Deserialize by automatically splitting for each element of a sublist in C#

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英文:

Deserialize by automatically splitting for each element of a sublist in C#

问题

以下是您要求的翻译：

有可能已经存在解决这个问题的答案，但我不太知道如何解释我的问题，所以我会在这里尝试一下。

假设我从一个 JSON 格式的 API 中获取数据。假设它的格式如下：

  1. {
  2.     "name": "Animals",
  3.     "className": "Mammals",
  4.     "speciesList": [
  5.         {
  6.             "species": "Panthera leo",
  7.             "subspecies": "leo"
  8.         },
  9.         {
  10.             "species": "Equus quagga",
  11.             "subspecies": "quagga"
  12.         }
  13.     ]
  14. }

我想将它们存储在以下类中：

  1. public class Animal
  2. {
  3.    public string name {get; set;}
  4.    public string className {get; set;}
  5.    public string species {get; set;}
  6.    public string subspecies {get; set;}
  7. }

如何使用这个类进行反序列化，以便创建两个对象，如下所示：

  1. Animal("Animals", "Mammals", "Panthera leo", "leo")
  2. Animal("Animals", "Mammals", "Equus quagga", "quagga")

我知道我可以提取所有数据，但想知道是否有一种直接的方法来实现这个！提前谢谢！

英文:

It is possible that an answer to this exist but I don't really know how to explain my problem so I'll do it here.

Let's imagine I am grabbing datas from an api in a JSON format. Let's say it has the following format :

  1.     {
  2.         "name": "Animals",
  3.         "className": "Mammals",
  4.         "speciesList": [
  5.             {
  6.                 "species": "Panthera leo",
  7.                 "subspecies": "leo"
  8.             },
  9.             {
  10.                 "species": "Equus quagga",
  11.                 "subspecies": "quagga"
  12.             }
  13.         ]
  14.     }

And I want to store them in the following class :

  1. public class Animal
  2. {
  3.    public string name {get; set;}
  4.    public string className {get; set;}
  5.    public string species {get; set;}
  6.    public string subspecies {get; set;}
  7. }

How can I deserialize it with a class so that it creates me 2 objects, as follows :

  1. Animal("Animals", "Mammals", "Panthera leo", "leo")
  2. Animal("Animals", "Mammals", "Equus quagga", "quagga")

I know that I can extract all but wanted to know if there was a direct way to create this !
Thanks in advance !

答案1

得分: 1

使用 Newtonsoft.Json 库，首先你要将 JSON 提取为 JObject。然后将 speciesList 数组提取为 JArray，这样你可以迭代每个元素并转换为 List<Animal>。

  1. using System.Linq;
  2. using System.Collections.Generic;
  3. using Newtonsoft.Json;
  4. using Newtonsoft.Json.Linq;
  6. JObject jObj = JObject.Parse(json);
  7. List<Animal> animals = (jObj.SelectToken("speciesList") as JArray)
  8. 	.Select(x => new Animal
  9. 			{
  10. 				Name = jObj.Value<string>("name"),
  11. 				ClassName = jObj.Value<string>("className"),
  12. 				Species = x.Value<string>("species"),
  13. 				Subspecies = x.Value<string>("subspecies")
  14. 			})
  15. 	.ToList();

与此同时，建议使用 Pascal 命名法为属性命名。

  1. public class Animal
  2. {
  3.    public string Name {get; set;}
  4.    public string ClassName {get; set;}
  5.    public string Species {get; set;}
  6.    public string Subspecies {get; set;}
  7. }
英文:

With Newtonsoft.Json library, first you extract the JSON as JObject. And extract the speciesList array as JArray, so you can iterate each element and convert into List<Animal>.

  1. using System.Linq;
  2. using System.Collections.Generic;
  3. using Newtonsoft.Json;
  4. using Newtonsoft.Json.Linq;
  6. JObject jObj = JObject.Parse(json);
  7. List<Animal> animals = (jObj.SelectToken("speciesList") as JArray)
  8. 	.Select(x => new Animal
  9. 			{
  10. 				Name = jObj.Value<string>("name"),
  11. 				ClassName = jObj.Value<string>("className"),
  12. 				Species = x.Value<string>("species"),
  13. 				Subspecies = x.Value<string>("subspecies")
  14. 			})
  15. 	.ToList();

Meanwhile, would suggest that to use Pascal case for the property name.

  1. public class Animal
  2. {
  3.    public string Name {get; set;}
  4.    public string ClassName {get; set;}
  5.    public string Species {get; set;}
  6.    public string Subspecies {get; set;}
  7. }

答案2

得分: 1

由于您想将所有代码放在一个类中，您可以考虑这样做：

  1. List<Animal> animals = JsonConvert.DeserializeObject<Animals>(json).SpeciesList;
  3. public class Animals
  4. {
  5. 	public List<Animal> SpeciesList { get; init; }
  7. 	public Animals(string name, string className, List<Animal> speciesList)
  8. 	{
  9. 		speciesList.ForEach(x =>
  10. 		{
  11. 			x.name = name;
  12. 			x.className = className;
  13. 		});
  14. 		SpeciesList = speciesList;
  15. 	}
  16. }
英文:

since you want to place all code inside of one class, you can consider this

  1. List<Animal> animals = JsonConvert.DeserializeObject<Animals>(json).SpeciesList;
  3. public class Animals
  4. {
  5. 	public List<Animal> SpeciesList { get; init; }
  7. 	public Animals(string name, string className, List<Animal> speciesList)
  8. 	{
  9. 		speciesList.ForEach(x =>
  10. 		{
  11. 			x.name = name;
  12. 			x.className = className;
  13. 		});
  14. 		SpeciesList = speciesList;
  15. 	}
  16. }
  18. </details>
  22. # 答案3
  23. **得分**: 0
  25. 你也可以使用 `JsonConvert.DeserializeObject` 将 `JSON` 数据转换为一个 `dynamic` 对象
  27. ```csharp
  28. public class Animal
  29. {
  30.    public string name {get; set;}
  31.    public string className {get; set;}
  32.    public string species {get; set;}
  33.    public string subspecies {get; set;}
  34. }
  36. // 将 JSON 数据反序列化为 Animal 对象列表
  37. string json = "{\"name\": \"Animals\", \"className\": \"Mammals\", \"speciesList\": [{\"species\": \"Panthera leo\",\"subspecies\": \"leo\"},{\"species\": \"Equus quagga\",\"subspecies\": \"quagga\"}]}";
  38. var data = JsonConvert.DeserializeObject<dynamic>(json);
  39. List<Animal> animals = new List<Animal>();
  40. foreach (var species in data.speciesList)
  41. {
  42.     Animal animal = new Animal
  43.     {
  44.         name = data.name,
  45.         className = data.className,
  46.         species = species.species,
  47.         subspecies = species.subspecies
  48.     };
  49.     animals.Add(animal);
  50. }
  51. ```
  53. <details>
  54. <summary>英文:</summary>
  56. You can also use `JsonConvert.DeserializeObject` to convert the `JSON` data into a `dynamic` object
  58.     public class Animal
  59.     {
  60.        public string name {get; set;}
  61.        public string className {get; set;}
  62.        public string species {get; set;}
  63.        public string subspecies {get; set;}
  64.     }
  65.     
  66.     // Deserialize the JSON data into a list of Animal objects
  67.     string json = &quot;{ \&quot;name\&quot;: \&quot;Animals\&quot;, \&quot;className\&quot;: \&quot;Mammals\&quot;, \&quot;speciesList\&quot;: [{\&quot;species\&quot;: \&quot;Panthera leo\&quot;,\&quot;subspecies\&quot;: \&quot;leo\&quot;},{\&quot;species\&quot;: \&quot;Equus quagga\&quot;,\&quot;subspecies\&quot;: \&quot;quagga\&quot;}]}&quot;;
  68.     var data = JsonConvert.DeserializeObject&lt;dynamic&gt;(json);
  69.     List&lt;Animal&gt; animals = new List&lt;Animal&gt;();
  70.     foreach (var species in data.speciesList)
  71.     {
  72.         Animal animal = new Animal
  73.         {
  74.             name = data.name,
  75.             className = data.className,
  76.             species = species.species,
  77.             subspecies = species.subspecies
  78.         };
  79.         animals.Add(animal);
  80.     }
  82. </details>

huangapple
  • 本文由 发表于 2023年5月10日 21:32:20
  • 转载请务必保留本文链接：https://go.coder-hub.com/76219075.html
  • c#
  • class
  • deserialization
  • json
  • oop
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