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

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];
    }
}

其他回答

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。

扩展类:

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

namespace MyProj.Extensions
{
    public static class XmlExtension
    {
        public static string Serialize<T>(this T value)
        {
            if (value == null) return string.Empty;

            var xmlSerializer = new XmlSerializer(typeof(T));

            using (var stringWriter = new StringWriter())
            {
                using (var xmlWriter = XmlWriter.Create(stringWriter,new XmlWriterSettings{Indent = true}))
                {
                    xmlSerializer.Serialize(xmlWriter, value);
                    return stringWriter.ToString();
                }    
            }
        }
    }
}

用法:

Foo foo = new Foo{MyProperty="I have been serialized"};

string xml = foo.Serialize();

只是引用名称空间持有您的扩展方法在文件中,你想使用它,它将工作(在我的例子中,它将是:使用myproject . extensions;)

请注意,如果您想使扩展方法只特定于一个特定的类(例如。, Foo),你可以在扩展方法中替换T参数。

序列化(这个Foo值){…}

必须使用XmlSerializer进行XML序列化。下面是一个示例代码片段。

 XmlSerializer xsSubmit = new XmlSerializer(typeof(MyObject));
 var subReq = new MyObject();
 var xml = "";

 using(var sww = new StringWriter())
 {
     using(XmlWriter writer = XmlWriter.Create(sww))
     {
         xsSubmit.Serialize(writer, subReq);
         xml = sww.ToString(); // Your XML
     }
 }

根据泛型类的@kiquenet请求:

public class MySerializer<T> where T : class
{
    public static string Serialize(T obj)
    {
        XmlSerializer xsSubmit = new XmlSerializer(typeof(T));
        using (var sww = new StringWriter())
        {
            using (XmlTextWriter writer = new XmlTextWriter(sww) { Formatting = Formatting.Indented })
            {
                xsSubmit.Serialize(writer, obj);
                return sww.ToString();
            }
        }
    }
}

用法:

string xmlMessage = MySerializer<MyClass>.Serialize(myObj);

我有一个简单的方法来序列化一个对象到XML使用c#,它的工作很棒,它是高度可重用的。我知道这是一个较老的帖子,但我想发布这个帖子,因为有人可能会发现这对他们有帮助。

下面是我如何调用该方法:

var objectToSerialize = new MyObject();
var xmlString = objectToSerialize.ToXmlString();

下面是完成这项工作的类:

注意:由于这些是扩展方法,它们需要在静态类中。

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

public static class XmlTools
{
    public static string ToXmlString<T>(this T input)
    {
        using (var writer = new StringWriter())
        {
            input.ToXml(writer);
            return writer.ToString();
        }
    }

    private static void ToXml<T>(this T objectToSerialize, StringWriter writer)
    {
        new XmlSerializer(typeof(T)).Serialize(writer, objectToSerialize);
    }
}

基于上述解决方案,这里有一个扩展类,您可以使用它序列化和反序列化任何对象。任何其他XML属性都由您决定。

就像这样使用它:

        string s = new MyObject().Serialize(); // to serialize into a string
        MyObject b = s.Deserialize<MyObject>();// deserialize from a string



internal static class Extensions
{
    public static T Deserialize<T>(this string value)
    {
        var xmlSerializer = new XmlSerializer(typeof(T));

        return (T)xmlSerializer.Deserialize(new StringReader(value));
    }

    public static string Serialize<T>(this T value)
    {
        if (value == null)
            return string.Empty;

        var xmlSerializer = new XmlSerializer(typeof(T));

        using (var stringWriter = new StringWriter())
        {
            using (var xmlWriter = XmlWriter.Create(stringWriter, new XmlWriterSettings { Indent = true }))
            {
                xmlSerializer.Serialize(xmlWriter, value);
                return stringWriter.ToString();
            }
        }
    }
}