如何在避免 n+1 查询的同时设置 JPA 双向一对一

Question

我正在尝试调试具有双向一对一关系的两个实体。 问题在于该关系导致发生 N+1 查询。 当我运行findAll()时，会针对我表中的每个项目执行单个查询。

这是一个例子：我有两个实体

package ...;
import ...;

@AllArgsConstructor(access = AccessLevel.PRIVATE)
@NoArgsConstructor
@EqualsAndHashCode
@Getter
@Builder

@Entity
@Table(name = "tracker")
public class TrackerEntity {
    @Id
    @Column(name = "vehicleid")
    private UUID vehicleId;

    @Column(name = "trackerid")
    private String trackerId;

    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "vehicleid", referencedColumnName = "id")
    private VehicleEntity vehicleEntity;
}

@AllArgsConstructor(access = AccessLevel.PRIVATE)
@NoArgsConstructor
@EqualsAndHashCode
@Getter
@Builder

@Entity
@Table(name = "vehicle")
public class VehicleEntity {
    @Id
    private UUID id;

    @Column(name = "vehiclename")
    private String vehicleName;

    @OneToOne(mappedBy = "vehicleEntity", cascade = CascadeType.REMOVE, orphanRemoval = true)
    private TrackerEntity trackerEntity;
}

当我运行vehicleRepository.findAll()时，我得到以下调试日志：

2022-05-17 19:16:45.071 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.097 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.104 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.106 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.109 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.110 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.113 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.115 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.117 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.119 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.123 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.125 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.128 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.130 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.133 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.135 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.138 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.141 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.144 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_0_, vehicleent0_.vehiclename as vehiclen2_1_0_ from vehicle vehicleent0_ where vehicleent0_.id=?
2022-05-17 19:16:45.146 DEBUG 62639 --- [           main] org.hibernate.SQL                        : insert into vehicle (vehiclename, id) values (?, ?)
2022-05-17 19:16:45.246 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select vehicleent0_.id as id1_1_, vehicleent0_.vehiclename as vehiclen2_1_ from vehicle vehicleent0_
2022-05-17 19:16:45.253 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.258 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.259 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.261 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.262 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.264 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.265 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.267 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.269 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?
2022-05-17 19:16:45.271 DEBUG 62639 --- [           main] org.hibernate.SQL                        : select trackerent0_.vehicleid as vehiclei1_0_0_, trackerent0_.trackerid as trackeri2_0_0_ from tracker trackerent0_ where trackerent0_.vehicleid=?

我搜索了 N+1 JPA 查询问题，希望找到解决方案，大多数建议使用 EntityGraph 或“字节码增强”。 我已经尝试使用实体图，但仍然有 n+1 个查询，我不想使用字节码增强，因为应用程序必须作为微服务在云上运行。

更重要的一件事是我希望在移除车辆时删除跟踪器，但不应在创建车辆时创建跟踪器。

有没有什么办法解决这一问题？

我试过使用 Query 和 join fetch：

public interface VehicleRepository extends JpaRepository<VehicleEntity, UUID> {

    @Override
    @Query("SELECT v FROM VehicleEntity v left outer join fetch v.trackerEntity t")
    <S extends VehicleEntity> List<S> findAll(Example<S> example);
}

这并没有改变任何东西，仍然有 n+1 个查询。 我试过 EntityGraph 也没有改变任何东西：

@Entity
@Table(name = "vehicle")
@NamedEntityGraph(name="vehicleGraph", attributeNodes = {@NamedAttributeNode("trackerEntity")})
public class VehicleEntity {
    @Id
    private UUID id;

    @Column(name = "vehiclename")
    private String vehicleName;

    @OneToOne(mappedBy = "vehicleEntity", cascade = CascadeType.REMOVE, orphanRemoval = true)
    private TrackerEntity trackerEntity;
}

public interface VehicleRepository extends JpaRepository<VehicleEntity, UUID> {

    @Override
    @EntityGraph(value = "vehicleGraph", type = EntityGraph.EntityGraphType.LOAD)
    <S extends VehicleEntity> List<S> findAll(Example<S> example);
}

Answer 1

您可以使用EntityGraph和延迟加载对其进行优化。 我假设VehicleEntity是这里的拥有实体。 在那种情况下， make fetch = FetchType.LAZY和cascade = CascadeType.ALL应该在拥有的实体上。 EntityGraph 可以通过两种方式定义

1. 命名实体图
2. AD-HOC 或未命名实体图

命名实体图：

您可以在拥有 class 时定义NamedEntityGraph并在VehicleEntityRepository接口中引用它

@NamedEntityGraph(name = "VehicleEntity.trackerEntity", attributeNodes = @NamedAttributeNode("trackerEntity"))
public class VehicleEntity {
}

VehicleEntityRepository.java

public interface VehicleEntityRepository extends JpaRepository<VehicleEntity, UUID> {
     @EntityGraph(value = "VehicleEntity.trackerEntity", type = EntityGraphType.LOAD)
    List<VehicleEntity> findAll();
}

AD-HOC 实体图：

您不必定义任何 EntityGraph 注释 VehicleEntity class 而是可以直接使用它 Repository 接口

VehicleEntityRepository.java

public interface VehicleEntityRepository extends JpaRepository<VehicleEntity, UUID> {
    @EntityGraph(attributePaths = { "trackerEntity" })
    List<VehicleEntity> findAll();
}

VehicleEntity.java

@AllArgsConstructor(access = AccessLevel.PRIVATE)
@NoArgsConstructor
@EqualsAndHashCode
@Getter
@Builder

@Entity
@Table(name = "vehicle")
@NamedEntityGraph(name = "VehicleEntity.trackerEntity", attributeNodes = @NamedAttributeNode("trackerEntity"))
public class VehicleEntity {
    @Id
    private UUID id;

    @Column(name = "vehiclename")
    private String vehicleName;

 
@OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true, mappedBy = "vehicleEntity")
    @JsonManagedReference
    private TrackerEntity trackerEntity;
}

TrackerEntity.java

package ...;
import ...;

@AllArgsConstructor(access = AccessLevel.PRIVATE)
@NoArgsConstructor
@EqualsAndHashCode
@Getter
@Builder

@Entity
@Table(name = "tracker")
public class TrackerEntity {
    @Id
    @Column(name = "id")
    private UUID id;

    @Column(name = "trackerid")
    private String trackerId;

    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "vehicleid", referencedColumnName = "id")
@JsonBackReference
    private VehicleEntity vehicleEntity;
}
    

在使用 Jackson 时，您还可以使用@JsonManagedReference和@JsonBackReference来防止嵌套获取或无限递归。