Question

我使用过Spring和Hibernate。现在看看带有JPA 2.2的Spring Data JPA（2.0.3）

AgencyTicketType

@Entity
@Table(name = "agency_ticket_type", catalog = "test")
public class AgencyTicketType implements java.io.Serializable {
   private Long id;
   private String name;
   private Agency agency;
   private Set<AgencyTicketCategory> agencyTicketCategories = new HashSet<AgencyTicketCategory>(0);

   @Id
   @GeneratedValue(strategy = IDENTITY)
   @Column(name = "id", unique = true, nullable = false)
   public Long getId() {
     return this.id;
   }
   public void setId(Long id) {
    this.id = id;
   }
   @Column(name = "name", nullable = false, length = 100)
   public String getName() {
    return this.name;
   }
   public void setName(String name) {
    this.name = name;
   }
   @ManyToOne(fetch = FetchType.LAZY)
   @JoinColumn(name = "agency_id", nullable = false)
   public Agency getAgency() {
     return this.agency;
   }
   public void setAgency(Agency agency) {
     this.agency = agency;
   }
   @OneToMany(fetch = FetchType.LAZY, mappedBy = "agencyTicketType")
   public Set<AgencyTicketCategory> getAgencyTicketCategories() {
      return this.agencyTicketCategories;
   }
   public void setAgencyTicketCategories(Set<AgencyTicketCategory> agencyTicketCategories) {
      this.agencyTicketCategories = agencyTicketCategories;
   }
}

AgencyTicketCategory

@Entity
@Table(name = "agency_ticket_category", catalog = "waytest")
public class AgencyTicketCategory implements java.io.Serializable {
    private Long id;
    private AgencyTicketType agencyTicketType;
    private String name;
    private BigDecimal price;
    private Set<TripTicket> tripTickets = new HashSet<TripTicket>(0);

    @Id
    @GeneratedValue(strategy = IDENTITY)

    @Column(name = "id", unique = true, nullable = false)
    public Long getId() {
        return this.id;
    }
    public void setId(Long id) {
        this.id = id;
    }
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "agency_ticket_type_id", nullable = false)
    public AgencyTicketType getAgencyTicketType() {
        return this.agencyTicketType;
    }
    public void setAgencyTicketType(AgencyTicketType agencyTicketType) {
        this.agencyTicketType = agencyTicketType;
    }
    @Column(name = "name", nullable = false, length = 100)
    public String getName() {
        return this.name;
    }
    public void setName(String name) {
        this.name = name;
    }
    @Column(name = "price", nullable = false, precision = 8)
    public BigDecimal getPrice() {
        return this.price;
    }
    public void setPrice(BigDecimal price) {
        this.price = price;
    }
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "agencyTicketCategory")
    public Set<TripTicket> getTripTickets() {
        return this.tripTickets;
    }
    public void setTripTickets(Set<TripTicket> tripTickets) {
        this.tripTickets = tripTickets;
    }
}

存储库

public interface TicketTypeRepository extends JpaRepository<AgencyTicketType, Long> {

  @EntityGraph(attributePaths={ "agencyTicketCategories" }, type=EntityGraphType.LOAD)
  @Query("select type from AgencyTicketType type where type.agency.code=?1")
  List<AgencyTicketType> findByAgency(String agencyCode);
}

服务

@Service
public class TicketServiceImpl implements TicketService {       
    @Autowired private TicketTypeRepository ticketType;

    @Transactional(readOnly=true)
    @Override
    public List<AgencyTicketType> findByName(String code) {
        return ticketType.findByAgency(code);
    }    
}

在服务上调试时，看起来，查询急切地获取所有延迟加载的属性 - agency，agencyTicketCategories - 以及所有内部延迟加载的属性，这会导致 JSON serilization error < /强>

只需要获取这些

AgencyTicketTypes [ { id, name, agencyTicketCategories [ {id,name,price},.... ] },..... ]

我可以使用@EntityGraph执行此操作吗？我缺少什么？

Answer 1

指定延迟加载只是JPA提供程序的提示。根据您使用的提供程序（Hibernate，EclipseLink等），可能会完全忽略它，并且可能会急切地获取依赖项。

您需要做的是配置类映射到json的方式。假设您使用的是Jackson，则可能需要使用@JsonIgnore或@JsonView等注释。您也可以映射只有您需要的字段的类。

Answer 2

您可以使用杰克逊注释@JsonBackReference / @JsonManagedReference。它们解决了对象模型中双向链接的无限递归问题。据我所知，这是你的情况。

有关详细信息，请参阅http://www.baeldung.com/jackson-bidirectional-relationships-and-infinite-recursion。

Answer 3

有一点需要注意的是，在事务处于打开状态时进行调试（触摸集合）会导致它被加载，即使实时它也没有...另一件事就是@Apokralipsa提到的，LAZY加载只是一个可以完全忽略的提示，不应该依赖于你正在使用的任何技术

Spring JPA EntityGraph获取所有延迟加载的属性

3 个答案: