将对象序列化为XML

我继承了一个c#类。我已经成功地“构建”了对象。但是我需要将对象序列化为XML。有什么简单的方法吗?

看起来类已经为序列化设置了，但我不确定如何获得XML表示。我的类定义是这样的:

[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.1")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://www.domain.com/test")]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://www.domain.com/test", IsNullable = false)]
public partial class MyObject
{
  ...
}

以下是我认为我能做的，但它不起作用:

MyObject o = new MyObject();
// Set o properties
string xml = o.ToString();

如何获得该对象的XML表示形式?

当前回答

以上所有点赞的答案都是正确的。这是最简单的版本:

private string Serialize(Object o)
{
    using (var writer = new StringWriter())
    {
        new XmlSerializer(o.GetType()).Serialize(writer, o);
        return writer.ToString();
    }
}

2018-12-17 13:30:16

其他回答

public string ObjectToXML(object input)
{
    try
    {
        var stringwriter = new System.IO.StringWriter();
        var serializer = new XmlSerializer(input.GetType());
        serializer.Serialize(stringwriter, input);
        return stringwriter.ToString();
    }
    catch (Exception ex)
    {
        if (ex.InnerException != null)
            ex = ex.InnerException;

        return "Could not convert: " + ex.Message;
    }
}

//Usage
var res = ObjectToXML(obj)

您需要使用以下类:

using System.IO;
using System.Xml;
using System.Xml.Serialization;

2019-12-18 11:30:54

这里有一个关于如何做到这一点的很好的教程

基本上应该使用System.Xml.Serialization.XmlSerializer类来做到这一点。

2010-11-08 12:03:29

可以将下面的函数复制到任何对象，以使用System.Xml名称空间添加XML保存函数。

/// <summary>
/// Saves to an xml file
/// </summary>
/// <param name="FileName">File path of the new xml file</param>
public void Save(string FileName)
{
    using (var writer = new System.IO.StreamWriter(FileName))
    {
        var serializer = new XmlSerializer(this.GetType());
        serializer.Serialize(writer, this);
        writer.Flush();
    }
}

要从保存的文件创建对象，添加以下函数并将[ObjectType]替换为要创建的对象类型。

/// <summary>
/// Load an object from an xml file
/// </summary>
/// <param name="FileName">Xml file name</param>
/// <returns>The object created from the xml file</returns>
public static [ObjectType] Load(string FileName)
{
    using (var stream = System.IO.File.OpenRead(FileName))
    {
        var serializer = new XmlSerializer(typeof([ObjectType]));
        return serializer.Deserialize(stream) as [ObjectType];
    }
}

2013-02-02 16:18:44

下面是一个基本代码，可以帮助将c#对象序列化为xml:

using System;

public class clsPerson
{
  public  string FirstName;
  public  string MI;
  public  string LastName;
}

class class1
{ 
   static void Main(string[] args)
   {
      clsPerson p=new clsPerson();
      p.FirstName = "Jeff";
      p.MI = "A";
      p.LastName = "Price";
      System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(p.GetType());
      x.Serialize(Console.Out, p);
      Console.WriteLine();
      Console.ReadLine();
   }
}

2018-02-06 12:33:14

现在可能太晚了，但是只有用户定义的命名空间是序列化的:

public static string XmlSerialize<T>(this T obj) where T : class
        {
            Type serialType = typeof(T);
            var xsSubmit = new XmlSerializer(serialType);

            XmlWriterSettings xws = new XmlWriterSettings() { OmitXmlDeclaration = true, Indent = true };
            

            var Namespaces = new XmlSerializerNamespaces(new XmlQualifiedName[] {           
            new XmlQualifiedName(string.Empty,  GetXmlNameSpace(serialType) ?? string.Empty )


        });
 private static string GetXmlNameSpace(Type target)
        {
            XmlRootAttribute attribute = (XmlRootAttribute)Attribute.GetCustomAttribute(target, typeof(XmlRootAttribute));
            return attribute == null ? null : attribute.Namespace;
        }

并在自定义类中定义命名空间

 [XmlRoot("IdentityTerminal",Namespace = "http://my-name-space/XMLSchema")]
   
    public class IdentityTerminal
    {
    }

这段代码允许只使用用户定义的名称空间，而忽略默认的名称空间。

2022-02-10 11:25:44

将对象序列化为XML

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