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

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);
    }
}

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。

我的工作代码。返回utf8 xml启用空命名空间。

// override StringWriter
public class Utf8StringWriter : StringWriter
{
    public override Encoding Encoding => Encoding.UTF8;
}

private string GenerateXmlResponse(Object obj)
{    
    Type t = obj.GetType();

    var xml = "";

    using (StringWriter sww = new Utf8StringWriter())
    {
        using (XmlWriter writer = XmlWriter.Create(sww))
        {
            var ns = new XmlSerializerNamespaces();
            // add empty namespace
            ns.Add("", "");
            XmlSerializer xsSubmit = new XmlSerializer(t);
            xsSubmit.Serialize(writer, obj, ns);
            xml = sww.ToString(); // Your XML
        }
    }
    return xml;
}

示例返回响应Yandex api支付Aviso url:

<?xml version="1.0" encoding="utf-8"?><paymentAvisoResponse xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" performedDatetime="2017-09-01T16:22:08.9747654+07:00" code="0" shopId="54321" invoiceId="12345" orderSumAmount="10643" />

我修改了我的返回一个字符串,而不是像下面这样使用一个ref变量。

public static string Serialize<T>(this T value)
{
    if (value == null)
    {
        return string.Empty;
    }
    try
    {
        var xmlserializer = new XmlSerializer(typeof(T));
        var stringWriter = new StringWriter();
        using (var writer = XmlWriter.Create(stringWriter))
        {
            xmlserializer.Serialize(writer, value);
            return stringWriter.ToString();
        }
    }
    catch (Exception ex)
    {
        throw new Exception("An error occurred", ex);
    }
}

它的用法是这样的:

var xmlString = obj.Serialize();

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);

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();
    }
}