为什么hibernate执行两个查询以急切加载@OneToOne双向关联？

Question

我有实体A有一个B实体,而B有一个带有@OneToOne双向关联的A.

现在,当我找到所有A记录时,hibernate在B上执行左外连接的两个查询,如下所示:

select a.id, a.id_b, a.field1, b.id, b.field1 from A as a, B as b left outer join b ON b.id=a.id_b;
select a.id, a.id_b, a.field1, b.id, b.field1 from A as a, B as b left outer join b ON b.id=a.id_b WHERE b.id=?

首先查询加载A和B字段,这没关系,但为什么要执行第二次查询来重新加载A？我认为这个查询加载了B中的A内容,但是这个A显然是包含B的A ......所以它已经加载了第一个查询,是不是真的？

- 编辑 -

实体A:

@Entity
public class A implements Serializable{
    // id and other ecc ecc
    @OneToOne
    @JoinColumn(name="id_b")
    B b;
}

实体B:

@Entity
public class B implements Serializable{
    // id and other ecc ecc
    @OneToOne(mappedBy="b")
    A a;
}

这就是情况,A上的findAll需要两个查询......为什么？

Answer 1

Blow,如果A和B 共享使用主键连接两个实体的同一主键列,则应使用@PrimaryKeyJoinColumn

@Entity
public class A implements Serializable {

    private MutableInt id = new MutableInt();

    private B b;

    public void setIdAsMutableInt(MutableInt id) {
        this.id = id;
    }

    @Id
    @GeneratedValue
    public Integer getId() {
        return id.intValue();
    }

    public void setId(Integer id) {
        this.id.setValue(id);
    }

    /**
      * Any ToOne annotation, such as @OneToOne and @ManyToOne, is EARGELY loaded, by default
      */
    @OneToOne(fetch=FetchType.LAZY)
    @PrimaryKeyJoinColumn
    @Cascade(CascadeType.SAVE_UPDATE)
    public B getB() {
        return b;
    }

    public void setB(B b) {
        b.setIdAsMutableInt(id);

        this.b = b;
    }

}

并且B请注意,由于@PrimaryKeyJoinColumn,您不需要mappedBy属性

@Entity
public class B implements Serializable {

    private MutableInt id = new MutableInt();

    private A a;

    public void setIdAsMutableInt(MutableInt id) {
        this.id = id;
    }

    @Id
    public Integer getId() {
        return id.intValue();
    }

    public void setId(Integer id) {
        this.id.setValue(id);
    }

    @OneToOne(fetch=FetchType.LAZY)
    @PrimaryKeyJoinColumn
    public A getA() {
        return a;
    }

    public void setA(A a) {
        this.a = a;
    }

}

让我们测试(你可以测试你想要的)

A a = new A();
B b = new B();

a.setB(b);

/**
  * b property will be saved because Cascade.SAVE_UPDATE
  */
Serializable id = session.save(a);

b = (B) session
        .createQuery("from B b left join fetch b.a where b.id = :id")
        .setParameter("id", id)
        .list()
        .get(0);

Assert.assertEquals(b.getId(), b.getA().getId());

注意我使用MutableInt字段(由Integer属性封装)而不是Integer,因为Integer 是一个不可变类型,A和B共享SAME分配的id

但是如果A和B 通过使用其他主键而加入,则应使用@JoinColumn和mappedBy(双向关系,右),如下所示

@Entity
public class A implements Serializable {

    private Integer id;

    private B b;

    @Id
    @GeneratedValue
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    /**
      * mappedBy="a" means: Look at "a" field / property at B Entity. If it has any assigned value, join us Through B_ID foreign key column
      */
    @OneToOne(fetch=FetchType.LAZY, mappedBy="a")
    /**
      * Table A has a foreign key column called "B_ID"
      */ 
    @JoinColumn(name="B_ID")
    @Cascade(CascadeType.SAVE_UPDATE)
    public B getB() {
        return b;
    }

    public void setB(B b) {
        this.b = b;
    }

}   

而B

@Entity
public class B implements Serializable {

    private Integer id;

    private A a;

    public void setIdAsMutableInt(MutableInt id) {
        this.id = id;
    }

    @Id
    @GeneratedValue
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    @OneToOne(fetch=FetchType.LAZY)
    public A getA() {
        return a;
    }

    public void setA(A a) {
        this.a = a;
    }

}

去测试

A a = new A();
B b = new B();

/**
  * Set up both sides
  * Or use some kind of add convenience method
  */ 
a.setB(b);
b.setA(a);

/**
  * b property will be saved because Cascade.SAVE_UPDATE
  */
Serializable id = session.save(a);

b = (B) session
        .createQuery("from B b left join fetch b.a where b.id = :id")
        .setParameter("id", id)
        .list()
        .get(0);

通过使用所有者B,您将获得两个select语句它是因为B Table 不包含指向表A的任何外键列但是通过使用

"来自A左连接获取ab,其中a.id =:id"

您将只获得一个select语句,因为A知道如何使用其B_ID外键列检索其连接的B.