有没有办法将ac＃class序列化为xml而不在类中放置属性？

Question

I would like to serialize ac# class but I do not want to put the attributes on the class. I've found FluentXML on codeplex, but it seems to be just an idea project and is not working anyway

Answer 1

Here is an example that worked for me. From Serialize object to string

I just googled "serialize class to xml string"

String XmlizedString = null;

XmlSerializer x = new XmlSerializer(objectNameHere.GetType());

MemoryStream memoryStream = new MemoryStream();

XmlTextWriter xmlTextWriter = new XmlTextWriter(memoryStream, Encoding.UTF8);

x.Serialize(xmlTextWriter, objectNameHere);

memoryStream = (MemoryStream)xmlTextWriter.BaseStream;

UTF8Encoding encoding = new UTF8Encoding();

XmlizedString = encoding.GetString(memoryStream.ToArray());

XmlizedString = XmlizedString.Substring(1);

It produced a string like this <?xml version="1.0" encoding="utf-8"?><Employee xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"><Name>StackOverflow User</Name><Title>User of StackOverflow.</Title><Salary>70000</Salary><Skills><string>Being Awesome</string><string>Being sweet</string></Skills></Employee>

From an Employee class as simple as this:

public class Employee
{
  public string Name { get; set; }
  public string Title { get; set; }
  public double Salary { get; set; }
  public Collection<string> Skills { get; set; }
}

Answer 2

There's nothing forcing you to put attributes on your class in order to use XmlSerializer . In this case the generated XML will simply reflect your object structure. If on the other hand you want to control the generated XML you could use a XmlWriter or an XDocument instead. But the attributes are designed exactly for that purpose and you should use them. If for some reason you cannot modify the class you are serializing (because for example you do not have the source code for it) you could always design a different class which you could decorate with attributes and then map the original class to this model that you would serialize.

Answer 3

您可以使用没有属性的XmlSerializer ，并使用它为您提供的默认名称，或者您也可以在类上实现IXmlSerializable接口，以获得更加自定义但手动输出，而无需使用属性。

Answer 4

假设你有一个很好的理由不使用属性，你可以使用XmlSerializer ，但要注意它只适用于公共成员，还需要一个公共类型和一个无参数构造函数。