Question

我有两个通过多对一关系连接的实体

@Entity
public class A {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "A_ID")
    public Long id;

    public String status;

    @OneToMany(mappedBy = "a")
    public Collection<B> b;
}

@Entity
public class B {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Long id;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="A_ID", nullable=false)
    public A a;

    // ... some other members ...
}

现在我希望得到所有具有“接受”状态A的B：

Criteria criteria = session.createCriteria(B.class);
criteria.createAlias("a.status", "status");
criteria.add(Restrictions.eq("status", "accepted"));
return criteria.list();

这导致hibernate抛出“org.hibernate.QueryException：not an association：status”。我做错了什么？我认为这些表是关系因为多对一的关系。

Answer 1

在@ManyToOne上，您忘了提及它正在加入的列。提供@JoinColumn并指定mappedBy ='a'/ inverse = true等，

检查this

<强>更新

Criteria criteria = session.createCriteria(B.class,"b");
criteria.createAlias("b.a", "a");
criteria.add(Restrictions.eq("a.status", "accepted"));
return criteria.list();

休眠多对一关联错误

1 个答案: