[英]Finding items with a set containing all elements of a given set with jpql
[英]HQL to get elements that possess all items in a set
当前,我有一个HQL查询,该查询返回一组指定奖励中所有拥有ANY奖励的会员:
from Member m left join m.awards as a where a.name in ("Trophy","Ribbon");
我现在需要的是HQL,它将退回所有拥有该奖项集中指定的所有奖项的会员。
因此,假设此数据:
Joe has Trophy, Medal
Sue has Trophy, Ribbon
Tom has Trophy, Ribbon, Medal
上面的查询将返回Joe,Sue和Tom,因为这三个人都至少拥有Trophy或Ribbon之一。 但是我只需要归还Sue和Tom,因为他们是唯一拥有所有指定奖项(奖杯和丝带)的人。
这是类结构(简化):
class Member {
private String name;
private Set<Award> awards;
}
class Award {
private String name;
}
select m from Member m left join m.awards as a where a.name in ("Trophy","Ribbon") group by m having count(a)=2
只是重复我自己...获取完全具有给定奖励集合的成员的代码:
from Member m
where not exists (
from Award a where a.name in {"Trophy", "Ribbon"}
and a not in(
select * from Award a2 where a2.owner = m
)
) and not exists (
from Award a3 where a3.owner = m and a3 not in {"Trophy", "Ribbon"}
)
您可以通过向查询调用IE中添加DISTINCT_ROOT_ENTITY结果转换器来强制执行不同的结果:
getSession().createQuery(hql).setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY)
我遇到了类似的问题,但我需要做的是(按照您的示例)选择所有成员,这些成员构成所有奖项,仅此而已。 因此,在您的示例中,唯一正确的结果是Sue。 有任何想法吗?
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