Question

我想从数据库加载一个UserReference对象，但是从verifier属性我想只加载id，firstName和lastName，这样userReference看起来像这样：

{
  "id": 1,
  "company": "company1",
  "companyContactName": "some name",
  "companyPosition": "programmer",
  "referenceDate": "02/04/2005",
  "verifier": {
        "id":1
        "firstName": "Jane",
        "lastName": "Smith"
        "email":null,
        "username":null,
        "office:null,
        "department":null
  }
}

我使用了UserReference类的实体图，但是我使用的实体图加载了用户拥有的所有信息，包括电子邮件，用户名，办公室和部门。有没有办法在子图中指定类似EntityGraphType.FETCH的东西，以便它只加载验证者的id，firstName和lastName？

这是我的UserReferenceRepository：

public interface UserReferenceRepository extends JpaRepository<UserReference, Long>{

    @EntityGraph(value = "userReferenceGraph" ,  type = EntityGraphType.FETCH )
    UserReference findOne(Long id);
}

UserReference类：

@Getter
@Setter
@EqualsAndHashCode (exclude = {"id", "verifier"})
@ToString(exclude = {"id"})
@Entity
@NamedEntityGraphs({
    @NamedEntityGraph(
        name = "userReferenceGraph",      
        attributeNodes = {
            @NamedAttributeNode(value = "verifier", subgraph = "verifierGraph")
    },
    subgraphs = {
        @NamedSubgraph( 
            name = "verifierGraph",
            type = User.class,
            attributeNodes = {
                @NamedAttributeNode(value = "id"),
                @NamedAttributeNode(value = "firstName"),
                @NamedAttributeNode(value = "lastName")})})
})

public class UserReference {

    @Id
    @GeneratedValue
    private Long id;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "user_id", referencedColumnName = "user_id", foreignKey = @ForeignKey (name = "FK_UserReference_UserHRDetails_user_id"))
    @JsonIgnore
    private UserHRDetails hrDetails;

    private String company;
    private String companyContactName;
    private String companyPosition;
    private Date referenceDate;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "verifier_id")
    private User verifier;
}

和用户：

@Getter @Setter
@EqualsAndHashCode(exclude = {"id", "department", "company", "authorities", "hrDetails"})
@ToString(exclude = {"password"})
@Entity
@AllArgsConstructor
@Builder
public class User implements Serializable{

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Access(value = AccessType.PROPERTY)
    private Long id;  

    @Size(max = 50)
    @Column(name = "first_name", length = 50)
    private String firstName;

    @Size(max = 50)
    @Column(name = "last_name", length = 50)
    private String lastName;

    @Column(length = 100, unique = true, nullable = false)
    private String email;

    @Column(length = 50, unique = true, nullable = false)
    private String username;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "department_id")
    private Department department;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "office_id")
    private Office office;
}

Answer 1

我想你使用Jackson生成JSON。在这种情况下，这是杰克逊与实体图的战斗，前者没有机会赢得这场战斗。实体图只是构建SQL查询的提示，您只能告诉Hibernate不加载某些属性。 Hibernate在加载基本实体字段时仍然不支持实体图，请参阅https://hibernate.atlassian.net/browse/HHH-9270。但主要的问题是，杰克逊会在JSON生成过程中调用你的实体中的每个getter，而Hibernate将懒得加载它们而不考虑你的实体图。我只能提出@JsonIgnore用法，但这可能不像你需要的那样灵活。

Answer 2

我遇到了相同的问题，并且看到了两种解决方法：

快速：您可以在您的实体中执行一些@PostLoad动作，并使不需要的字段无效。

@PostLoad
private void postLoad() {
    if (verifier != null) {
        verifier.email = null;
        verifier.office = null;
        verifier.department = null;
    }
}

正确： 另一种方法是通过将实体转换为DTO来保护您的实体。创建单独的POJO，并将您的User和UserReference转换为该DTO POJO类。在那里，您肯定会更好地控制自己的反应。

如何只加载@NamedEntityGraph中子图中的指定属性

2 个答案: