SQL选择在哪里作为交集而不是联合

Question

I'll explain the problem with examples so it is easy to understand, given the following data structure:

id  userId  
1   1    
1   2    
2   2    
3   2    
1   3     
2   3 

I can get a list of id s for a set of users as follows:

declare @tmp table (id int, userId int)
insert into @tmp values(1,1), (1,2), (2,2), (3,2), (1,3), (2,3)


select id from @tmp
where userId in (1,2,3)
group by id

This will return the following as expected:

id
1
2
3

My question is, how can I only get the ids that have mapping for EVERY userId in the where clause? eg the result for userId in (1,2,3) should be 1 and for userId in (2,3) should be 1,2

I've tried going through each id and then merging those but so far had no luck on finding an actual solution.

NOTE The solution must work for larger data sets, imagine millions of rows and thousands of userIds, efficiency of the solution is not as important (as it does not have to run very often)

Second NOTE I just noticed that counting the result does not actually guarantee correctness, because two different userIds may have the same count of mappings but mapped to different Items. In that scenario it is not an intersection anymore

Answer 1

You could use a count of user IDs to check... Like this

SELECT Id
From table
Where userid in (1,2,3)
Group by id
Having count(userid) = (select count(distinct userid) from table where userid in(1,2,3))

Ideally the 2 conditions get parameterized, but that is outside the scope of this question.

Answer 2

I have used temp table to store userids

declare @tmp table (id int, userId int)
insert into @tmp values(1,1), (1,2), (2,2), (3,2), (1,3), (2,3)

declare @userid table (id int)
insert into @userid values (1), (2), (3)

select
    t.id
from 
    (select *, cnt = count(*) over () from @userid) u
    join @tmp t on u.id = t.userId
group by t.id, u.cnt
having u.cnt = count(distinct u.id)

Answer 3

If your only problem is that you want to specify the user IDs just once, use your user table:

with u as (select userid from users where userid in (1,2,3))
select id 
from mytable
where userid in (select userid from u)
group by id
having count(distinct userid) = (select count(*) from u);

If you wanted to react on invalid user IDs with an empty result set, you'd replace the users table with a values clause:

with u as (select userid from (values (1), (2), (3)) AS ids(userid))

Answer 4

You can avoid a count distinct as well as having to enter your userId values more than once - or at all if you populate the equivalent of the @i dataset from a dynamic query:

declare @t table (id int, userId int);
insert into @t values(1,1), (1,2), (2,2), (3,2), (1,3), (2,3);

declare @i table (i int);
insert into @i values(1),(2),(3);

select t.id
from @i as i
    join @t as t
        on i.i = t.userId
group by t.id
having count(i.i) = (select count(1) from @i);

Answer 5

One method is using HAVING :

SELECT id
FROM YourTable
GROUP BY id
HAVING COUNT(CASE WHEN UserID = 1 THEN 1 END) > 0
   AND COUNT(CASE WHEN UserID = 2 THEN 1 END) > 0
   AND COUNT(CASE WHEN UserID = 3 THEN 1 END) > 0;