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 的样子;
<Response>
<CustId></CustId>
<CustName></CustName>
</Response>
但是我怎样才能让响应看起来像;
<Response>
<Customer>
<CustId></CustId>
<CustName></CustName>
</Customer>
</Response>
MyClass
上的XmlElementAttribute
(根据 http://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlelementattribute(v=vs.110).aspx,它实际上在那里无效)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是(与问题中的序列化方法相同,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;
}
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;}
}
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")]
对您的示例无效...