Question

我正在使用Hibernate作为ORM和Jackson作为JSON序列化程序的Spring启动应用程序。

我有三个模型对象和所有三个模型的CRUD操作。

Class Student{
     private Teacher teacher;  // Teacher of the student — to be fetched eagerly
    +Getter/Setter
}

class Teacher {
      private List<Subject> subject;  // List of subjects associated to that user— to be fetched eagerly
      +Getter/Setter 
}

class Subject {
     private long subjectId 
    //Other subject properties
    + Getter/Setter
}

每当我触发学生信息的获取请求时，我都会获得正确的教师信息，因为我也收到了主题信息，这对我来说是不必要的。在我请求教师信息的同时，我需要主题信息与此相关联。如果我使用@JsonBackReference作为主题我会一直失去它。我不知道如何实现这一目标。

提前感谢您的帮助!!

Answer 1

您可以使用JSON Views

来自春季博客：

public class View {
  interface Summary {}
}

public class User {

  @JsonView(View.Summary.class)
  private Long id;

  @JsonView(View.Summary.class)
  private String firstname;

  @JsonView(View.Summary.class)
  private String lastname;

  private String email;
  private String address;
  private String postalCode;
  private String city;
  private String country;
}

public class Message {

  @JsonView(View.Summary.class)
  private Long id;

  @JsonView(View.Summary.class)
  private LocalDate created;

  @JsonView(View.Summary.class)
  private String title;

  @JsonView(View.Summary.class)
  private User author;

  private List<User> recipients;

  private String body;
}

并在控制器中

@RestController
public class MessageController {

  @Autowired
  private MessageService messageService;

  @JsonView(View.Summary.class)
  @RequestMapping("/")
  public List<Message> getAllMessages() {
    return messageService.getAll();
  }

  @RequestMapping("/{id}")
  public Message getMessage(@PathVariable Long id) {
    return messageService.get(id);
  }
}

PS：目前没有链接到http://fasterxml.com/。

Answer 2

您也可以这样注释

Class Student{
    @JsonIgnoreProperties("subject")
    private Teacher teacher;  // Teacher of the student — to be fetched eagerly
}

忽略Jackson OnDemand中的嵌套属性

2 个答案: