如何以类名作为根元素序列化C#类

如何以类名作为根元素序列化C#类,c#,xml,serialization,C#,Xml,Serialization,我有下面的C班 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 m

我有下面的C班

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);
    }
}
我当前对XML的输出如下:

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

但我如何才能得到这样的回应

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

您可以创建一个包含客户的响应对象,因为这也是您想要的xml显示的内容

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

MyClass
上的
xmlementAttribute
(根据)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是(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):


您可以这样定义它们:

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; }
}

顺便说一下,
[XmlElement(“Customer”)]
在您的示例中无效…

如果是这样的话,应该是[XmlRoot(“Response”)],但他已经在其他地方处理过了。虽然我确实认为这会给出正确的输出。您不能将XmlElementAttribute与类一起使用。我刚刚注意到,在我写这篇文章时,发布了两个答案:)…我将我的问题从[XmlElement(“客户”)改为[XmlRoot(“客户”)],这是应该的
    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;
    }
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; }
}