全部为真时仅返回行

Question

在轻微的心理纠结中,我希望这比我想象的容易.得到以下表格:

create table #x
(
handid int,
cardid int
)
insert into #x
values
(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),
(2,2),(2,3),(2,4),(2,300),(2,400),(2,500),(2,8),
(3,2),(3,3),(3,4),(3,300),(3,400),(3,7),(3,8),
(4,2),(4,300),(4,400),(4,500),(4,6),(4,7),(4,8)


create table #winners(cardid int)
insert into #winners values(300),(400),(500)

select a.* 
from 
        #x a 
        inner join #winners b
            on
            a.cardid = b.cardid 

这将返回以下内容:

我只希望此查询在a的所有三个cardid存在时返回行handid.因此,期望的结果集将不包括handid3.

这是现实的典范.实际上,#x包含500个记录.

编辑

好的 - 实际上有一些获胜者由数据集组成,#winners其中包含可变数量的记录.因此,将原始代码修改为以下结果集不应包含handId1或handId3.我还在结果集中获得了一些不需要的重复记录:

create table #x
(
handid int,
cardid int
)
insert into #x
values
(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8000),
(2,2),(2,3),(2,4),(2,300),(2,400),(2,500),(2,8),
(3,2),(3,3),(3,4),(3,300),(3,400),(3,7),(3,8),
(4,2),(4,300),(4,400),(4,500),(4,6),(4,7),(4,8)


create table #winners(winningComb char(1), cardid int)
insert into #winners values('A',300),('A',400),('A',500),('B',8000),('B',400)

select a.* 
from 
        #x a 
        inner join #winners b
            on
            a.cardid = b.cardid 

Answer 1

你可以使用这样的东西:

select handid
from #x  
where cardid in (select cardid from #winners)
group by handid
having count(handid) = (select count(distinct cardid)
                         from #winners);

请参阅SQL Fiddle with Demo

结果:

| HANDID |
----------
|      2 |
|      4 |

根据您的编辑,这是一个返回正确结果的尝试,但我不确定它是否适用于您拥有的较大数据集:

;with cte as
(   
    select w1.cardid, w1.winningComb, w2.ComboCardCount
    from winners w1
    inner join
    (
        select COUNT(*) ComboCardCount, winningComb
        from winners
        group by winningComb
    ) w2
        on w1.winningComb = w2.winningComb
) 
select a.handid
from x a
inner join cte b
    on a.cardid = b.cardid
where a.cardid in (select cardid from cte)
group by handid, b.ComboCardCount
having COUNT(a.handid) = b.ComboCardCount

请参阅SQL Fiddle with Demo

结果:

| HANDID |
----------
|      2 |
|      4 |