Question

这与this question有关，但下面的例子较短，所以我认为另一个问题是有道理的。

我有两个实体，A和B，一对一的关系。对于A，B是可选的，并且每个B必须具有A.我想级联从A到B的删除。这是我的第一次尝试：

@Entity
public class A extends Model {

    @Id
    private Long id;

    @OneToOne(optional = true, mappedBy = "a", cascade = CascadeType.REMOVE, orphanRemoval = true)
    private B b;

}

@Entity
public class B extends Model {

    @Id
    private Long id;

    @OneToOne(optional = false)
    private A a;

}

但是，似乎Ebean忽略了“可选”注释，因为当我为id为1的已保存A执行查找时，将执行以下SQL：

select t0.id c0, t1.id c1 
from a t0
join b t1 on t1.a_id = t0.id 
where t0.id = 1

换句话说，它执行内部而不是左连接，这会导致查找在没有关联的B时失败。我尝试了@JoinColumn等各种组合无效。我发现的唯一有点令人满意的解决方法是将A-to-B模型化为“假的”一对多关系。有更好的解决方案吗？这是一个错误还是Ebean已知/明确的限制？

Answer 1

我找到了解决方案。我改变了这个映射的方向。因此，我从mappedBy = "a"课程中删除了A，并在mappedBy = "b"课程中添加了B。所以代码现在看起来像这样：

@Entity
public class A extends Model {

    @Id
    private Long id;

    @OneToOne(optional = true, cascade = CascadeType.REMOVE, orphanRemoval = true)
    private B b;

 ...
}


@Entity
public class B extends Model {

    @Id
    private Long id;

    @OneToOne(optional = false, mappedBy = "b")
    private A a;

    private String name;

    ...
}

我在B类中添加了name字段，以使此测试更有趣。

我的测试方法：

@Test
public void abTest () {
    FakeApplication app = Helpers.fakeApplication(Helpers.inMemoryDatabase());
    Helpers.start(app);

    A a = new A();
    B b = new B();
    a.setId(1L);
    b.setId(2L);
    a.setB(b);
    b.setA(a);
    b.setName("bbb");

    Ebean.save(b);
    Ebean.save(a);

    A fa = Ebean.find(A.class, 1L);
    System.out.println("a.id: "+fa.getId());
    System.out.println("a.b.id: "+fa.getB());
    System.out.println("a.b.name: "+fa.getB().getName());

    A a1 = new A();
    a1.setId(3L);
    Ebean.save(a1);
    A fa1 = Ebean.find(A.class, 3L);
    System.out.println("a1.id: "+fa1.getId());
    System.out.println("a1.b.id: "+fa1.getB());

    B fb = Ebean.find(B.class, 2L);
    System.out.println("b.id: "+fb.getId());
    System.out.println("b.name: "+fb.getName());
    System.out.println("b.a.id: "+fb.getA().getId());
}

此测试的结果是：

[debug] c.j.b.PreparedStatementHandle - insert into b (id, name) values (2,'bbb')
[debug] c.j.b.PreparedStatementHandle - insert into a (id, b_id) values (1,2)
[debug] c.j.b.PreparedStatementHandle - select t0.id c0, t0.b_id c1 from a t0 where t0.id = 1
a.id: 1
a.b.id: models.B@2
[debug] c.j.b.PreparedStatementHandle - select t0.id c0, t0.name c1, t1.id c2 from b t0 left outer join a t1 on t1.b_id = t0.id  where t0.id = 2
a.b.name: bbb
[debug] c.j.b.PreparedStatementHandle - insert into a (id, b_id) values (3,'[SQL NULL of type -5]')
[debug] c.j.b.PreparedStatementHandle - select t0.id c0, t0.b_id c1 from a t0 where t0.id = 3
a1.id: 3
a1.b.id: null
[debug] c.j.b.PreparedStatementHandle - select t0.id c0, t0.name c1, t1.id c2 from b t0 left outer join a t1 on t1.b_id = t0.id  where t0.id = 2
b.id: 2
b.name: bbb
b.a.id: 1

因此，无论A.b是否为null，此代码都能正常运行。正如我们在日志中看到的那样，现在left outer join代替join。

Play + Ebean + JPA：在OneToOne映射上级联删除

1 个答案: