Question

我遇到了下一个序列化问题。我有一个我要序列化的MyObject类的列表，但不能在所需的视图中执行。问题如下。我有这个类结构：

[XmlRoot(ElementName = "response")]
public class Response : List<MyObject> {}

[XmlElement("myObject")]
public class MyObject { ... }

在序列化之后我得到了这个：

<response>
    <myObject>
        ...
    </myObject>
    <myObject>
        ...
    </myObject>
    <myObject>
        ...
    </myObject>
</response>

但我需要这个：

<response>
    <myObejcts>
        <myObject>
            ...
        </myObject>
        <myObject>
            ...
        </myObject>
        <myObject>
            ...
        </myObject>
    </myObejcts>
</response>

我可以以任何方式获得XML结构而不添加属性

public List<MyObject> {get; set;}

到Response课程。我不能使用List属性的原因是因为我使用了Entity Framework，在这种情况下，EF在数据库中添加了一些额外的不需要的表。

Answer 1

您可以包装完整的Response对象，然后使用此对象进行序列化吗？

[XmlRoot("response")]
public class ResponseWrapper<T>
{
    [XmlArray("MyObjects")]
    [XmlArrayItem]
    public List<T> Items { get; set; }

    public ResponseWrapper()
    {
    }

    public ResponseWrapper(List<T> items)
    {
        this.Items = items;
    }
}

然后生成如下输出：

<?xml version="1.0"?>
<response xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http:
//www.w3.org/2001/XMLSchema">
  <MyObjects>
    <MyObject>
      <ID>1</ID>
      <Name>Buck</Name>
    </MyObject>
    <MyObject>
      <ID>2</ID>
      <Name>Rogers</Name>
    </MyObject>
    <MyObject>
      <ID>3</ID>
      <Name>Flash</Name>
    </MyObject>
    <MyObject>
      <ID>4</ID>
      <Name>Gordon</Name>
    </MyObject>
  </MyObjects>
</response>

只是，它仍然需要你做一些额外的工作，是的:)

Answer 2

@Vsevywniy再次问好，虽然我仍然相信创建一个Wrapper类是实现这一目标的最佳方法。另一个选项可以是实现IXmlSerializable来控制序列化并添加自定义元素。我能够将您的类结构序列化为您期望的XML。请看下面的代码：

public class Program
    {
        public static void Main(string[] args)
        {
            var products = new MyObjects()
            {
                new MyObject() {Id = 1, Name = "Dell"},
                new MyObject() {Id = 2, Name = "Sony"},
                new MyObject() {Id = 3, Name = "HP"}
            };

            var serializeObject = SerializeObject<MyObjects>(products);
            Console.WriteLine(serializeObject);
            Console.ReadLine();
        }

        public static string SerializeObject<T>(Object data)
        {
            string result;
            try
            {
                var xmlData = new MemoryStream();
                var xmlEncodedData = new XmlTextWriter(xmlData, Encoding.UTF8);
                var serial = new XmlSerializer(typeof(T));
                serial.Serialize(xmlEncodedData, data);
                xmlData.Position = 0;

                var reader = new StreamReader(xmlData);

                result = reader.ReadToEnd();

            }
            catch (Exception)
            {
                return string.Format("Error in return data");
            }

            return result;
        }
    }

    [XmlRoot("Response")]
    [Serializable]
    public class MyObjects : List<MyObject>, IXmlSerializable
    {

        public XmlSchema GetSchema()
        {
            return null;
        }

        public void ReadXml(XmlReader reader)
        {
            throw new NotImplementedException();
        }

        public void WriteXml(XmlWriter writer)
        {
            writer.WriteStartElement("MyObjects");
            foreach (var myObject in this)
            {
                new XmlSerializer(typeof(MyObject)).Serialize(writer, myObject);
            }
            writer.WriteEndElement();
        }
    }

    public class MyObject
    {
        public int Id { get; set; }
        public string Name { get; set; }


       }
}

序列化列表<t>问题</t>

2 个答案: