:

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

[TestMethod]
public void GenerateXsdFromXmlTest()
{
    string folder = @"C:\mydir\mydata\xmlToCSharp";
    XmlReader reader = XmlReader.Create(folder + "\some_xml.xml");
    XmlSchemaSet schemaSet = new XmlSchemaSet();
    XmlSchemaInference schema = new XmlSchemaInference();

    schemaSet = schema.InferSchema(reader);


    foreach (XmlSchema s in schemaSet.Schemas())
    {
        XmlWriter xsdFile = new XmlTextWriter(folder + "\some_xsd.xsd", System.Text.Encoding.UTF8);
        s.Write(xsdFile);
        xsdFile.Close();
    }
}

// now from the visual studio command line type: xsd some_xsd.xsd /classes

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);
}

试试这个用于Xml序列化和反序列化的泛型类。

public class SerializeConfig<T> where T : class
{
    public static void Serialize(string path, T type)
    {
        var serializer = new XmlSerializer(type.GetType());
        using (var writer = new FileStream(path, FileMode.Create))
        {
            serializer.Serialize(writer, type);
        }
    }

    public static T DeSerialize(string path)
    {
        T type;
        var serializer = new XmlSerializer(typeof(T));
        using (var reader = XmlReader.Create(path))
        {
            type = serializer.Deserialize(reader) as T;
        }
        return type;
    }
}

Kevin的回答很好，但事实上，在现实世界中，您通常无法修改原始XML以满足您的需要。

对于原始XML也有一个简单的解决方案:

[XmlRoot("Cars")]
public class XmlData
{
    [XmlElement("Car")]
    public List<Car> Cars{ get; set; }
}

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

然后你可以简单地调用:

var ser = new XmlSerializer(typeof(XmlData));
var data = (XmlData)ser.Deserialize(XmlReader.Create(PathToCarsXml));