Question

我有以下C＃类

[XmlRoot("Customer")]
public class MyClass
{
    [XmlElement("CustId")]
    public int Id {get;set;}

    [XmlElement("CustName")]
    public string Name {get;set;}
}

然后我使用以下函数将类对象序列化为Xml

 public static XmlDocument SerializeObjectToXML(object obj, string sElementName)
 {
    XmlSerializer serializer = 
          new XmlSerializer(obj.GetType(), new XmlRootAttribute("Response"));

    using (MemoryStream ms = new MemoryStream())
    {
       XmlDocument xmlDoc = new XmlDocument();
       serializer.Serialize(ms, obj);
       ms.Position = 0;
       xmlDoc.Load(ms);
    }
}

我当前输出到XML就像;

<Response>
  <CustId></CustId>
  <CustName></CustName>
</Response>

但我怎样才能得到响应;

<Response>
  <Customer>
     <CustId></CustId>
     <CustName></CustName>
  </Customer>
</Response>

Answer 1

将XmlElementAttribute上的MyClass（根据http://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlelementattribute(v=vs.110).aspx实际上并非有效）更改为XmlTypeAttribute：

    [XmlType("Customer")]
    public class MyClass
    {
        [XmlElement("CustId")]
        public int Id { get; set; }

        [XmlElement("CustName")]
        public string Name { get; set; }
    }

序列化方法现在可以（与问题中的方法相同，但在XmlSerializer的构造函数中没有第二个参数）：

    public static XmlDocument SerializeObjectToXML(object obj, string sElementName)
    {
        XmlSerializer serializer = new XmlSerializer(obj.GetType());
        XmlDocument xmlDoc = new XmlDocument();
        using (MemoryStream ms = new MemoryStream())
        {

            serializer.Serialize(ms, obj);
            ms.Position = 0;
            xmlDoc.Load(ms);
        }

        return xmlDoc;
    }

Answer 2

您可以创建一个包含您的客户的响应对象，因为这也是您所需的xml显示的内容。

[XmlRoot("Response")] 
public class ResponseClass
{
  [XmlElement("Customer")]
  public Myclass Customer {get;set;}
}

Answer 3

你可以像这样定义它们：

public class MyClass
{
    [XmlElement("Customer")]
    public Customer cust { get; set; } 

}
public class Customer
{
    [XmlElement("CustId")]
    public int Id { get; set; }

    [XmlElement("CustName")]
    public string Name { get; set; }
}

顺便说一句，[XmlElement("Customer")]在您的示例中无效......

如何使用类名作为根元素序列化C＃类

3 个答案: