导入具有非典型格式的XML到一个类中

huangapple
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英文:

Import XML with a non typical format to a class

问题

我需要导入一个看起来像这样的XML文件：

  1.     <Carpark>
  2.        <Trucks>
  3.           <Truck brand='Chrysler' kg='2100'/>
  4.        </Trucks>
  5.        <Cars>
  6.           <Car brand='Mercedes' kg='1100'/>
  7.        </Cars>
  8.        <Tractors>
  9.           <Tractor brand='John Deere' kg='1500'/>
  10.        </Tractors>
  11.     </Carpark>

基本上，我想要有一个这样的Carpark列表：

  1.     public class Carpark
  2.     {
  3.         // Type can be for example Truck, Car
  4.         public string Type { get; set; }
  5.         public string Brand { get; set; }
  6.         public int Weight { get; set; }
  7.     }

我不明白如何将这个XML文件转换成我想要的类。

我尝试过：

  1.     XElement xElement = XElement.Load(filePath);
  2.     IEnumerable<Carpark> carpark = xElement.Elements();
  4.     foreach(var vehicle in carpark)
  5.     {
  6.         // 一旦我写出这行，我对接下来该怎么做感到困惑
  7.         Console.WriteLine(vehicle.ToString());
  8.     }
英文:

I need to import a XML file looking like this:

  1. <Carpark>
  2.    <Trucks>
  3.       <Truck brand='Chrysler' kg='2100'/>
  4.    </Trucks>
  5.    <Cars>
  6.       <Car brand='Mercedes' kg='1100'/>
  7.    </Cars>
  8.    <Tractors>
  9.       <Tractor brand='John Deere' kg='1500'/>
  10.    </Tractors>
  11. </Carpark>

Essentially I want to have a list of Carpark like this:

  1. public class Carpark
  2. {
  3.     // Type can be for example Truck, Car
  4.     public string Type { get; set; }
  5.     public string Brand { get; set; }
  6.     public int Weight { get; set; }
  7. }

I don't understand how to convert this XML file to the class I want.

What I have tried is:

  1. XElement xElement = XElement.Load(filePath);
  2. IEnumerable<Carpark> carpark = xElement.Elements();
  4. foreach(var vehicle in carpark)
  5. {
  6.     // Once I wrote out the line I was confused as to how to proceed
  7.     Console.WriteLine(vehicle.ToString());
  8. }

答案1

得分: 1

以下是您要翻译的内容：

使用Linq_To_XML和我相信还可以使用NewtonsoftJSON来完成这个操作：

  1. void Main()
  2. {
  3.     var xml = XElement.Parse(x)
  4.         .DescendantsAndSelf("Carpark")
  5.         .Descendants()
  6.         .Where(xe => xe.HasAttributes)
  7.         .Select(xe => new Carpark {
  8.             Type = xe.Name.ToString(),
  9.             Brand = (string)xe.Attribute("brand"),
  10.             Weight = (int)xe.Attribute("kg")
  11.         });
  13.     foreach (var cp in xml)
  14.     {
  15.         Console.WriteLine($"{cp.Type}, {cp.Brand}, {cp.Weight}");
  16.     }
  17. }
  19. static readonly string x = @"<Carpark>
  20.    <Trucks>
  21.       <Truck brand='Chrysler' kg='2100'/>
  22.    </Trucks>
  23.    <Cars>
  24.       <Car brand='Mercedes' kg='1100'/>
  25.    </Cars>
  26.    <Tractors>
  27.       <Tractor brand='John Deere' kg='1500'/>
  28.    </Tractors>
  29. </Carpark>";
  31. public class Carpark
  32. {
  33.     // Type can be for example Truck, Car
  34.     public string Type { get; set; }
  35.     public string Brand { get; set; }
  36.     public int Weight { get; set; }
  37. }
英文:

You can do that with Linq_To_XML and I believe with NewtonsoftJSON:

  1. void Main()
  2. {
  3. 	var xml = XElement.Parse(x)
  4. 		.DescendantsAndSelf(&quot;Carpark&quot;)
  5. 		.Descendants()
  6. 		.Where(xe =&gt; xe.HasAttributes)
  7. 		.Select(xe =&gt; new Carpark {
  8. 			Type = xe.Name.ToString(),
  9. 			Brand = (string)xe.Attribute(&quot;brand&quot;),
  10. 			Weight = (int)xe.Attribute(&quot;kg&quot;)
  11. 		})
  12. 		;
  13. 		
  14. 	foreach (var cp in xml)
  15. 	{
  16. 		Console.WriteLine($&quot;{cp.Type}, {cp.Brand}, {cp.Weight}&quot;);
  17. 	}
  18. 	
  19. }
  21. static readonly string x = @&quot;&lt;Carpark&gt;
  22.    &lt;Trucks&gt;
  23.       &lt;Truck brand=&#39;Chrysler&#39; kg=&#39;2100&#39;/&gt;
  24.    &lt;/Trucks&gt;
  25.    &lt;Cars&gt;
  26.       &lt;Car brand=&#39;Mercedes&#39; kg=&#39;1100&#39;/&gt;
  27.    &lt;/Cars&gt;
  28.    &lt;Tractors&gt;
  29.       &lt;Tractor brand=&#39;John Deere&#39; kg=&#39;1500&#39;/&gt;
  30.    &lt;/Tractors&gt;
  31. &lt;/Carpark&gt;&quot;;
  33. public class Carpark
  34. {
  35. 	// Type can be for example Truck, Car
  36. 	public string Type { get; set; }
  37. 	public string Brand { get; set; }
  38. 	public int Weight { get; set; }
  39. }

答案2

得分: 0

你可以使用以下类来自动使用 XmlSerializer 进行反序列化：

  1. public class Carpark
  2. {
  3.     public List<Truck> Trucks { get; set; }
  4.     public List<Car> Cars { get; set; }
  5.     public List<Tractor> Tractors { get; set; }
  6. }
  8. [XmlInclude(typeof(Truck))]
  9. [XmlInclude(typeof(Car))]
  10. [XmlInclude(typeof(Tractor))]
  11. public class Vehicle
  12. {
  13.     [XmlIgnore]
  14.     public string Type => this.GetType().Name;
  16.     [XmlAttribute("brand")]
  17.     public string Brand { get; set; }
  19.     [XmlAttribute("kg")]
  20.     public int Weight { get; set; }
  21. }
  23. public class Truck : Vehicle
  24. {
  25. }
  27. public class Car : Vehicle
  28. {
  29. }
  31. public class Tractor : Vehicle
  32. {
  33. }

然后，你可以使用以下代码来进行反序列化：

  1. var cp = (Carpark)serializer.Deserialize(someStreamReader);

如果需要更多帮助，请告诉我。

英文:

You can use the following classes to deserialize this with XmlSerializer automatically

  1. public class Carpark
  2. {
  3. 	public List&lt;Truck&gt; Trucks { get; set; }
  4. 	public List&lt;Car&gt; Cars { get; set; }
  5. 	public List&lt;Tractor&gt; Tractors { get; set; }
  6. }
  8. [XmlInclude(typeof(Truck))]
  9. [XmlInclude(typeof(Car))]
  10. [XmlInclude(typeof(Tractor))]
  11. public class Vehicle
  12. {
  13. 	[XmlIgnore]
  14. 	public string Type =&gt; this.GetType().Name;
  15. 	
  16. 	[XmlAttribute(&quot;brand&quot;)]
  17. 	public string Brand { get; set; }
  18. 	
  19. 	[XmlAttribute(&quot;kg&quot;)]
  20. 	public int Weight { get; set; }
  21. }
  23. public class Truck : Vehicle
  24. {
  25. }
  27. public class Car : Vehicle
  28. {
  29. }
  31. public class Tractor : Vehicle
  32. {
  33. }
  1. var cp = (Carpark)serializer.Deserialize(someStreamReader);

dotnetfiddle

huangapple
  • 本文由 发表于 2023年6月22日 00:46:09
  • 转载请务必保留本文链接：https://go.coder-hub.com/76525495.html
  • c#
  • linq-to-xml
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