如何将使用密钥的 XML 反序列化为对象

Question

我有一个 ASP.NET 核心 2.1 MVC 项目，我正在从 Serlilog MSSqlServr 接收器检索数据，该接收器将值作为 XML 数据类型存储在属性字段中。我想将该数据反序列化到视图模型中，以便我可以将各个元素作为视图中的数据呈现。

这里是数据库日志 table 中属性字段的 XML 示例。

<properties>
  <property key="EventId">
    <structure type="">
      <property key="Id">404</property>
    </structure>
  </property>
  <property key="ActionId">0592d9e8-f4fd-459f-96b3-2b787d01a754</property>
  <property key="ActionName">API.Controllers.CompletionsController.GetCompletion (PS.API)</property>
  <property key="RequestId">0HLJ2IL5A9:00000001</property>
  <property key="RequestPath">/api/completions/0</property>
  <property key="CorrelationId" />
  <property key="ConnectionId">0HLJ2IL59</property>
  <property key="MachineName">RD0003FF1</property>
  <property key="ThreadId">117</property>
</properties>

我设置了一个class用于解码如下；

using System.Xml.Serialization;

namespace PS.Models.ApiLogs
{
    [XmlRoot("properties")]
    public class LogProperties
    {

        [XmlElement("SourceContext")]
        public string SourceContext { get; set; }

        [XmlElement("ActionId")]
        public string ActionId { get; set; }

        [XmlElement("ActionName")]
        public string ActionName { get; set; }

        [XmlElement("RequestId")]
        public string RequestId { get; set; }

        [XmlElement("RequestPath")]
        public string RequestPath { get; set; }

        [XmlElement("CorrelationId")]
        public string CorrelationId { get; set; }

        [XmlElement("ConnectionId")]
        public string ConnectionId { get; set; }

        [XmlElement("MachineName")]
        public string MachineName { get; set; }

        [XmlElement("ThreadId")]
        public string ThreadId { get; set; }

    }
}

在我的控制器中，我有以下代码；

        var serializer = new XmlSerializer(typeof(LogProperties));

        LogProperties logProperties;

        using (TextReader reader = new StringReader(log.Properties))
        {
            logProperties = (LogProperties)serializer.Deserialize(reader);
        }

但是 logProperties 变量不包含任何内容，所以我假设我在 LogProperties class.

中的 XML 属性不正确

我花了很多时间寻找解决方案，并且在输入此问题时查看了所有相关帖子，但我找不到 XML 使用 [=51 的示例=] 或如何处理 "key=" 属性属性（如果这是正确的术语）

有什么想法吗？

[2019 年 2 月 21 日更新]

我最终使用了@jdweng 的建议，因为它最不复杂，并且给了我想要的东西。

我创建了 2 个 classes（因为我喜欢将我的 class 文件分开作为个人偏好）。 classes 在下面；

using System.Collections.Generic;
using System.Xml.Serialization;

namespace PS.Models.ApiLogs
{
    [XmlRoot("properties")]
    public class LogProperties
    {
        [XmlElement("property")]
        public List<LogProperty> Property { get; set; }

    }
}

和

using System.Xml.Serialization;

namespace PS.Models.ApiLogs
{
    [XmlRoot("property")]
    public class LogProperty
    {
        [XmlAttribute("key")]
        public string Key { get; set; }
        [XmlText]
        public string Value { get; set; }
    }
}

然后在我的控制器中，我有以下 Detail 方法；

        var response = await _client.GetLogAsync(id, $"api/logs", token);
        if (response == null)
        {
            return NotFound($"Unable to find a record for Log ID [{id}].");
        }

        var log = _mapper.Map<DetailLogViewModel>(response.Record);

        var serializer = new XmlSerializer(typeof(LogProperties));

        LogProperties logProperties;

        using (TextReader reader = new StringReader(log.Properties))
        {
            logProperties = (LogProperties)serializer.Deserialize(reader);
        }

        var logWithProperties = new DetailLogWithPropertiesViewModel
        {
            Id = log.Id,
            Message = log.Message,
            TimeStamp = log.TimeStamp,
            Exception = log.Exception,
            XmlProperties = logProperties 
        };


        return View(logWithProperties);

下面是我的 DetailLogWithPropertiesViewModel；

public class DetailLogWithPropertiesViewModel
{
    public int Id { get; set; }

    [Display(Name = "Message")]
    public string Message { get; set; }

    [Display(Name = "Level")]

    public string Level { get; set; }

    [Display(Name = "Time Stamp")]
    public DateTimeOffset TimeStamp { get; set; }

    [Display(Name = "Exception")]
    public string Exception { get; set; }

    [Display(Name = "Properties")]
    public string Properties { get; set; }

    public LogProperties XmlProperties { get; set; }

}

下面是我的 Detail.cshtml 的相关部分；

<div class="card-body ml3 mr3">


    @foreach (var logProperty in Model.XmlProperties.Property)
    {
        <div class="row">
            <div class="col-4 bg-light border border-primary">
                <span class="font-weight-bold">@logProperty.Key</span>
            </div>
            <div class="col-8 bg-secondary border border-left-0 border-primary">
                <span>@logProperty.Value</span>
            </div>
        </div>


    }

</div>

由 Serilog 生成并存储在 MS SQL 数据库中的 XML 具有可变数量的属性，我现在理解它们表示为 key/value 对。所以这个方法可以让我确保所有提供的属性都显示在网站的日志查看器中。

Answer 1

尝试以下操作：

    [XmlRoot("properties")]
    public class LogProperties
    {

        [XmlElement("property")]
        public List<LogProperty> property { get; set; }

    }
    [XmlRoot("property")]
    public class LogProperty
    {
        [XmlAttribute("key")]
        public string key { get; set; }
        [XmlText]
        public string value { get; set; }
    }

Answer 2

1) 将您的 XML 复制到剪贴板...
2) ...打开 Visual Studio 并创建一个空的 cs 文件...
3) ...转到编辑 > 选择性粘贴 > XML 到类（ 可能需要安装 ASP.NET webdevelopment）...
3) ...将导致此代码：

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
public partial class properties
{

    private propertiesProperty[] propertyField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("property")]
    public propertiesProperty[] property
    {
        get
        {
            return this.propertyField;
        }
        set
        {
            this.propertyField = value;
        }
    }
}

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class propertiesProperty
{

    private propertiesPropertyStructure structureField;

    private string[] textField;

    private string keyField;

    /// <remarks/>
    public propertiesPropertyStructure structure
    {
        get
        {
            return this.structureField;
        }
        set
        {
            this.structureField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlTextAttribute()]
    public string[] Text
    {
        get
        {
            return this.textField;
        }
        set
        {
            this.textField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string key
    {
        get
        {
            return this.keyField;
        }
        set
        {
            this.keyField = value;
        }
    }
}

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class propertiesPropertyStructure
{

    private propertiesPropertyStructureProperty propertyField;

    private string typeField;

    /// <remarks/>
    public propertiesPropertyStructureProperty property
    {
        get
        {
            return this.propertyField;
        }
        set
        {
            this.propertyField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string type
    {
        get
        {
            return this.typeField;
        }
        set
        {
            this.typeField = value;
        }
    }
}

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class propertiesPropertyStructureProperty
{

    private string keyField;

    private ushort valueField;

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string key
    {
        get
        {
            return this.keyField;
        }
        set
        {
            this.keyField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlTextAttribute()]
    public ushort Value
    {
        get
        {
            return this.valueField;
        }
        set
        {
            this.valueField = value;
        }
    }
}

4) 是的，代码很多o_O。使用 XmlDeserializer(typeof(properties))

Answer 3

如果您创建 LogProperties 的示例实例 class 并像这样对其进行序列化：

var serializer = new XmlSerializer(typeof(LogProperties));

LogProperties logProperties = new LogProperties() 
{ 
    SourceContext = "MySourceContext",
    ActionId = "MyActionId",
    ActionName = "MyActionName"
};

StringBuilder sb = new StringBuilder();
using (StringWriter writer = new StringWriter(sb))
{
    serializer.Serialize(writer, logProperties);
}

Console.WriteLine(sb.ToString());

这是你得到的：

<?xml version="1.0" encoding="utf-16"?>
<properties xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <SourceContext>MySourceContext</SourceContext>
  <ActionId>MyActionId</ActionId>
  <ActionName>MyActionName</ActionName>
</properties>

所以您拥有的 class 与您拥有的 XML 并不相配。双向尝试序列化总是有用的。

下面是一种可以将您拥有的 XML 序列化为您拥有的 class 的方法（明确区分对象模型和持久性格式）。它使用 System.Xml.Linq 命名空间中的 XDocument class。

// Must escape all quotes !
string xmlSample = @"
<properties>
    <property key=""EventId"">
        <structure type = """">
            <property key=""Id"">404</property>
        </structure>
    </property>
    <property key=""ActionId""> 0592d9e8 - f4fd - 459f - 96b3 - 2b787d01a754</property>
    <property key=""ActionName""> API.Controllers.CompletionsController.GetCompletion(PS.API)</property>
    <property key=""RequestId""> 0HLJ2IL5A9: 00000001</property>
    <property key=""RequestPath"">/api/completions/0</property>
    <property key=""CorrelationId"" />
    <property key=""ConnectionId"">0HLJ2IL59</property>
    <property key=""MachineName"">RD0003FF1</property>
    <property key=""ThreadId"">117</property>
</properties>";

StringReader reader = new StringReader(xmlSample);
XDocument xdoc = XDocument.Parse(xmlSample);

LogProperties log = new LogProperties();
foreach (XElement prop in xdoc.Root.Elements())
{
    switch (prop.Attribute("key").Value)
    {
    case "ActionId" : log.ActionId = prop.Value; break;
    case "ActionName" : log.ActionName = prop.Value; break;

    // and so on

    }
}

请注意，使用此方法：

您的 class
你不需要塑造你的 class 来适应 XML（或相反）
您可以添加自己的 LogProperties 自定义初始化 class（不再需要默认构造函数）
您可以添加自己的自定义错误处理

如何将使用密钥的 XML 反序列化为对象

How do I deserialize XML that uses a key into an object

c#

xml

xml-parsing

xmlserializer

asp.net-core-2.1