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        C# Xml 序列化列表&lt;T&gt;具有 Xml 属性的后代

        时间:2023-07-27

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                  本文介绍了C# Xml 序列化列表&lt;T&gt;具有 Xml 属性的后代的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  早安,

                  我有一个来自 List 的集合,并且有一个公共属性.Xml 序列化程序不接受我的财产.列表项可以很好地序列化.我试过 XmlAttribute 属性无济于事.各位有解决办法吗?

                  I have a collection that descends from List and has a public property. The Xml serializer does not pick up my proeprty. The list items serialize fine. I have tried the XmlAttribute attribute to no avail. Do you guys have a solution?

                      public partial class MainWindow : Window
                  {
                      public MainWindow()
                      {
                          InitializeComponent();
                      }
                      private void button1_Click(object sender, RoutedEventArgs e)
                      {
                          var people = new PersonCollection
                          {
                              new Person { FirstName="Sue", Age=17 },
                              new Person { FirstName="Joe", Age=21 }
                          };
                          people.FavoritePerson = "Sue";
                  
                          var x = new XmlSerializer(people.GetType());
                          var b = new StringBuilder();
                          var w = XmlTextWriter.Create(b, new XmlWriterSettings { NewLineChars = "
                  ", Indent = true });
                          x.Serialize(w, people);
                          var s = b.ToString();
                      }
                  }
                  
                  [XmlRoot(ElementName="People")]
                  public class PersonCollection : List<Person>
                  {
                      //DOES NOT WORK! ARGHHH
                      [XmlAttribute]
                      public string FavoritePerson { get; set; }    
                  }
                  
                  public class Person
                  {
                      [XmlAttribute]
                      public string FirstName { get; set; }
                      [XmlAttribute]
                      public int Age { get; set; }
                  }
                  

                  我得到以下 xml

                  <?xml version="1.0" encoding="utf-16"?>
                          <People xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
                            <Person FirstName="Sue" Age="17" />
                            <Person FirstName="Joe" Age="21" />
                          </People>
                  

                  我想要这个

                  <?xml version="1.0" encoding="utf-16"?>
                          <People FavoritePerson="Sue" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
                            <Person FirstName="Sue" Age="17" />
                            <Person FirstName="Joe" Age="21" />
                          </People>
                  

                  推荐答案

                  我继续通过实现 IXmlSerializable 解决了这个问题.如果存在更简单的解决方案,请发布!

                  I went ahead and solved the problem by implementing IXmlSerializable. If a simpler solution exists, post it!

                      [XmlRoot(ElementName="People")]
                  public class PersonCollection : List<Person>, IXmlSerializable
                  {
                      //IT WORKS NOW!!! Too bad we have to implement IXmlSerializable
                      [XmlAttribute]
                      public string FavoritePerson { get; set; }
                  
                      public System.Xml.Schema.XmlSchema GetSchema()
                      {
                          return null;
                      }
                      public void ReadXml(XmlReader reader)
                      {
                          FavoritePerson = reader[0];            
                          while (reader.Read())
                          {
                              if (reader.Name == "Person")
                              {
                                  var p = new Person();
                                  p.FirstName = reader[0];
                                  p.Age = int.Parse( reader[1] ); 
                                  Add(p);
                              }
                          }
                      }
                      public void WriteXml(XmlWriter writer)
                      {
                          writer.WriteAttributeString("FavoritePerson", FavoritePerson);
                          foreach (var p in this)
                          {
                              writer.WriteStartElement("Person");
                              writer.WriteAttributeString("FirstName", p.FirstName);
                              writer.WriteAttributeString("Age", p.Age.ToString());
                              writer.WriteEndElement();            
                          }
                      }
                  }
                  

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