Hibernate 本机查询也获得 JPA 选择

Question

我有两个数据库表，“A”和“B”，@OneToMany(mappedBy = "a") 在实体 A 的 List 字段上，@ManyToOne 在字段 B.a 上。我在对 A 进行默认查询时遇到了“N+1”问题，所以我正在尝试一个本机查询，例如：

@Query(value="select * from A as a left join B as b " +
            "on a.ID = b.b ",
            nativeQuery=true)

这在数据按预期映射回实体的意义上是有效的。

我的问题是我可以看到 Hibernate 正在为每个 B 进行单独的选择，而不是使用连接的结果。也就是说，我在控制台中看到了以下序列：

• 我指定的选择
• 对于 A 的每个实例，使用 A 中的 ID 为 B 选择另一个实例

换句话说，我还有“n+1”的问题。

我认为 @OneToMany 和 @ManyToOne 注释可能会导致 Hibernate 执行这些额外的选择，但是当我将它们取出时，我的 IDE (IntelliJ) 会说：

'Basic' attribute should not be a container

...在 A 中的 List 属性上。

如何使用连接将结果映射回单个选择？我应该放弃 Hibernate 和 JPA 吗？

我正在使用 spring-boot-start-data-jpa.2.5.4

Answer 1

native@Query没有足够的映射能力，看来必须要Hibernate原生查询。

import java.util.ArrayList;
import java.util.List;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import org.hibernate.Session;
import org.hibernate.transform.BasicTransformerAdapter;
import org.springframework.stereotype.Repository;

// https://docs.spring.io/spring-data/jpa/docs/2.5.6/reference/html/#repositories.custom-implementations
@Repository
public class CustomizedARepositoryImpl implements CustomizedARepository {

    @PersistenceContext
    private EntityManager entityManager;

    @Override
    public List<A> getAll() {
        // https://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html#sql-entity-associations-query
        final Session sess = (Session) entityManager.getDelegate();
        final List<A> res = sess
            // If no duplicate column names, original sql can be used, too.
            .createNativeQuery("select {a.*},{b.*} from A as a left join B as b on a.ID = b.a ")
            .addEntity("a", A.class)
            .addJoin("b", "a.bs")
            .setResultTransformer(DistinctResultTransformer.INSTANCE)
            .list();

        return res;
    }

    // https://stackoverflow.com/q/12071014/4506703
    static class DistinctResultTransformer extends BasicTransformerAdapter {
        private static final long serialVersionUID = 1L;

        static final DistinctResultTransformer INSTANCE = new DistinctResultTransformer();

        @Override
        public List transformList(final List collection) {
            final List<Object> res = new ArrayList<>();
            for (final Object[] obj : (List<Object[]>) collection) {
                if (!res.contains(obj[0])) {
                    res.add(obj[0]);
                }
            }
            return res;
        }
    }
}

以上代码执行 1 个查询：

select a.id as id1_0_0_, a.name as name2_0_0_,b.a as a3_1_0__, b.id as id1_1_0__, b.id as id1_1_1_, b.a as a3_1_1_, b.name as name2_1_1_
from A as a left join B as b on a.ID = b.a

full sample code

---

您可以使用一些方法来避免 N+1 问题。

使用 JPQL 获取，而不是原生查询：

    @Query("select distinct a from A a left join fetch a.bs")
    List<A> getAllJpqlFetch();

以上代码执行 1 个查询：

select distinct a0_.id as id1_0_0_, bs1_.id as id1_1_1_, a0_.name as name2_0_0_, bs1_.a as a3_1_1_, bs1_.name as name2_1_1_, bs1_.a as a3_1_0__, bs1_.id as id1_1_0__ 
from a a0_ left outer join b bs1_ on a0_.id=bs1_.a

diff

---

使用JPA Criteria fetch，相当于上面的JPQL：

@Repository
public class CustomizedARepositoryImpl implements CustomizedARepository {

    @PersistenceContext
    private EntityManager entityManager;

    @Override
    public List<A> getAllCriteria() {
        // https://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html#criteria-from-fetch
        final CriteriaBuilder builder = entityManager.getCriteriaBuilder();
        final CriteriaQuery<A> criteria = builder.createQuery(A.class);
        final Root<A> root = criteria.from(A.class);
        root.fetch("bs", JoinType.LEFT);
        criteria.select(root).distinct(true);
        return entityManager.createQuery(criteria).getResultList();
    }

以上代码执行 1 个查询：

select distinct a0_.id as id1_0_0_, bs1_.id as id1_1_1_, a0_.name as name2_0_0_, bs1_.a as a3_1_1_, bs1_.name as name2_1_1_, bs1_.a as a3_1_0__, bs1_.id as id1_1_0__ 
from a a0_ left outer join b bs1_ on a0_.id=bs1_.a

diff

---

使用@Fetch(FetchMode.SUBSELECT)：

import org.hibernate.annotations.Fetch;
import org.hibernate.annotations.FetchMode;
// ...

@Entity
public class A {

    @OneToMany(mappedBy = "a")
    @Fetch(FetchMode.SUBSELECT)
    private List<B> bs;

    // ...
}

// findAll() method implementation is auto-generated by Spring Data JPA
// https://docs.spring.io/spring-data/jpa/docs/2.5.6/reference/html/#repositories.core-concepts
repository.findAll();

以上代码执行 2 个查询（根实体及其关系实体）：

select a0_.id as id1_0_, a0_.name as name2_0_ from a a0_

select bs0_.a as a3_1_1_, bs0_.id as id1_1_1_, bs0_.id as id1_1_0_, bs0_.a as a3_1_0_, bs0_.name as name2_1_0_ 
from b bs0_ where bs0_.a in (select a0_.id from a a0_)

diff

Answer 2

我最终使用了上面 DEWA Kazuyuki 提供的以下解决方案。我在这里复制它是因为 DEWA 提出了几个答案，我认为确定对我有用的特定答案很有用。谢谢，DEWA。

@Repository
public class CustomizedARepositoryImpl implements CustomizedARepository {

@PersistenceContext
private EntityManager entityManager;

@Override
public List<A> getAllCriteria() {
    // https://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html#criteria-from-fetch
    final CriteriaBuilder builder = entityManager.getCriteriaBuilder();
    final CriteriaQuery<A> criteria = builder.createQuery(A.class);
    final Root<A> root = criteria.from(A.class);
    root.fetch("bs", JoinType.LEFT);
    criteria.select(root).distinct(true);
    return entityManager.createQuery(criteria).getResultList();
}