java - 带别名的 Hibernate 查询

Question

怎么了

session.createCriteria(Composed.class, "main")
.createAlias("main.id.branch", "b1")
.add(Restrictions.eq("b1.owner", user))
.list();

?对应的HQL可以正常工作

String hql = "select main from Composed as main"
        + "left join main.id.branch as b1 where b1.owner = ?";
session.createQuery(hql)
.setInteger(0, user.id().intValue())
.list();

根据条件，Hibernate 不创建联接 并使用 where b1x1_.owner_id=?，但任何地方都没有 b1x1_，因此失败带有“无法准备声明”。

类(class)很简单

@Entity class Composed {
    @Id ComposedId id; // also tried @EmbeddedId
    ... irrelevant stuff
}

@Embeddable class ComposedId {
    @ManyToOne(optional=false) Branch branch;
    ... irrelevant stuff
}

@Entity class Branch {
    @Id Integer id;
    @ManyToOne(optional=false) User owner;
    ... irrelevant stuff
}

@Entity class User {
    @Id Integer id;
    ... irrelevant stuff
}

更新

我终于创建了一个 SSCCE并提交了issue .抱歉这个令人困惑的问题，没有 SSCCE，很难重现。

Answer 1

我没试过，但也许用左连接创建两个别名对你有帮助。我的意思是:

session.createCriteria(Composed.class, "main")
   .createAlias("main.id", "id1", JoinType.LEFT_OUTER_JOIN)
   .createAlias("id1.branch", "b1", JoinType.LEFT_OUTER_JOIN)
   .add(Restrictions.eq("b1.owner", user))

希望对您有所帮助!