如何将JSON转换为具有特定字段的POJO。

huangapple
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英文:

How to convert json to pojo with specific fields

问题

  1. import com.fasterxml.jackson.annotation.JsonProperty;
  3. public class CustomPerson {
  5.     @JsonProperty("names")
  6.     private Names names;
  8.     public CustomPerson() {
  9.     }
  11.     public String getGivenName() {
  12.         return this.names == null ? null : this.names.getGivenName();
  13.     }
  15.     public String getFamilyName() {
  16.         return this.names == null ? null : this.names.getFamilyName();
  17.     }
  19.     private static class Names {
  20.         @JsonProperty("displayName")
  21.         private String displayName;
  23.         @JsonProperty("familyName")
  24.         private String familyName;
  26.         @JsonProperty("givenName")
  27.         private String givenName;
  29.         public String getGivenName() {
  30.             return givenName;
  31.         }
  33.         public String getFamilyName() {
  34.             return familyName;
  35.         }
  36.     }
  37. }

Please make sure that you also have the necessary dependencies, such as Jackson, properly configured in your project to handle JSON serialization and deserialization.

英文:

I tried to convert this json (it's a response from google people api) to pojo. i get null instead of object with data. i think my pojo is not correct and jackson cant convert it

  1. {
  2. "resourceName": "people/113175645456469629209",
  3. "etag": "%EgoBAj0DBgk+NTcuGgQBAgnfvtrUH",
  4. "names": [
  5.     {
  6.         "metadata": {
  7.             "primary": true,
  8.             "source": {
  9.                 "type": "PROFILE",
  10.                 "id": "113175645456469629209"
  11.             }
  12.         },
  13.         "displayName": "firstName lastName",
  14.         "familyName": "lastName",
  15.         "givenName": "firstName",
  16.         "displayNameLastFirst": "firstName, lastName",
  17.         "unstructuredName": "firstName, lastName"
  18.     }
  19. ]
  20. }

I want to get only pojo with first name and last name

  1. import com.fasterxml.jackson.annotation.JsonProperty;
  2. import org.springframework.social.google.api.plus.Person;
  4. public class CustomPerson {
  6. 	@JsonProperty("names")
  7. 	private CustomPerson.Names names;
  9. 	public CustomPerson() {
  10. 	}
  12. 	public String getGivenName() {
  13. 		return this.names == null ? null : this.names.givenName;
  14. 	}
  16. 	public String getFamilyName() {
  17. 		return this.names == null ? null : this.names.familyName;
  18. 	}
  20. 	private static class Names {
  21. 		@JsonProperty
  22. 		private String givenName;
  23. 		@JsonProperty
  24. 		private String familyName;
  26. 		private Names() {
  27. 		}
  28. 	}
  29. }

any advice would help me

答案1

得分: 1

  1. public class CustomPerson {
  3.     @JsonProperty
  4.     private List<CustomPerson.Names> names;
  6.     public CustomPerson() {
  7.     }
  9.     private static class Names {
  10.         @JsonProperty
  11.         private String givenName;
  12.         @JsonProperty
  13.         private String familyName;
  15.         private Names() {
  16.         }
  18.         public String getGivenName() {
  19.             return this.names == null ? null : this.names.givenName;
  20.         }
  22.         public String getFamilyName() {
  23.             return this.names == null ? null : this.names.familyName;
  24.         }
  25.     }
  26. }
英文: 
  1. public class CustomPerson {
  3.     @JsonProperty
  4.     private List&lt;CustomPerson.Names&gt; names;
  6.     public CustomPerson() {
  7.     }
  9.     private static class Names {
  10.         @JsonProperty
  11.         private String givenName;
  12.         @JsonProperty
  13.         private String familyName;
  15.         private Names() {
  16.         }
  18.         public String getGivenName() {
  19.             return this.names == null ? null : this.names.givenName;
  20.         }
  22.         public String getFamilyName() {
  23.             return this.names == null ? null : this.names.familyName;
  24.         }
  25.     }
  26. }

huangapple
  • 本文由 发表于 2020年9月18日 19:51:51
  • 转载请务必保留本文链接：https://go.coder-hub.com/63955242.html
  • jackson
  • java
  • json
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