将动态的 JSON 键映射到一个类字段，使用 JPA。

huangapple
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英文:

Map a dynamic json key to a class field using JPA

问题

我正在尝试将一个 JSON 对象映射到一个实体类（Entity class）。我的 JSON 如下所示：

  1. {
  2.     "availabilities": {
  3.         "2020-07-14T00:00:00Z": "Not Available",
  4.         "2020-08-14T00:00:00Z": "Not Available",
  5.         "2020-06-14T00:00:00Z": "Not Available",
  6.         "2020-05-14T00:00:00Z": "Not Available",
  7.         "2020-00-14T00:00:00Z": "Not Available"
  8.     }
  9. }

我使用 Spring Data 创建了一个实体类：

  1. @Entity
  2. public class Availabilities {
  4.     @Id
  5.     @GeneratedValue(strategy = GenerationType.AUTO)
  6.     private Long id;
  8.     private String date;
  9.     private String availability;
  11.     public String getDate() {
  12.         return date;
  13.     }
  15.     public void setDate(String date) {
  16.         this.date = date;
  17.     }
  19.     public String getAvailability() {
  20.         return availability;
  21.     }
  23.     public void setAvailability(String availability) {
  24.         this.availability = availability;
  25.     }
  26. }

但我得到了以下错误：

  1. com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "2020-07-14T00:00:00Z" (class org.abc.hotel.booking.model.Availabilities)

请帮助。

英文:

I am trying to map a json object to an Entity class. My Json is as follows:

  1.     {
  2.    
  3.     "availabilities": {
  4.     "2020-07-14T00:00:00Z": "Not Available",
  5.     "2020-08-14T00:00:00Z": "Not Available",
  6.     "2020-06-14T00:00:00Z": "Not Available",
  7.     "2020-05-14T00:00:00Z": "Not Available",
  8.     "2020-00-14T00:00:00Z": "Not Available"
  10.       }
  11.     }

I created a entity class using spring data :

  1. @Entity
  2. public class Availabilities {
  5. @Id
  6. @GeneratedValue(strategy = GenerationType.AUTO)
  7. private Long id;
  10. private String date;
  11. private String availibility;
  15. public String getDate() {
  16. 	return date;
  17. }
  18. public void setDate(String date) {
  19. 	this.date = date;
  20. }
  21. public String getAvailibility() {
  22. 	return availibility;
  23. }
  24. public void setAvailibility(String availibility) {
  25. 	this.availibility = availibility;
  26. }
  28. }

But i am getting the following error:

  1. com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "2020- 
  2. 07-14T00:00:00Z" (class org.abc.hotel.booking.model.Availabilities)

Please help.

答案1

得分: 1

你的代码正在尝试按键值映射，这意味着你的变量是键。在你的情况下，JSON 应该如下工作：

  1. // 映射到可用性
  2. {
  3.   "date": "2020-07-14T00:00:00Z",
  4.   "availibility": "Not Available"
  5. }

如果你想解析成列表 => https://stackoverflow.com/questions/44589381/how-to-convert-json-string-into-list-of-java-object

英文:

your code is trying map by key-value, that mean your variable is the key.
in your case. json look like this will work

  1. // map to Availabilities
  2.       {
  3.         "date": "2020-07-14T00:00:00Z",
  4.         "availibility": "Not Available",
  5.       }

if you want to parse into a list => https://stackoverflow.com/questions/44589381/how-to-convert-json-string-into-list-of-java-object

huangapple
  • 本文由 发表于 2020年7月22日 09:56:35
  • 转载请务必保留本文链接：https://go.coder-hub.com/63025625.html
  • java
  • jpa
  • spring
  • spring-data-jpa
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