如何仅返回在 mysql 连接中找到所有提供的值的行？

Question

I have an inner join like this:

SELECT
    p.id
    , p.date
    , pu.user_id
FROM projections p
INNER JOIN projection_user AS pu ON p.id = pu.projection_id
WHERE pu.user_id IN ('1', '3')

This returns the following results:

id | date       | user_id 
--------------------------
16 | 2020-03-05 | 1
17 | 2020-03-05 | 1
17 | 2020-03-05 | 3

I would like to only return rows where the projection ( p.id ) has the exact pu.user_id values supplied in my WHERE clause -- in this case the projection with a p.id of 17, since it has both pu.user_id 1 and 3, and no others. In the full code I am actually grouping by p.id for the final result:

SELECT
    p.id
    , p.date
FROM projections p
INNER JOIN projection_user AS pu ON p.id = pu.projection_id
WHERE pu.user_id IN ('1', '3')
GROUP BY p.id

id | date
--------------------------
16 | 2020-03-05
17 | 2020-03-05

How can I only return the projection with p.id of 17?

Answer 1

Use aggregation:

SELECT p.id, p.date, GROUP_CONCAT(pu.user_id)
FROM projections p INNER JOIN
     projection_user pu
     ON p.id = pu.projection_id
GROUP BY p.id, p.date
HAVING SUM(pu.user_id IN (1, 3)) = COUNT(*) AND
       COUNT(*) = 2;

Note: This assumes no duplicates in projection_user , which is typically a reasonable assumption. The query can be tweaked if such duplicates are possible.

Or, if you prefer:

SELECT p.id, p.date,
       GROUP_CONCAT(pu.user_id ORDER BY pu.user_id) as user_ids
FROM projections p INNER JOIN
     projection_user pu
     ON p.id = pu.projection_id
GROUP BY p.id, p.date
HAVING user_ids = '1,3';

Answer 2

This amendment to Gordon Linoff's answer works as needed:

SELECT
    p.id
    , p.date
FROM projections p
INNER JOIN projection_user AS pu ON p.id = pu.projection_id
GROUP BY p.id
HAVING SUM(pu.user_id IN (1, 3)) = COUNT(*) AND COUNT(*) = 2