2017-04-27 53 views
0

我有以下代码中我想获得骑手的id,但我不知道如何做到这一点,其实骑手是链接的用户类型到RiderLocation表,当我改变骑手用户的类型,然后我不能发送的ID在URL的参数这样如何获得JPA存储库中的链接表的ID

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

代码如下

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); 

} 
+0

也是骑士对象的一部分? – sAm

+0

'id'是RiderLOcation id @sAm的一部分 –

回答

0

的问题现在已经解决了 唯一的错误是我写的

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

代替:

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