Spring JPA一对多加入

Question

@Entity
@Getter @Setter
public class IndieApp {
    @Id
    @Column(name = "indie_app_id")
    private Long id;

    @Column(name = "name")
    private String name;

    @OneToMany(mappedBy = "indieApp")
    private List<Genre> genres = new ArrayList<>();
}

@Entity
@Getter @Setter
public class Genre {
    @Id
    @Column(name = "genre_id")
    private Long genreId;

    @Column(name = "description")
    private String description; //like "RPG", "Action"

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "indie_app_id")
    private IndieApp indieApp;
}

@Data
public class RandomRecDto {
    private final Long id;
    private final String name;
    private final String genres;
}

@Repository
@RequiredArgsConstructor
public class RandomRecRepository {

    private final EntityManager em;

    public List<RandomRecDto> findRandomApps() {
        return em.createQuery(
                "select new study.weba.studyJPA.dto.RandomRecDto(i.id, i.name, g.description)" +
                " from IndieApp i join i.genres g", RandomRecDto.class)
                .setMaxResults(12)
                .getResultList();
    }
}

前辈您好！当我如下所示放置虚拟数据时，

indie_app_id	name
1	App1
2	App2

genre_id	description	indie_app_id
1	Action	1
2	RPG	1
3	FPS	2
4	Sport	2

我可以得到这样的结果。

randomRecDto = RandomRecDto(id=1, name=App1, genres=Action)
randomRecDto = RandomRecDto(id=1, name=App1, genres=RPG)
randomRecDto = RandomRecDto(id=2, name=App2, genres=FPS)
randomRecDto = RandomRecDto(id=2, name=App2, genres=Sport)

然而，我想要的结果是这样的

randomRecDto = RandomRecDto(id=1, name=App1, genres=Action, RPG)
randomRecDto = RandomRecDto(id=2, name=App2, genres=FPS, Sport)

我想通过数组获取描述。我该怎么办？

Answer 1

为什么不获取 IndieApp 实体，然后将它们转换为 RandomRecDto DTO？这甚至可以简化处理 IndieApp 而不是 RandomRecDto:

的存储库

public interface IndieAppRepository extends JpaRepository<IndieApp, Long> {
    List<IndieApp> findTop12();
}

您可能想在以下链接中阅读有关 Spring 数据的更多详细信息：

Answer 2

JpaResultMapper jpaResultMapper = new JpaResultMapper();

String sql = "SELECT i.indie_app_id, i.name, group_concat(g.description separator ',') FROM indie_app AS i" +
                " JOIN genre AS g ON i.indie_app_id = g.indie_app_id" +
                " group by i.indie_app_id, i.name";

Query nativeQuery = em.createNativeQuery(sql);

List<RandomRecDto> results = jpaResultMapper.list(nativeQuery, RandomRecDto.class);

Spring JPA一对多加入

Spring JPA one to many join

spring

jpa

jpql

spring-boot