模型到XML文件的映射-如何在源XML文件中找到反序列化对象的位置

本文关键字：文件 XML 反序列化 对象 位置 映射 模型 | 更新日期: 2023-09-27 18:18:31

这是使用c#序列化器将我的模型映射到XML文件的最佳方法。我的意思是，如果我选择一个反序列化的对象，我可以在xml文件中找到xml源文本

模型到XML文件的映射-如何在源XML文件中找到反序列化对象的位置

我为你准备了一个工作示例，你可以进一步探索它。

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();
    }
  }
}

和它会给出像

这样的结果

Name Arnold,Age35, XML文件2中对象行号

Name Tom,Age36, XML文件7中对象行号

你可以试试这个扩展方法:

public static string ToXml<T>(this object obj)
{
    using (var memoryStream = new MemoryStream())
    {
        using (TextWriter streamWriter = new StreamWriter(memoryStream))
        {
            var xmlSerializer = new XmlSerializer(typeof(T));
            xmlSerializer.Serialize(streamWriter, obj);
            return Encoding.ASCII.GetString(memoryStream.ToArray());
        }
    }
}
public static void ToXmlFile<T>(this object obj, string fileName)
{
    using (TextWriter streamWriter = new StreamWriter(fileName))
    {
        var xmlSerializer = new XmlSerializer(typeof(T));
        xmlSerializer.Serialize(streamWriter, obj);
    }
}

用法:

// you will get this on a string variable
var xmlString = yourModel.ToXml<YourModel>();
// you will save our object in a file.
yourModel.ToXmlFile<YourModel>(@"C:'yourModelDump.xml");

请注意在你的类中添加SerializableAttribute 

[Serializable]
public class YourModel
{
  //...
}

应该可以了

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