hibernate - 了解 Hibernate/JPA 对 findAll 和 findOne 的不同行为

Question

我正在使用 findAll 方法(来自 Spring Data JPA 存储库)调试性能问题，问题是正在使用的提取计划。 (Hibernate 正在执行额外的 Selects 来检索额外的实体，而不是使用 fetch join)

我的实体与另一个实体有一些关系，我面临的问题与 *ToOne 关系有关。

在谷歌上花了一些时间后，我意识到有一些方法可以管理 Hibernate 将执行的查询(使用 JPA Criteria、NamedEntityGraph、自定义查询等)。

我创建了一个小例子来演示不同的行为，但我想 理解为什么 Hibernate 默认这样做 .我花了一些时间在文档中搜索，但找不到任何对此默认行为的引用。

https://github.com/pmvilaca/jpa-test

区别:

Hibernate: 
SELECT
  user0_.id                 AS id1_1_0_,
  userdetail1_.id           AS id1_21_,
  contactdet2_.id           AS id1_0_2_,
  user0_.contact_details_id AS contact_3_1_0_,
  user0_.name               AS name2_1_0_,
  userdetail1_.nationality  AS national2_2_1_,
  userdetail1_.user_id      AS user_id3_2_1_,
  contactdet2_.email        AS email2_0_2_,
  contactdet2_.mobile       AS mobile3_0_2_
FROM user user0_
  LEFT OUTER JOIN user_details userdetail1_ ON user0_.id = userdetail1_.user_id
  LEFT OUTER JOIN contact_details contactdet2_ ON user0_.contact_details_id = contactdet2_.id

和

Hibernate: select user0_.id as id1_1_, user0_.contact_details_id as contact_3_1_, user0_.name as name2_1_ from user user0_
Hibernate: select contactdet0_.id as id1_0_0_, contactdet0_.email as email2_0_0_, contactdet0_.mobile as mobile3_0_0_ from contact_details contactdet0_ where contactdet0_.id=?
Hibernate: select userdetail0_.id as id1_2_2_, userdetail0_.nationality as national2_2_2_, userdetail0_.user_id as user_id3_2_2_, user1_.id as id1_1_0_, user1_.contact_details_id as contact_3_1_0_, user1_.name as name2_1_0_, contactdet2_.id as id1_0_1_, contactdet2_.email as email2_0_1_, contactdet2_.mobile as mobile3_0_1_ from user_details userdetail0_ left outer join user user1_ on userdetail0_.user_id=user1_.id left outer join contact_details contactdet2_ on user1_.contact_details_id=contactdet2_.id where userdetail0_.user_id=?
Hibernate: select contactdet0_.id as id1_0_0_, contactdet0_.email as email2_0_0_, contactdet0_.mobile as mobile3_0_0_ from contact_details contactdet0_ where contactdet0_.id=?
Hibernate: select userdetail0_.id as id1_2_2_, userdetail0_.nationality as national2_2_2_, userdetail0_.user_id as user_id3_2_2_, user1_.id as id1_1_0_, user1_.contact_details_id as contact_3_1_0_, user1_.name as name2_1_0_, contactdet2_.id as id1_0_1_, contactdet2_.email as email2_0_1_, contactdet2_.mobile as mobile3_0_1_ from user_details userdetail0_ left outer join user user1_ on userdetail0_.user_id=user1_.id left outer join contact_details contactdet2_ on user1_.contact_details_id=contactdet2_.id where userdetail0_.user_id=?
Hibernate: select userdetail0_.id as id1_2_2_, userdetail0_.nationality as national2_2_2_, userdetail0_.user_id as user_id3_2_2_, user1_.id as id1_1_0_, user1_.contact_details_id as contact_3_1_0_, user1_.name as name2_1_0_, contactdet2_.id as id1_0_1_, contactdet2_.email as email2_0_1_, contactdet2_.mobile as mobile3_0_1_ from user_details userdetail0_ left outer join user user1_ on userdetail0_.user_id=user1_.id left outer join contact_details contactdet2_ on user1_.contact_details_id=contactdet2_.id where userdetail0_.user_id=?

任何的想法？

谢谢

Answer 1

@OneToOne 的默认获取类型是 FetchType.EAGER .
因此，在没有任何关于如何优化查询的提示的情况下，Hibernate 将遵循以下步骤:

全选 User的:

Hibernate: 
select
    user0_.id as id1_1_,
    user0_.contact_details_id as contact_3_1_,
    user0_.name as name2_1_ 
from
    user user0_

现在急切地加载每个 User的 ContactDetails和 UserDetails

Hibernate: 
select
    contactdet0_.id as id1_0_0_,
    contactdet0_.email as email2_0_0_,
    contactdet0_.mobile as mobile3_0_0_ 
from
    contact_details contactdet0_ 
where
    contactdet0_.id=?
Hibernate: 
select
    userdetail0_.id as id1_2_2_,
    userdetail0_.nationality as national2_2_2_,
    userdetail0_.user_id as user_id3_2_2_,
    user1_.id as id1_1_0_,
    user1_.contact_details_id as contact_3_1_0_,
    user1_.name as name2_1_0_,
    contactdet2_.id as id1_0_1_,
    contactdet2_.email as email2_0_1_,
    contactdet2_.mobile as mobile3_0_1_ 
from
    user_details userdetail0_ 
left outer join
    user user1_ 
        on userdetail0_.user_id=user1_.id 
left outer join
    contact_details contactdet2_ 
        on user1_.contact_details_id=contactdet2_.id 
where
    userdetail0_.user_id=?

...

这通常称为 n + 1问题。见 here例如。