SQL 仅选择第二个表中的行，包含从第一个表收到的确切结果

Question

The answer must be rather simple, but I stuck with it for a while. 2 tables

1. BUILDING-TYPE Some types of work are located in some buildings

2. WORKER-TYPE Worker can do some types of work

TYPE_BLD_ID	BLD	TYPE
2	1	2
6	2	5
7	2	6
8	3	6
9	4	2
4	5	3
3	5	4
5	5	7
13	6	1
14	6	7
12	7	5

WRK_TYPE_ID	WORKER	TYPE
0	0	0
5	1	3
4	1	4
6	1	7
7	2	2
11	2	4
10	2	5
9	2	6
12	2	7
15	3	5
14	3	6
19	3	7
17	4	1
16	4	2
18	4	7

WHEN I select eg Building #4, I expect getting Worker #1, who can handle types of work 1&7. It looks rather stupidly simple, but

1. I select types of work from the building: SELECT TYPE FROM TYPES_IN_BUILDING WHERE BLD = 1; Get 1 and 7
2. NEXT I do SELECT WORKER FROM WORKER_TYPES WHERE TYPE IN (1,7);

And, obviously receive all workers, who can do 1 and 7. But I want to get ONLY WORKERS, who has BOTH types, 1 AND 7.

The issue in the question - JOINs in the SQL version, used in my system, something extremely outdated.

THE FOLLOWING WUERY DID NOT WORK:

'''

SELECT * FROM WORKER W
WHERE W.WORKER IN (SELECT WA.WORKER
    FROM WORKER_ATYPE WA
    WHERE WA.ATYPE IN 
    (SELECT AB.ATYPE FROM ATYPE_IN_BUILD AB WHERE AB.BLD = 6)
    GROUP BY WA.WORKER
    HAVING COUNT(DISTINCT ATYPE) = 
        (SELECT COUNT(ATYPE) FROM TYPE_IN_BUILD WHERE BLD = 6)
 ;

'''

Answer 1

You want the worker types to be a superset of the building types. You can do this using:

select w.worker
from (select b.*, count(*) over (partition by building) as cnt
      from building b
     ) b join
     workers w
     on b.type = w.type
where b.building = 1
group by w.worker, b.cnt
having count(*) = b.cnt;

This matches workers to the building based on types. Only rows that match are included. Then, the workers are aggregated and a worker matches if the total number of rows matches the number of types for the building.