如何在Unity中获取具有许多变量的JSON数据？

huangapple
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英文:

How in Unity get JSON data that has many variables in it?

问题

我制作了这个代码，可以在API文本中有一个变量及其值时正常工作，但是当文本如下时，我的脚本不起作用，如何修复？我尝试只获取一个变量，但它不起作用，我得到了null。

我的代码可以使用的API文本：https://catfact.ninja/fact

我的代码不能使用的API文本：https://randomuser.me/api

我编辑了代码以从多个变量的JSON中获取数据，但再次得到了null。

英文:

I made this code that works when API text has one variable and it's value, but when text is like this my script doesn't work, how to fix it? I tryed to get only one variable but it didn't work and i got null.

API text that my code work with: https://catfact.ninja/fact

API text that my code does not work with: https://randomuser.me/api

I edited code to get data from many variables JSON, but again got null:

  1. using UnityEngine;
  2. using TMPro;
  3. using UnityEngine.Networking;
  4. using System.Collections;
  5. using System;
  6. using Newtonsoft.Json;
  8. public class Service : MonoBehaviour
  9. {
  10. [SerializeField] private TextMeshProUGUI _text;
  12. private void Start()
  13. {
  14. 	StartCoroutine(GetRequest("https://randomuser.me/api"));
  15. }
  17. private IEnumerator GetRequest(String url)
  18. {
  19. 	using (UnityWebRequest webRequest = UnityWebRequest.Get(url))
  20. 	{
  21. 		yield return webRequest.SendWebRequest();
  23. 		switch (webRequest.result)
  24. 		{
  25. 			case UnityWebRequest.Result.ConnectionError:
  26. 			case UnityWebRequest.Result.DataProcessingError:
  27. 			_text.text = "Error: " + webRequest.error;
  28. 			break;
  30. 			case UnityWebRequest.Result.ProtocolError:
  31. 			_text.text = "HTTP Error: " + webRequest.error;
  32. 			break;
  34. 			case UnityWebRequest.Result.Success:
  35. 			if (CheckAPI(webRequest))
  36. 			{
  37. 				Name name = JsonConvert.DeserializeObject<Name>(webRequest.downloadHandler.text);
  39. 				if (name.title == null || name.first == null || name.last == null)
  40. 				{
  41. 					_text.text = "Error: no information with such variables";
  42. 				}
  43. 				else
  44. 				{
  45. 					_text.text = name.title + " " + name.first + " " + name.last;
  46. 				}
  47. 			}
  48. 			else
  49. 			{
  50. 				_text.text = "Error: there are no API data";
  51. 			}
  52. 			break;
  53. 		}
  54. 	}
  55. }
  57. private bool CheckAPI(UnityWebRequest data)
  58. {
  59. 	string contentType = data.GetResponseHeader("Content-Type");
  61. 	if (!string.IsNullOrEmpty(contentType) && contentType.Contains("application/json"))
  62. 	{
  63. 		return true;
  64. 	}
  65. 	else
  66. 	{
  67. 		return false;
  68. 	}
  69. }
  71. public class Name
  72. {
  73. 	public string title { get; set; }
  74. 	public string first { get; set; }
  75. 	public string last { get; set; }
  76. }
  78. public void NewRequest()
  79. {
  80. 	Start();
  81. }
  82. }

答案1

得分: 2

在第一个示例中，您正在使用名为“Fact”的类（我假设）具有以下结构（或类似结构）：

  1. public class Fact
  2. {
  3.     public string fact { get; set; }
  4.     public int length { get; set; }
  5. }

您将JSON反序列化为该对象：

  1. Fact fact = JsonConvert.DeserializeObject<Fact>(webRequest.downloadHandler.text);

但在第二个示例中，响应不同，因此您需要以下类：

  1. public class myNewClass
  2. {
  3.     public List<Result> results { get; set; }
  4.     public Info info { get; set; }
  5. }
  7. public class Coordinates
  8. {
  9.     public string latitude { get; set; }
  10.     public string longitude { get; set; }
  11. }
  13. // 其他类的翻译...

并使用以下方式进行反序列化：

  1. myNewClass myDeserializedClass = JsonConvert.DeserializeObject<myNewClass>(webRequest.downloadHandler.text);

当然，将您的主类的名称更改为您想要的任何名称。

英文:

In the first example you are working with a class "Fact" with (I suppose) this structure (or similar):

  1. public class Fact
  2. {
  3.     public string fact { get; set; }
  4.     public int length { get; set; }
  5. }

and you are deserializating the json to the object with:

  1.  Fact fact = JsonConvert.DeserializeObject&lt;Fact&gt;(webRequest.downloadHandler.text);

but in the second example the response is different, so you need these classes:

  1. public class myNewClass
  2. {
  3.     public List&lt;Result&gt; results { get; set; }
  4.     public Info info { get; set; }
  5. }
  7. public class Coordinates
  8. {
  9.     public string latitude { get; set; }
  10.     public string longitude { get; set; }
  11. }
  13. public class Dob
  14. {
  15.     public DateTime date { get; set; }
  16.     public int age { get; set; }
  17. }
  19. public class Id
  20. {
  21.     public string name { get; set; }
  22.     public string value { get; set; }
  23. }
  25. public class Info
  26. {
  27.     public string seed { get; set; }
  28.     public int results { get; set; }
  29.     public int page { get; set; }
  30.     public string version { get; set; }
  31. }
  33. public class Location
  34. {
  35.     public Street street { get; set; }
  36.     public string city { get; set; }
  37.     public string state { get; set; }
  38.     public string country { get; set; }
  39.     public int postcode { get; set; }
  40.     public Coordinates coordinates { get; set; }
  41.     public Timezone timezone { get; set; }
  42. }
  44. public class Login
  45. {
  46.     public string uuid { get; set; }
  47.     public string username { get; set; }
  48.     public string password { get; set; }
  49.     public string salt { get; set; }
  50.     public string md5 { get; set; }
  51.     public string sha1 { get; set; }
  52.     public string sha256 { get; set; }
  53. }
  55. public class Name
  56. {
  57.     public string title { get; set; }
  58.     public string first { get; set; }
  59.     public string last { get; set; }
  60. }
  62. public class Picture
  63. {
  64.     public string large { get; set; }
  65.     public string medium { get; set; }
  66.     public string thumbnail { get; set; }
  67. }
  69. public class Registered
  70. {
  71.     public DateTime date { get; set; }
  72.     public int age { get; set; }
  73. }
  75. public class Result
  76. {
  77.     public string gender { get; set; }
  78.     public Name name { get; set; }
  79.     public Location location { get; set; }
  80.     public string email { get; set; }
  81.     public Login login { get; set; }
  82.     public Dob dob { get; set; }
  83.     public Registered registered { get; set; }
  84.     public string phone { get; set; }
  85.     public string cell { get; set; }
  86.     public Id id { get; set; }
  87.     public Picture picture { get; set; }
  88.     public string nat { get; set; }
  89. }
  93. public class Street
  94. {
  95.     public int number { get; set; }
  96.     public string name { get; set; }
  97. }
  99. public class Timezone
  100. {
  101.     public string offset { get; set; }
  102.     public string description { get; set; }
  103. }

and deserializate with:

  1. myNewClass myDeserializedClass = JsonConvert.DeserializeObject&lt;myNewClass&gt;(webRequest.downloadHandler.text);

of course, change the name of your main class to whatever you want.

You can work with this or any other case using the following tool:

https://json2csharp.com/

答案2

得分: 1

如果你只需要一个名称，你可以解析 JSON 字符串并反序列化一个名称属性：

  1.  Name name = JObject.Parse(webRequest.downloadHandler.text)["results"][0]["name"]
  2.                     .ToObject<Name>();
英文:

if you need just a name you can parse a json string and deserilaize a name property

  1.  Name name = JObject.Parse(webRequest.downloadHandler.text)[&quot;results&quot;][0][&quot;name&quot;]
  2.                     .ToObject&lt;Name&gt;();

huangapple
  • 本文由 发表于 2023年4月6日 23:36:12
  • 转载请务必保留本文链接：https://go.coder-hub.com/75951308.html
  • c#
  • json
  • unity-game-engine
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