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

要序列化一个对象,请执行:

 using (StreamWriter myWriter = new StreamWriter(path, false))
 {
     XmlSerializer mySerializer = new XmlSerializer(typeof(your_object_type));
     mySerializer.Serialize(myWriter, objectToSerialize);
 }

还要记住,要使XmlSerializer工作,需要一个无参数构造函数。

其他回答

要序列化一个对象,请执行:

 using (StreamWriter myWriter = new StreamWriter(path, false))
 {
     XmlSerializer mySerializer = new XmlSerializer(typeof(your_object_type));
     mySerializer.Serialize(myWriter, objectToSerialize);
 }

还要记住,要使XmlSerializer工作,需要一个无参数构造函数。

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);
    }
}

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。

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;