c# - 模型到 XML 文件映射 - 如何在 soucre XML 文件中找到反序列化对象的位置。

Question

这是使用 C# 序列化程序将我的模型映射到 XML 文件的最佳方式。我的意思是，例如，如果我选择一个反序列化对象，我可以在 XML 文件中找到 xml 源文本。

Answer 1

我为您准备了一个工作示例，您可以进一步探索它。

using System;
using System.Linq;
using System.Xml;
using System.Xml.Linq;
using System.Xml.Serialization;
using System.Collections.Generic;
using System.IO;


namespace ConsoleApplication5
{
  public class Person
  {
    public int Age { get; set; }
    public string Name { get; set; }
    public int XMLLine { get; set; }
  }
  public class Persons : List<Person> { }

  class Program
  {

    static void Main(string[] args)
    {
      //create your objects

      Person p = new Person();
      p.Age = 35;
      p.Name = "Arnold";

      Person p2 = new Person();
      p2.Age = 36;
      p2.Name = "Tom";

      Persons ps = new Persons();
      ps.Add(p);
      ps.Add(p2);

      //Serialize them to XML

      XmlSerializer xs = new XmlSerializer(typeof(Persons));

      XDocument d = new XDocument();

      using (XmlWriter xw = d.CreateWriter())
        xs.Serialize(xw, ps);

      //print xml
      //System.Diagnostics.Debug.WriteLine(d.ToString());

      // it will produce following xml. You can save it to file.
      //I have saved it to variable xml for demo



      string xml = @"<ArrayOfPerson>
                      <Person>
                        <Age>35</Age>
                        <Name>Arnold</Name>
                        <XMLLine>0</XMLLine>
                     </Person> 
                     <Person>
                       <Age>36</Age>
                       <Name>Tom</Name>
                       <XMLLine>0</XMLLine>
                      </Person>
                    </ArrayOfPerson>";




      XDocument xdoc = XDocument.Parse(xml, LoadOptions.SetLineInfo);

      // A little trick to get xml line
      xdoc.Descendants("Person").All(a => { a.SetElementValue("XMLLine", ((IXmlLineInfo)a).HasLineInfo() ? ((IXmlLineInfo)a).LineNumber : -1); return true; });
      
      
      //deserialize back to object

      Persons pplz = xs.Deserialize((xdoc.CreateReader())) as Persons;

      pplz.All(a => { Console.WriteLine(string.Format("Name {0} ,Age{1} ,Line number of object in XML File {2}", a.Name, a.Age, a.XMLLine)); return true; });

      Console.ReadLine();

    }
  }
}

它会给你这样的结果

Name Arnold ,Age35 ,Line number of object in XML File 2

Name Tom ,Age36 ,Line number of object in XML File 7

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