使 in 子句将所有项目与任何替代项相匹配？

我有一张桌子hotel [hotelid,hotelname,etc]

和另一张桌子facilities[facilityid,facilityname]

这两个表通过 table 链接hotel_to_facilities_map[hotelid,facility_id]

所以桌子hotel_to_facilities_map可能包含如下值

hotelid facility_id
-------------------
1         3
1         5
1         6
2         6
2         2
2         5

现在我想检索符合所有要求的设施的所有酒店

select * from hotel_to_facilities_map where facility_id IN (3,5,2)

但这会导致比赛OR当我需要的时候表达AND.

有什么解决方法或解决方案吗？

select hotelid
from hotel_to_facilities_map
where facility_id in (3,5,2)
group by hotelid
having count(*) = 3