Question

我使用Spring Data JPA并且有两个参与者：

@Entity
@Data
@Table(name="vehicle_entity")
public class Vehicle {
    @Id
    @GeneratedValue
    private Integer id;
    private String type;
    private String vehicleRegNumber;
    @OneToMany(targetEntity = com.transport.model.BookedTime.class, fetch=FetchType.EAGER)
    @JoinColumn(name="bookingTime", referencedColumnName="id")
    Set<BookingTime> bookingTime;
}

和

@Entity
@Data
@Table(name="booked_time")
public class BookedTime {
    @Id
    @GeneratedValue
    Integer id;
    Long startPeriod;
    Long finishPeriod;

}

和存储库：

public interface VehicleRepository extends JpaRepository<Vehicle, Integer> {

   @Query("correct query")
    List<Vehicle> findAllByPeriod(@Param("startPeriod") int startPeriod, @Param("endPeriod") int endPeriod);

}

我需要找到availbale车辆，没有按时间预订。有谁知道我该怎么做。

Answer 1

我会选择类似的东西：

 @Query(nativeQuery=true, value = "select * from vehicle_entity v join booked_time b on b.vehicle = v.id where not (b.startPeriod > :startPeriod and b.endPeriod < :endPeriod)"

BTW我认为您可能想尝试将EAGER的FetchType更改为LAZY bookingTime，然后在选择查询中使用join fetch。

Answer 2

所以我找到了几个解决我问题的方法。第一个是由Enigo提供的，第二个是由

提供的

   @Query(nativeQuery = false, value = "select v from Vehicle v right join fetch v.bookedTime b \n" +
           "Where b.vehicle = v.id " +
           "AND (b.startPeriod < :startPeriod OR b.startPeriod > :finishPeriod) " +
           "AND (b.finishPeriod < :startPeriod OR b.finishPeriod > :finishPeriod)")
   Set<Vehicle> findAllByPeriod(@Param("startPeriod") Long startPeriod, @Param("finishPeriod") Long finishPeriod);

如何在Spring Data JPA中找到具有OneToMany关系的cortege

2 个答案: