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() 对我的 table.

中的每个项目执行一个查询

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() 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=?

我已经搜索了 N+1 个 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，因为应用程序必须运行在云上作为微服务。

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

有什么办法可以解决这个问题吗？

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);
}

没有任何改变，仍然有 n+1 个查询。我已经尝试过 EntityGraph，它也没有任何改变：

@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 是这里的所有者实体。在那种情况下，make fetch = FetchType.LAZY 和 cascade = CascadeType.ALL 应该在拥有实体上。 EntityGraph 可以通过两种方式定义

命名实体图
AD-HOC 或未命名实体图

命名实体图：

你可以定义 NamedEntityGraph 拥有 class 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 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;
}

您还可以使用 @JsonManagedReference 和 @JsonBackReference 来防止在使用 Jackson 时出现嵌套提取或无限递归。

如何在避免 n+1 查询的同时设置 JPA 双向一对一

How to setup JPA bi-directional one to one while avoiding n+1 queries

java

spring

spring-boot

hibernate

jpa