9

我有以下实体，item其中最多可以有两个类别，主要和次要。这两个类别ManyToOne都category使用JoinColumnsOrFormulas. 第一个按预期获取EAGER，但第二个不会出现在 SQL 语句中并被延迟加载。这种延迟加载会导致经典的 n+1 问题。

这是我的项目实体，其中包含应该加入的两个类别实体：

@Entity
@Table(name = "item", schema = "public", catalog = "stackoverflow_question")
@DynamicUpdate
public class Item extends StackOverflowQuestionEntity {

    @Id
    @Column(name = "id")
    protected Long id;

    @Column(name = "site")
    private String site;

    @ManyToOne
    @JoinColumnsOrFormulas({
            @JoinColumnOrFormula(formula = @JoinFormula(value = "site", referencedColumnName = "site")),
            @JoinColumnOrFormula(formula = @JoinFormula(value = "primary_category_id", referencedColumnName = "category_id"))
    })
    private Category primaryCategory;

    @Column(name = "primary_category_id")
    private Long primaryCategoryId;

    @ManyToOne
    @JoinColumnsOrFormulas({
            @JoinColumnOrFormula(formula = @JoinFormula(value = "site", referencedColumnName = "site")),
            @JoinColumnOrFormula(formula = @JoinFormula(value = "secondary_category_id", referencedColumnName = "category_id"))
    })
    private Category secondaryCategory;

    @Column(name = "secondary_category_id")
    private Long secondaryCategoryId;
}

这是类别实体：

@Entity
@Table(name = "category", schema = "public", catalog = "stackoverflow_question")
public class Category extends StackOverflowQuestionEntity {

    @Column(name = "category_id")
    private Long categoryId;

    @Column(name = "name")
    private String name;

    @Column(name = "site")
    private String site;
}

结果查询仅包含主要类别：

SELECT this_.id AS id1_9_9_,
       this_.inserted AS inserted2_9_9_,
       this_.updated AS updated3_9_9_,
       this_.primary_category_id AS formula174_9_,
       this_.secondary_category_id AS formula176_9_,
       category2_.id AS id1_0_0_,
       category2_.inserted AS inserted2_0_0_,
       category2_.updated AS updated3_0_0_,
       category2_.name AS name7_0_0_
FROM public.item this_
LEFT OUTER JOIN public.category category2_ ON this_.site=category2_.site
AND this_.primary_category_id=category2_.category_id
WHERE True;

因此，次要类别变得懒惰：

SELECT category0_.id AS id1_0_0_,
       category0_.inserted AS inserted2_0_0_,
       category0_.updated AS updated3_0_0_,
       category0_.name AS name4_0_0_,
       category0_.site AS site5_0_0_
FROM public.category category0_
WHERE category0_.site=?
  AND category0_.category_id=?;

为什么 Hibernate 加入次要类别懒惰，注解似乎是一样的。

我使用的休眠版本是 5.0.10.Final。

这是基本实体的样子：

@MappedSuperclass
abstract public class StackOverflowQuestionEntity implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", unique = true, insertable = true, updatable = false, nullable = false)
    protected Long id;

    @Type(type="LocalDateTime")
    @Column(name = "created", nullable = false, insertable = true, updatable = false)
    protected LocalDateTime created;

    @Type(type="LocalDateTime")
    @Column(name = "refreshed", nullable = false, insertable = true, updatable = true)
    protected LocalDateTime refreshed;

    @PreUpdate
    protected void onUpdate() {
        refreshed = now();
    }

    @PrePersist
    protected void onCreate() { 
        created = refreshed = now();
    }
}

这是一个示例“查询”，正如我所说我使用的是休眠标准和 HQL，这两种方法都会出现问题。

session
    .createCriteria(Item.class)
    .add(eq("id", id))
    .uniqueResult();
4

3 回答 3

4

使用标准 JPA 注释，它看起来像这样（更新）：

@ManyToOne
@JoinColumns({
    @JoinColumn(name="site", referencedColumnName="site", insertable = false, updatable = false),
    @JoinColumn(name="primary_category_id", referencedColumnName="category_id", insertable = false, updatable = false)
})
private Category primaryCategory;

@ManyToOne
@JoinColumns({
    @JoinColumn(name="site", referencedColumnName="site", insertable = false, updatable = false),
    @JoinColumn(name="secondary_category_id", referencedColumnName="category_id", insertable = false, updatable = false)
})
private Category secondaryCategory;

更新：我发现select只有当您使用复合键时才会生成第二条语句join：Hibernate 尝试解析{site=site, id=null}使用的关联TwoPhaseLoad。但是如果你写

@ManyToOne
@JoinColumn(name="secondary_category_id")
private Category secondaryCategory;

然后secondary_category_id将生成null唯一一条select语句，其secondaryCategory值为null. 也许它会以某种方式帮助你。例如，您可以site在构建条件时对字段添加约束：

Category c = (Category) session.createCriteria(Category.class)
    .add(Restrictions.eq("id", 1L)) // for example
    // here you define additional restriction on site field
    .createAlias("secondaryCategory", "sc", JoinType.LEFT_OUTER_JOIN, Restrictions.sqlRestriction("this_.site = {alias}.site"))
    .uniqueResult();
于 2017-12-11T21:18:31.663 回答
1

尝试以下解决方案：

@Entity
@Table(name = "item", schema = "public", catalog = "stackoverflow_question")
  @DynamicUpdate
  public class Item {

    @ManyToOne
    @JoinColumn(name="site")
    private Category primaryCategory;

    @ManyToOne
    @JoinColumn(name="site")
    private Category primaryCategory;
  }

  @Entity
  @Table(name = "category", schema = "public", catalog = "stackoverflow_question")
  public class Category {

     @OneToMany(targetEntity=Item.class, mappedBy="primaryCategory", cascade=CascadeType.ALL)
      private List<Item> primaryCategoryList;

    @OneToMany(targetEntity=Item.class, mappedBy="secondaryCategory", cascade=CascadeType.ALL)
       private List<Item> secondaryCategoryList;

  }
于 2017-12-15T06:29:32.487 回答
1

我使用您的类和以下查询代码进行了快速测试（使用 JPA 条件查询而不是本机 Hibernate）

CriteriaQuery<Item> cq = em.getCriteriaBuilder().createQuery(Item.class);
EntityGraph<Item> entityGraph = em.createEntityGraph(Item.class);
entityGraph.addSubgraph("primaryCategory", Category.class);
entityGraph.addSubgraph("secondaryCategory", Category.class);
List<Item> items = em.createQuery(cq.select(cq.from(Item.class)))
    .setHint("javax.persistence.loadgraph", entityGraph)
    .getResultList();

导致生成以下 SQL（为便于阅读而格式化）：

select item0_.id as id1_1_0_, 
    category1_.id as id1_0_1_, 
    category2_.id as id1_0_2_, 
    item0_.site as site4_1_0_, 
    item0_.primary_category_id as primary_2_1_0_,
    item0_.secondary_category_id as secondar3_1_0_, 
    category1_.category_id as category2_0_1_, 
    category1_.name as name3_0_1_, 
    category1_.site as site4_0_1_, 
    category2_.category_id as category2_0_2_, 
    category2_.name as name3_0_2_, 
    category2_.site as site4_0_2_ 
from item item0_ 
left outer join category category1_
    on item0_.site=category1_.site
    and item0_.secondary_category_id=category1_.category_id 
left outer join category category2_
    on item0_.site=category2_.site
    and item0_.primary_category_id=category2_.category_id

如您所见，两个类别表都在同一个 SELECT 中连接

于 2017-12-15T13:41:21.297 回答