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Question

我必须XML（de）序列化以下类：

enter image description here

这给出了以下输出：

<ArrayOfPropertyFilter xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                       xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <PropertyFilter>
    <AndOr>And</AndOr>
    <LeftBracket>None</LeftBracket>
    <Property>17</Property>
    <Operator>Equal</Operator>
    <Value xsi:type="xsd:string">lll</Value>
    <RightBracket>None</RightBracket>
  </PropertyFilter>
</ArrayOfPropertyFilter>

并且，在反序列化之后它给出了 enter image description here

如何“告诉”Serializer保持Value“按原样”，没有任何XML节点....（在具体情况下，Value应该是“lll”而不是包含Text“lll”的XMLNode）？

修改

Bellow是C＃中的完整工作示例。输出是

Value is ='System.Xml.XmlNode []'

using System;
using System.IO;
using System.Text;
using System.Windows.Forms;
using System.Xml;
using System.Xml.Serialization;

namespace WindowsFormsApplication13
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_Load(object sender, EventArgs e)
        {
            PropertyFilter filter = new PropertyFilter();
            filter.AndOr = "Jora";
            filter.Value = "haha";
            filter.Property = 15;

            var xml = filter.SerializeToString();
            XmlDocument xmlDoc = new XmlDocument();
            xmlDoc.LoadXml(xml);

            PropertyFilter cloneFilter = xmlDoc.Deserialize<PropertyFilter>();

            Console.WriteLine("Value is = '{0}'", cloneFilter.Value);
        }
    }

    public class PropertyFilter
    {
        public string AndOr { get; set; }
        public string LeftBracket { get; set; }
        public int Property { get; set; }
        public string Operator { get; set; }
        public object Value { get; set; }
        public string RightBracket { get; set; }
    }

    public static class Utils
    {
        public static string SerializeToString(this object instance)
        {
            if (instance == null)
                throw new ArgumentNullException("instance");
            StringBuilder sb = new StringBuilder();
            StringWriter sw = new StringWriter(sb);
            XmlSerializer serializer = new XmlSerializer(
                instance.GetType(), null, new Type[0], null, null);
            serializer.Serialize(sw, instance);
            return sb.ToString();
        }

        public static T Deserialize<T>(this XmlDocument xmlDoc)
        {
            XmlNodeReader reader = new XmlNodeReader(xmlDoc.DocumentElement);
            XmlSerializer serializer = new XmlSerializer(typeof(T));
            object obj = serializer.Deserialize(reader);
            T myT = (T)obj;
            return myT;
        }

    }
}

编辑2

强调Anton的答案，第二个例子（用Groo的评论更新）：

using System;
using System.IO;
using System.Text;
using System.Windows.Forms;
using System.Xml;
using System.Xml.Serialization;

namespace WindowsFormsApplication13
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_Load(object sender, EventArgs e)
        {
            PropertyFilter filter = new PropertyFilter();
            filter.AndOr = "Jora";
            var obj = new Hehe();
            obj.Behehe = 4526;
            filter.Value = obj;
            filter.Property = 15;

            var xml = filter.SerializeToString();
            PropertyFilter cloneFilter = xml.Deserialize<PropertyFilter>();

            Console.WriteLine("Value is = '{0}'", cloneFilter.Value);
        }
    }

    public class Hehe
    {
        public int Behehe { get; set; }
        public override string ToString()
        {
            return string.Format("behehe is '{0}'", Behehe);
        }
    }

    public class PropertyFilter
    {
        public string AndOr { get; set; }
        public string LeftBracket { get; set; }
        public int Property { get; set; }
        public string Operator { get; set; }
        //[XmlElement(typeof(Hehe))]
        public object Value { get; set; }
        public string RightBracket { get; set; }
    }

    public static class Utils
    {
        public static string SerializeToString(this object instance)
        {
            if (instance == null)
                throw new ArgumentNullException("instance");
            StringBuilder sb = new StringBuilder();

            XmlSerializer serializer = new XmlSerializer(instance.GetType(), null, new Type[0], null, null);
            using (StringWriter sw = new StringWriter(sb))
            {
                serializer.Serialize(sw, instance);
            }
            return sb.ToString();
        }

        public static T Deserialize<T>(this string xmlString)
        {
            XmlSerializer serializer = new XmlSerializer(typeof(T));
            using (StringReader sr = new StringReader(xmlString))
            {
                return (T)serializer.Deserialize(sr);
            }
        }

