Question

我想使用序列化创建一些这样的XML：

<?xml version="1.0" encoding="utf-8"?>
<Person>
  <Name>Bob</Name>
  <Sex>Male</Sex>
  <links rel="relations">
    <link rel="self" href="/Persons/Bob" />
    <link rel="child" href="/Persons/Lisa" />
  </links>
<Person>

现在我知道我可以使用以下方式设置link项目的属性：

Public Class link    
   <System.Xml.Serialization.XmlAttribute("rel")>
   Public rel As String = ""

   <System.Xml.Serialization.XmlAttribute("href")>
   Public href As String = ""    
End Class

但我无法确定如何在rel元素上设置links属性。这将是Person类设置links元素的代码：

Public Class Person
   ...
   <System.Xml.Serialization.XmlArray("links")>
   Public links As New List(Of link)
End Class

我能以正确的方式解决这个问题吗？

Answer 1

自定义序列化是一个选项，只需在Person的类上实现IXmlSerializable接口，然后在IXmlSerializable.WriteXml（）方法中控制自己如何编写xml。

   public class Link
    {
        public string rel { get; set; }

        public string href { get; set; }
    }


    [Serializable]    
    public class Person : IXmlSerializable
    {
        public Person()
        {
        }

        public string Name { get; set; }

        public string Sex { get; set; }

        public Link[]  Links {get; set;}

        public XmlSchema GetSchema()
        {
            return null;
        }

        public void ReadXml(XmlReader reader)
        {
        }

        public void WriteXml(XmlWriter writer)
        {
            writer.WriteElementString("Name", this.Name);
            writer.WriteElementString("Sex", this.Sex);

            writer.WriteStartElement("links");
            writer.WriteAttributeString("rel", "relations");

            foreach(Link link in this.Links)
            {
                writer.WriteStartElement("link");
                writer.WriteAttributeString("rel", link.rel);
                writer.WriteAttributeString("href", link.href);
                writer.WriteEndElement();
            }

            writer.WriteEndElement();
        }
    }

这是调用者代码，它将创建一个Person类，然后在你想要的xml中序列化它

public SerializePerson()
{

    Link link1 = new Link() { rel = "self", href = "/Persons/Bob" };
    Link link2 = new Link() { rel = "child", href = "/Persons/Lisa" };

    Person person = new Person() { Name = "Bob", Sex = "Male"};
    person.Links = new Link[] { link1, link2 };

    XmlSerializer serializer = new XmlSerializer(typeof(Person));
    StreamWriter streamWriter = new StreamWriter(@"C:\temp\serializedBob.txt");
    serializer.Serialize(streamWriter, person);
    streamWriter.Close();
}

如何将属性添加到XmlArray（不是元素）

1 个答案: