9

我有以下代码：

基本内容.cs

public class BaseContent
{
   // Some auto properties
}

新闻.cs

public class News : BaseContent
{
   // Some more auto properties
}

事件.cs

public class Event : BaseContent
{
   // Some more auto properites
}

通用响应.cs

public class GenericResponse<T> 
{
  [XmlArray("Content")]
  [XmlArrayItem("NewsObject", typeof(News)]
  [XmlArrayItem("EventObject", typeof(Event)]
  public List<T> ContentItems { get; set; }
}

新闻响应.cs

public class NewsResponse : GenericResponse<News> {}

事件响应.cs

public class EventResponse : GenericResponse<Event> {}

如您所见，我有一个基类BaseContent和两个派生自它的类。接下来我有一个通用响应类，因为 xml 文件的结构始终相同，但它们在某些属性上有所不同。

我想我可以指定[XmlArrayItem]用于特定类的名称。但现在我得到了错误：

System.InvalidOperationException：无法生成临时类（结果 = 1）。错误 CS0012：类型“System.Object”在未引用的程序集中定义。您必须添加对程序集“System.Runtime, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a”的引用。

我无法添加此参考，因为我正在使用 Windows 8 应用程序。

如果我注释掉其中一个，[XmlArrayItem]它运行良好。

有人有解决这个问题的想法吗？

更新 我不能使用 DataContractSerializer，因为我必须使用 XmlAttributes

4

4 回答 4

11

编辑：随意下载演示项目

您没有提供对象的所有属性，所以请允许我添加一些 - 仅作为示例：

public class BaseContent
{
    [XmlAttribute("Name")]
    public string Name { get; set; }
}

[XmlType(TypeName = "EventObject")]
public class Event : BaseContent
{
    [XmlAttribute("EventId")]
    public int EventId { get; set; }
}

[XmlType(TypeName = "NewsObject")]
public class News : BaseContent
{
    [XmlAttribute("NewsId")]
    public int NewsId { get; set; }
}

GenericResponse.cs 可以这样定义——不需要为数组项指定 typeof：

public class GenericResponse<T>
{
    [XmlArray("Content")]
    public List<T> ContentItems { get; set; }

    public GenericResponse()
    {
        this.ContentItems = new List<T>();
    }
}

然后你有响应类：

public class EventResponse : GenericResponse<Event>
{
}

public class NewsResponse : GenericResponse<News>
{
}

示例 1：序列化 EventResponse 对象

var response = new EventResponse
{
    ContentItems = new List<Event> 
    {
        new Event {
            EventId = 1,
            Name = "Event 1"
        },

        new Event {
            EventId = 2,
            Name = "Event 2"
        }
    }
};

string xml = XmlSerializer<EventResponse>.Serialize(response);

输出 XML：

<?xml version="1.0" encoding="utf-8"?>
<EventResponse xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <Content>
        <EventObject Name="Event 1" EventId="1" />
        <EventObject Name="Event 2" EventId="2" />
    </Content>
</EventResponse>

如果您尝试对 NewsResponse 进行相同的操作，它将正常工作。顺便说一句，我正在使用我的通用 XmlSerializer，单击链接以了解更多信息。

XmlSerializer.cs：

/// <summary>
/// XML serializer helper class. Serializes and deserializes objects from/to XML
/// </summary>
/// <typeparam name="T">The type of the object to serialize/deserialize.
/// Must have a parameterless constructor and implement <see cref="Serializable"/></typeparam>
public class XmlSerializer<T> where T: class, new()
{
    /// <summary>
    /// Deserializes a XML string into an object
    /// Default encoding: <c>UTF8</c>
    /// </summary>
    /// <param name="xml">The XML string to deserialize</param>
    /// <returns>An object of type <c>T</c></returns>
    public static T Deserialize(string xml)
    {
        return Deserialize(xml, Encoding.UTF8, null);
    }

    /// <summary>
    /// Deserializes a XML string into an object
    /// Default encoding: <c>UTF8</c>
    /// </summary>
    /// <param name="xml">The XML string to deserialize</param>
    /// <param name="encoding">The encoding</param>
    /// <returns>An object of type <c>T</c></returns>
    public static T Deserialize(string xml, Encoding encoding)
    {
        return Deserialize(xml, encoding, null);
    }

    /// <summary>
    /// Deserializes a XML string into an object
    /// </summary>
    /// <param name="xml">The XML string to deserialize</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlReaderSettings"/></param>
    /// <returns>An object of type <c>T</c></returns>
    public static T Deserialize(string xml, XmlReaderSettings settings)
    {
        return Deserialize(xml, Encoding.UTF8, settings);
    }

    /// <summary>
    /// Deserializes a XML string into an object
    /// </summary>
    /// <param name="xml">The XML string to deserialize</param>
    /// <param name="encoding">The encoding</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlReaderSettings"/></param>
    /// <returns>An object of type <c>T</c></returns>
    public static T Deserialize(string xml, Encoding encoding, XmlReaderSettings settings)
    {
        if (string.IsNullOrEmpty(xml))
            throw new ArgumentException("XML cannot be null or empty", "xml");

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

        using (MemoryStream memoryStream = new MemoryStream(encoding.GetBytes(xml)))
        {
            using (XmlReader xmlReader = XmlReader.Create(memoryStream, settings))
            {
                return (T) xmlSerializer.Deserialize(xmlReader);
            }
        }
    }

    /// <summary>
    /// Deserializes a XML file.
    /// </summary>
    /// <param name="filename">The filename of the XML file to deserialize</param>
    /// <returns>An object of type <c>T</c></returns>
    public static T DeserializeFromFile(string filename)
    {
        return DeserializeFromFile(filename, new XmlReaderSettings());
    }

    /// <summary>
    /// Deserializes a XML file.
    /// </summary>
    /// <param name="filename">The filename of the XML file to deserialize</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlReaderSettings"/></param>
    /// <returns>An object of type <c>T</c></returns>
    public static T DeserializeFromFile(string filename, XmlReaderSettings settings)
    {
        if (string.IsNullOrEmpty(filename))
            throw new ArgumentException("filename", "XML filename cannot be null or empty");

        if (! File.Exists(filename))
            throw new FileNotFoundException("Cannot find XML file to deserialize", filename);

        // Create the stream writer with the specified encoding
        using (XmlReader reader = XmlReader.Create(filename, settings))
        {
            System.Xml.Serialization.XmlSerializer xmlSerializer = new System.Xml.Serialization.XmlSerializer(typeof(T));
            return (T) xmlSerializer.Deserialize(reader);
        }
    }

    /// <summary>
    /// Serialize an object
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <returns>A XML string that represents the object to be serialized</returns>
    public static string Serialize(T source)
    {
        // indented XML by default
        return Serialize(source, null, GetIndentedSettings());
    }

    /// <summary>
    /// Serialize an object
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="namespaces">Namespaces to include in serialization</param>
    /// <returns>A XML string that represents the object to be serialized</returns>
    public static string Serialize(T source, XmlSerializerNamespaces namespaces)
    {
        // indented XML by default
        return Serialize(source, namespaces, GetIndentedSettings());
    }

    /// <summary>
    /// Serialize an object
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlWriterSettings"/></param>
    /// <returns>A XML string that represents the object to be serialized</returns>
    public static string Serialize(T source, XmlWriterSettings settings)
    {
        return Serialize(source, null, settings);
    }

    /// <summary>
    /// Serialize an object
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="namespaces">Namespaces to include in serialization</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlWriterSettings"/></param>
    /// <returns>A XML string that represents the object to be serialized</returns>
    public static string Serialize(T source, XmlSerializerNamespaces namespaces, XmlWriterSettings settings)
    {
        if (source == null)
            throw new ArgumentNullException("source", "Object to serialize cannot be null");

        string xml = null;
        XmlSerializer serializer = new XmlSerializer(source.GetType());

        using (MemoryStream memoryStream = new MemoryStream())
        {
            using (XmlWriter xmlWriter = XmlWriter.Create(memoryStream, settings))
            {
                System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(typeof(T));
                x.Serialize(xmlWriter, source, namespaces);

                memoryStream.Position = 0; // rewind the stream before reading back.
                using (StreamReader sr = new StreamReader(memoryStream))
                {
                    xml = sr.ReadToEnd();
                } 
            }
        }

        return xml;
    }

    /// <summary>
    /// Serialize an object to a XML file
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="filename">The file to generate</param>
    public static void SerializeToFile(T source, string filename)
    {
        // indented XML by default
        SerializeToFile(source, filename, null, GetIndentedSettings());
    }

    /// <summary>
    /// Serialize an object to a XML file
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="filename">The file to generate</param>
    /// <param name="namespaces">Namespaces to include in serialization</param>
    public static void SerializeToFile(T source, string filename, XmlSerializerNamespaces namespaces)
    {
        // indented XML by default
        SerializeToFile(source, filename, namespaces, GetIndentedSettings());
    }

    /// <summary>
    /// Serialize an object to a XML file
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="filename">The file to generate</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlWriterSettings"/></param>
    public static void SerializeToFile(T source, string filename, XmlWriterSettings settings)
    {
         SerializeToFile(source, filename, null, settings);
    }

    /// <summary>
    /// Serialize an object to a XML file
    /// </summary>
    /// <param name="source">The object to serialize</param>
    /// <param name="filename">The file to generate</param>
    /// <param name="namespaces">Namespaces to include in serialization</param>
    /// <param name="settings">XML serialization settings. <see cref="System.Xml.XmlWriterSettings"/></param>
    public static void SerializeToFile(T source, string filename, XmlSerializerNamespaces namespaces, XmlWriterSettings settings)
    {
        if (source == null)
            throw new ArgumentNullException("source", "Object to serialize cannot be null");

        XmlSerializer serializer = new XmlSerializer(source.GetType());

        using (XmlWriter xmlWriter = XmlWriter.Create(filename, settings))
        {
            System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(typeof(T));
            x.Serialize(xmlWriter, source, namespaces);
        }
    }

    #region Private methods


    private static XmlWriterSettings GetIndentedSettings()
    {
        XmlWriterSettings xmlWriterSettings = new XmlWriterSettings();
        xmlWriterSettings.Indent = true;
        xmlWriterSettings.IndentChars = "\t";

        return xmlWriterSettings;
    }

    #endregion
}
于 2012-12-30T23:33:37.053 回答
3

我找到了我的解决方案。不是最好的，但它正在工作。

我实现了 IXmlSerializable 并自己处理了这些东西：

public void ReadXml(System.Xml.XmlReader reader)
{
    reader.Read();
    reader.MoveToContent();

    if (reader.LocalName == "AnotherNode")
    {
        var innerXml = Serializer<AnotherClass>.CreateSerializer();
        Remove = (AnotherClass) innerXml.Deserialize(reader);
        reader.MoveToContent();
    }

    reader.Read();

    // Here is the trick
    if (reader.IsStartElement())
    {
        do
        {
            var innerXml = Serializer<T>.CreateSerializer();

            var obj = (T) innerXml.Deserialize(reader);
            Updates.Add(obj);
        } while (reader.MoveToContent() == XmlNodeType.Element);
    }
}

public void WriteXml(System.Xml.XmlWriter writer)
{
    var removeSerializer = Serializer<RemoveElement>.CreateSerializer();
    removeSerializer.Serialize(writer, Remove);

    if (Updates.Any())
    {
        var innerXml = Serializer<T>.CreateSerializer();
        writer.WriteStartElement("ContentUpdates");
        foreach (var update in Updates)
        {
            innerXml.Serialize(writer, update);
        }
        writer.WriteEndElement();
    }
}
于 2012-12-27T08:44:27.823 回答
0

考虑使用 DataContractSerializer 而不是 XmlSerializer。

XmlSerializer 在运行时生成一个临时程序集，以加快序列化 - 稍后的反序列化过程。因此它需要代码编译。

另一方面，DataContractSerializer 没有。如果您使用 DataContractSerializer 方法，则必须使用适当的属性，以便控制“继承”问题。

于 2012-12-23T11:25:39.677 回答
0

这听起来像是专门针对 Windows Store 平台的代码发射器中的一个错误XmlSerializer，但它有点没有实际意义，因为它在常规 .NET 上仍然失败，但有不同的消息：

无法生成临时类（结果=1）。

错误 CS0030：无法将类型“新闻”转换为“事件”

错误 CS1502：“System.Collections.Generic.List.Add(News)”的最佳重载方法匹配有一些无效参数

错误 CS1503：参数 1：无法从“事件”转换为“新闻”

所以基本上，它看起来像是一个不受支持的场景，在两种场景中表现不同。从根本上说，问题在于您的注释在说（for T= News）“如果这News是 a News，则调用元素'NewsObject'；如果这News是 an Event，则调用元素'EventObject'”-这很奇怪，因为 a从来都不News是， ETC。Event

然而，好消息是你想做的事情可以通过使用来完成[XmlType(...)]：

[XmlType("NewsObject")]
public class News : BaseContent
{
    // Some more auto properties
    public int B { get; set; }
}
[XmlType("EventObject")]
public class Event : BaseContent
{
    // Some more auto properites
    public int C { get; set; }
}
...
public class GenericResponse<T>
{
    [XmlArray("Content")]
    public List<T> ContentItems { get; set; }
}

注意：[XmlArrayItem(...)]以上没有。

然后输出（在格式化、清理等之后）：

<NewsResponse>
    <Content>
        <NewsObject><A>1</A><B>2</B></NewsObject>
    </Content>
</NewsResponse>

通过测试代码：

var obj = new NewsResponse { ContentItems = new List<News> {
    new News { A = 1, B = 2 } } };

var sw = new StringWriter();
using (var xw = System.Xml.XmlWriter.Create(sw))
{
    var ser = new XmlSerializer(obj.GetType());
    ser.Serialize(xw, obj);
}
string xml = sw.ToString();

如果您需要更多的控制权，那么XmlAttributeOverrides就是要考虑的事情；但是您必须缓存用 , 来创建的序列化XmlAttributeOverrides程序，以避免程序集出血和内存泄漏。

于 2013-01-02T08:11:33.360 回答