    }
}

Answer 1

好的，这是有道理的，你使用XmlDocument进行反序列化。只需使用String（或任何流阅读器，如@Seb already pointed out），它就可以使用：

public static class Utils
{
    public static string SerializeToString(this object instance)
    {
        if (instance == null)
            throw new ArgumentNullException("instance");

        StringBuilder sb = new StringBuilder();
        XmlSerializer serializer = new XmlSerializer(instance.GetType());

        using (StringWriter sw = new StringWriter(sb))
        {
            serializer.Serialize(sw, instance);
        }

        return sb.ToString();
    }

    public static T Deserialize<T>(this string xmlString)
    {
        XmlSerializer serializer = new XmlSerializer(typeof(T));
        using (StringReader sr = new StringReader(xmlString))
        {
            return (T)serializer.Deserialize(sr);
        }
    }
}

用法：

// serialize
var xml = filter.SerializeToString();

// deserialize
var cloneFilter = xml.Deserialize<PropertyFilter>();

<强> [编辑]

另外，作为旁注：永远不要忘记处置实现IDisposable的对象。这就是为任何实例化的using添加Stream构造的原因。

<强> [EDIT2]

正如Anton所说，在这种情况下，您需要明确指定其他类型，因为XmlSerializer不会搜索所有可能的类型来查找匹配的类。

稍微更好的解决方案可能是使用接受这些类型的XmlSerializer重载（这样您就不需要手动添加属性）：

public static T Deserialize<T>(this string xmlString)
{
    Type[] typesToInclude = GetAllPossibleTypes();

    XmlSerializer serializer = new XmlSerializer(typeof(T), typesToInclude);
    using (StringReader sr = new StringReader(xmlString))
    {
        return (T)serializer.Deserialize(sr);
    }
}

这可以在应用程序启动时完成一次，但是您需要提供适当的程序集（或多个程序集）以确保涵盖所有可能的类型：

public static class Utils
{
    private static readonly Type[] _typesToInclude = GetPossibleUserTypes();

    private static Type[] GetPossibleUserTypes()
    {
        // this part should be changed to load types from the assembly
        // that contains your potential Value candidates
        Assembly assembly = Assembly.GetAssembly(typeof(PropertyFilter));

        // get public classes only
        return assembly.GetTypes().Where(t => t.IsPublic && !t.IsAbstract).ToArray();
    }

    public static string SerializeToString(this object instance)
    {
        if (instance == null)
            throw new ArgumentNullException("instance");

        var sb = new StringBuilder();
        var serializer = new XmlSerializer(instance.GetType(), _typesToInclude);

        using (StringWriter sw = new StringWriter(sb))
        {
            serializer.Serialize(sw, instance);
        }

        return sb.ToString();
    }

    public static T Deserialize<T>(this string xmlString)
    {
        var serializer = new XmlSerializer(typeof(T), _typesToInclude);
        using (StringReader sr = new StringReader(xmlString))
        {
            return (T)serializer.Deserialize(sr);
        }
    }
}

Answer 2

只要Value获取原始，XSD定义类型（如string和int）的值，或T中某处提到的用户定义类型，Gree的解决方案就会有效的定义（T本身，其属性的类型等）。一旦您需要反序列化与这些类型不同的类型的值，您必须声明所有可能类型的{ {1}} Value，例如

XmlElementAttribute

Answer 3

好的，试试这个，但我不确定你想要达到的目的。

public class PropertyFilter
{
    public string AndOr {get; set;}
    public string LeftBracket {get; set;}
    public int Property {get; set;}
    public string Operator {get; set;}
    public object Value {get; set;}
    public string RightBracket {get; set;}
}
    public void MyMethod()
    {
        using (System.IO.StreamReader reader = new System.IO.StreamReader(@"Input.xml"))
        {
            System.Xml.Serialization.XmlSerializer serializer = new XmlSerializer(typeof(PropertyFilter[]));
            PropertyFilter[] deserialized = (PropertyFilter[])serializer.Deserialize(reader);
        }
    }

我只是将您的示例XML放在Input.xml文件中。我希望这会有所帮助。

（De）序列化“对象”

修改

编辑2

3 个答案: