JpaRepository 返回列表中第一项的子项，然后仅返回其余项的 id

Question

我有以下Post 类：

@Entity
@Table(name = "posts")
@Getter
@Setter
@JsonIdentityInfo(      generator = ObjectIdGenerators.PropertyGenerator.class, 
                        property  = "id", 
                        scope     = Long.class)
public class Post {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    
    private String title;
    
    private String subtitle;
    
    private String content;
    
    private String img_url;
    
    @CreationTimestamp
    private Timestamp created_on;
    
    @UpdateTimestamp
    private Timestamp last_updated_on;
    
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "owner_id", nullable=false)
    @JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
    private User creator;
    
}

以及以下扩展 JpaRepository 的存储库

@Repository
public interface PostRepository extends JpaRepository<Post, Long> {

    Optional<Post> findById(Long id);
    
    List<Post> findAll();
    
}

当在下面的控制器中返回findAll()的结果时，只有第一个创建者项目被完全发送，其余的只包含id：

@GetMapping("/news")
    public List<Post> getNews() {
        return postRepository.findAll();
    }

这是我得到的 JSON：

[
{"id":15,"title":"Title example #1","subtitle":"Subtitle example #1","content":"Lorem #1 ipsum dolor sit amet","img_url":null,"created_on":"2021-12-01T00:00:00.000+00:00","last_updated_on":"2021-12-01T00:00:00.000+00:00","creator":{"id":1,"username":"user-example","email":"blablabla@gmail.com","roles":[{"id":1,"name":"ROLE_USER"}]}}

,{"id":25,"title":"Title example #2","subtitle":"Subtitle example #2","content":"Lorem #2 ipsum dolor sit amet","img_url":null,"created_on":"2021-12-01T00:00:00.000+00:00","last_updated_on":"2021-12-01T00:00:00.000+00:00","creator":1}
]

为什么会这样？有没有办法为 JSON 数组中的每个元素获取整个子对象？

谢谢

编辑：添加了User 类

@Entity
@Table( name = "users", 
        uniqueConstraints = { 
            @UniqueConstraint(columnNames = "username"),
            @UniqueConstraint(columnNames = "email") 
        })
@DiscriminatorValue(value="USER")
public class User extends OwnerEntity {

    @NotBlank
    @NotNull
    @Size(max = 20)
    private String username;

    @NotBlank
    @NotNull
    @Size(max = 50)
    @Email
    private String email;

    @NotBlank
    @Size(max = 120)
    @JsonIgnore
    private String password;
    
    @CreationTimestamp
    private Timestamp created_on;
    
    @UpdateTimestamp
    private Timestamp last_updated_on;

    @ManyToMany(fetch = FetchType.LAZY)
    @JoinTable( name = "user_roles", 
                joinColumns = @JoinColumn(name = "user_id"), 
                inverseJoinColumns = @JoinColumn(name = "role_id"))
    private Set<Role> roles = new HashSet<>();

    @ManyToMany(fetch = FetchType.LAZY)
    private Set<Institution> institutions;
    
    @OneToMany(mappedBy="creator", fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    protected Set<Post> posts;
    
    @ManyToMany(fetch = FetchType.LAZY)
    private Set<Institution> following;
}

编辑 2：添加 OwnerEntity 类

@Entity
@Table(name = "entities")
@Inheritance(strategy = InheritanceType.JOINED)
@DiscriminatorColumn
@Getter
@Setter
@JsonIdentityInfo(      generator = ObjectIdGenerators.PropertyGenerator.class, 
                        property  = "id", 
                        scope     = Long.class)
public class OwnerEntity {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    protected Long id;
    
}

Answer 1

您的OwnerEntity 也有@JsonIdentityInfo。在其reference documentation 中，我们可以阅读以下内容：

Annotation 用于指示带注释类型的值或 属性应该序列化，以便实例包含 附加对象标识符（除了实际对象属性）， 或作为由引用完整的对象 id 组成的引用 序列化。在实践中，这是通过序列化第一个 实例作为完整的对象和对象标识，以及其他对 对象作为参考值。

这完美地解释了为什么你会得到这样的 JSON。如果您不想这样做，只需删除 @JsonIdentityInfo，但它可能会在序列化双向关系时修复无限递归（您可以在以下在线资源 https://www.baeldung.com/jackson-bidirectional-relationships-and-infinite-recursion 中阅读更多相关信息）。