Question

我有以下代码，其中我想获得骑手的id，但我不知道如何做到这一点，实际上rider是链接到RiderLocation表的用户类型，当我改变时用户的骑手类型然后我不能像这样

在URL的参数中发送id

http://localhost:3000/api/riderLocations/search/findByRider?rider_id=3

代码低于

RiderLocation.java

@Entity
public class RiderLocation implements Serializable {

    private static final long serialVersionUID = 1L;
        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        private Long id;

        @ManyToOne
        private User rider;

        private Double latitude;
        private Double longitude;

        public Long getId() {
            return id;
        }

        public void setId(Long id) {
            this.id = id;
        }
        public User getRider() {
            return rider;
        }
        public void setRider(User rider) {
            this.rider = rider;
        }
        public Double getLatitude() {
            return latitude;
        }

        public void setLatitude(Double latitude) {
            this.latitude = latitude;
        }

        public Double getLongitude() {
            return longitude;
        }
      public void setLongitude(Double longitude) {
            this.longitude = longitude;
        }
    }

RiderLocationRepository

public interface RiderLocationRepository extends JpaRepository<RiderLocation, Long>{

    @Query("select r.latitude,r.longitude from RiderLocation r ")
    ArrayList<Object[]>   findAllRidersLocation();
    RiderLocation findByRider(@Param("rider_id") Long rider);

}

Answer 1

问题现在解决了唯一的错误是我写的

findByRider(@Param("rider_id") Long rider);

而不是：

findByRiderId(@Param("rider_id") Long rider);

如何在JPA存储库中获取链接表的ID

1 个答案: