Spring JPA 一对多加入

Posted 2023-03-16

【中文标题】Spring JPA 一对多加入

【英文标题】：Spring JPA one to many join

【发布时间】：2021-10-19 09:40:27

【问题描述】：

@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();
    

前辈们好！ 当我将虚拟数据如下所示时，

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)

但是，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？

【参考方案1】：

您为什么不获取IndieApp 实体，然后将它们转换为您的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();

您可能希望在以下链接中阅读有关 Spring Data 的更多详细信息：

Spring Data JPA Spring Data Repositories Limiting query results with Spring Data JPA Defining Spring Data Query Methods Spring Data Repositories Query Keywords

【讨论】：

感谢您的建议！但我使用本机查询 + qlrm 解决了这个问题。

【参考方案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);