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 

返回以下内容：

我只希望此查询在cardid存在三个handid的所有时返回行。因此，期望的结果集不包括handid 3。

这是现实的典范。 实际上，＃x包含500个记录。

修改

好的 - 实际上获胜者由来自#winners的数据集组成，这些数据集具有可变数量的记录。因此，将原始代码修改为以下结果集不应包括handId 1或handId 3.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

您可以使用以下内容：

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，这是一次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 |

Answer 2

试试这个：

with cte as
(select a.* 
from 
        #x a 
        inner join #winners b
            on
            a.cardid = b.cardid ),
cte1 as 
     (select *,ROW_NUMBER() over(partition by handid order by cardid)  as row_num
       from cte),
cte2 as
     (select handid from cte1 where row_num=(select COUNT(*) from #winners) )
select * from cte where handid in (select handid from cte2)

SQL fiddle demo

Answer 3

您可以将查询更改为

select a.*, count(a.*) as a_count 
from 
    #x a 
    inner join #winners b
        on
        a.cardid = b.cardid 
group by a.handid 
having a_count = 3

Answer 4

@Bluefleets（+1）方法在数据集的性能方面看起来很好;我想有5亿条记录，性能配置文件会发生变化。

我认为OP希望输出格式略有不同，稍微适应@ Bluefleet的代码会产生：

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

我还会考虑恐惧游标解决方案 - 因为它可能会在大量记录上表现更好，具体取决于数据结构，索引，获胜者数量等。

但是没有完整的数据集，我真的不能说。