Question

我正在尝试实现这样的示例：Person类有一个它喜欢的地方列表。但是当我想查询它时，我希望结果为每个人只有最喜欢的地方（只是第一个不是全部）。所以我这样做了：

@Entity
class Person{

    ...

    @ManyToMany(cascade = {CascadeType.REFRESH,CascadeType.PERSIST}, fetch = FetchType.EAGER)
    @JoinTable(name = "person_favorite_place",
            joinColumns = @JoinColumn(name = "person_id", referencedColumnName = "id"),
            inverseJoinColumns = @JoinColumn(name = "place_id", referencedColumnName = "id")
    )
    @OrderColumn(name="favorite_place_order")
    List<Place> favoritePlaces;
}

在存储库中，我做了：

public interface PersonRepository extends JpaRepository<Person, Long> {

    @Query(value = "select person0_.id as id1_0_0_, person0_.age as age2_0_0_, person0_.name as name3_0_0_, favoriteme1_.person_id as person_id1_1_1_, place2_.id as place_id2_1_1_, favoriteme1_.favorite_place_order as favorite3_1_, place2_.id as id1_3_2_, place2_.invented as invented2_3_2_, place2_.name as name3_3_2_ from person person0_ left outer join person_favorite_place favoriteme1_ on person0_.id=favoriteme1_.person_id left outer join place place2_ on favoriteme1_.place_id=place2_.id where person0_.id=:personId  and favoriteme1_.favorite_place_order = 0", nativeQuery = true)
    Person getPersonWithFavoritePlace(@Param("personId") Long personId);

}

但是看起来，有2个sql查询正在运行。第一个是：

select person0_.id as id1_0_0_, person0_.age as age2_0_0_, person0_.name as name3_0_0_, favoriteme1_.person_id as person_id1_1_1_, place2_.id as place_id2_1_1_, favoriteme1_.favorite_place_order as favorite3_1_, place2_.id as id1_3_2_, place2_.invented as invented2_3_2_, place2_.name as name3_3_2_ from person person0_ left outer join person_favorite_place favoriteme1_ on person0_.id=favoriteme1_.person_id left outer join place place2_ on favoriteme1_.place_id=place2_.id where person0_.id=:personId  and favoriteme1_.favorite_place_order = 0

第二个：

select favoriteme0_.person_id as person_id1_1_0_, favoriteme0_.place_id as place_id2_1_0_, favoriteme0_.favorite_place_order as favorite3_0_, place1_.id as id1_3_1_, place1_.invented as invented2_3_1_, place1_.name as name3_3_1_ from person_favorite_place favoriteme0_ inner join place place1_ on favoriteme0_.place_id=place1_.id where favoriteme0_.person_id=?

我能理解第一个，这完全是我想要执行的查询，但第二个，我不知道它来自哪里。所以我认为，因为我拥有一个人最喜欢的地方但不是最想要的地方。

有什么想法吗？

BR

PS：我已经在＆＃34; findOne＆＃34;的原生输出上写了查询。方法。刚添加＆＃34;和favoriteme1_.favorite_place_order = 0＆＃34;到底。此外，我试图使用确切的查询而不进行修改，它就像一个魅力!!

Answer 1

第二个查询用于加载您使用FetchType.EAGER定义的favoritePlaces

Answer 2

您的存储库正在查询人员而不是地点，因为FetchType.EAGER也会使用第二个查询加载所有地方;查询Places使用PlacesRepository。

带注释的JPA本机查询执行不同的查询

2 个答案: