Question

由于MultipleBagFetchException，我无法执行简单的fetchjoin。

@Entity
public class Person {

    @OneToMany(mappedBy="person",fetch=FetchType.LAZY)
    private List<Auto> autos;    
}


@Entity
public class Auto {

    @ManyToOne
    @JoinColumn(name = "person_id", nullable = false)
    private Person person;

   @OneToMany(mappedBy="auto",fetch=FetchType.LAZY)
   private List<Tool> tools;

}

@Entity
@Table(name="tool")
public class Tool {

    @ManyToOne
    @JoinColumn(name = "auto_id", nullable = false)
    private Auto auto;
}

您可以看到我所有的关联都使用默认的访存类型。

 @Query("SELECT p FROM Person p JOIN FETCH p.autos a JOIN FETCH a.tools")
 List<Person>findAll();

结果：

Caused by: org.hibernate.loader.MultipleBagFetchException: cannot simultaneously fetch multiple bags: [com.example.entities.Person.autos, com.example.entities.Auto.tools]

我已经阅读了有关此异常的信息，但是在这些情况下，导致此异常的原因是对集合使用EAGER提取类型。那这个呢？这是最简单的实体关系。

最重要的是，假设我们不允许接触实体。如何仅在查询端解决此问题？

Answer 1

有一种避免n + 1条查询而不接触实体的方法，仅更改对findAll的查询。我们可以编写一个包装器函数，该函数将首先为人加载汽车，然后他们一次选择即可获取所有工具。

PersonRepository

    @Query("SELECT distinct p FROM Person p JOIN FETCH p.autos a")
    List<Person> findAll();

包装代码

    List<Person> persons = personRepository.findAll();
    Session session = (Session) entityManager.getDelegate();
    List<Auto> autos = new ArrayList<>();
    for (Person person : persons) {
        if(!CollectionUtils.isEmpty(person.getAutos())) {
            autos.addAll(person.getAutos());
        }
    }
    try{
        autos = session.createQuery("select distinct a from Auto a Join fetch a.tools " +
                " where a in :autos", Auto.class)
                .setParameter("autos", autos)
                .setHint(QueryHints.PASS_DISTINCT_THROUGH, false)
                .getResultList();
    } catch (Exception ex) {
        ex.printStackTrace();
    }

第一个查询将是：

SELECT DISTINCT
          person0_.id       AS id1_6_0_,
          autos1_.id        AS id1_0_1_,
          person0_.name     AS name2_6_0_,
          autos1_.name      AS name2_0_1_,
          autos1_.person_id AS person_i3_0_1_,
          autos1_.person_id AS person_i3_0_0__,
          autos1_.id        AS id1_0_0__
FROM
          Person person0_
          INNER JOIN
                    Auto autos1_
          ON
                    person0_.id=autos1_.person_id

生成的第二个查询将是：

    SELECT
          auto0_.id        AS id1_0_0_,
          tools1_.id       AS id1_8_1_,
          auto0_.name      AS name2_0_0_,
          auto0_.person_id AS person_i3_0_0_,
          tools1_.auto_id  AS auto_id3_8_1_,
          tools1_.name     AS name2_8_1_,
          tools1_.auto_id  AS auto_id3_8_0__,
          tools1_.id       AS id1_8_0__
FROM
          Auto auto0_
          INNER JOIN
                    Tool tools1_
          ON
                    auto0_.id=tools1_.auto_id
WHERE
          auto0_.id IN (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)

除此之外，我认为我们的选择是有限的，我们将不得不更改 Tool 实体FetchMode或为默认的FetchMode.SELECT添加BatchSize以便在单独的查询中获得Tools。

    @OneToMany(mappedBy = "auto", fetch = FetchType.LAZY)
    @Fetch(FetchMode.SUBSELECT)
    private List<Tool> tools;

查询将

SELECT
          tools0_.auto_id AS auto_id3_8_1_
        , tools0_.id      AS id1_8_1_
        , tools0_.id      AS id1_8_0_
        , tools0_.auto_id AS auto_id3_8_0_
        , tools0_.name    AS name2_8_0_
FROM
          Tool tools0_
WHERE
          tools0_.auto_id IN
          (
                    SELECT
                              autos1_.id
                    FROM
                              Person person0_
                              INNER JOIN
                                        Auto autos1_
                              ON
                                        person0_.id=autos1_.person_id
          )

无法获取连接懒惰的初始化的集合

1 个答案: