How to make a JPA query use primitive ID instead of managed instance?-CodePudding

I tried but I couldn't find a better title so it looks weird but the question is actually very simple...

I have a JPA query which uses a foreigner key, by default JPA takes the managed entity as foreigner key to execute the queries, is there any way I can do it passing the long id instead?

EXAMPLE:

@Entity
@Table(name = "user")
public class User   {

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

     


    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER, orphanRemoval = true, mappedBy = "user")
    private Set<Item> entries = new HashSet<>();
    
    }
    
    
    @Entity
@Table(name = "item") 
public class Item implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

 
    @ManyToOne(targetEntity = User.class, optional = false, fetch = FetchType.LAZY)
    private User user;
    
}


@Repository
public interface ItemRepository extends PagingAndSortingRepository<User, Long> {


    // @formatter:off
    @Query("SELECT item FROM Item item WHERE (:user is null OR :user = item.user)")
    Page<Item> findByUser(@Param("user")User user, Pageable pageable);
    //@formatter:on

}

What I want is instead of passing the user entity to the repository class, pass only its id, this way i can take id as parameter and avoid one more query on database just to retrieve the user from its id

how can i do that

CodePudding user response：

Well simply make the parameter as Long and then

@Query("SELECT item FROM Item item WHERE (:userId is null OR :userId = item.user.id)")
Page<Item> findByUser(@Param("userId")Long userId, Pageable pageable);