SELECT clause with MINUS INTERSECT and UNION - sql

Question

In my database, i created a table name DEPTLOC
//DEPTLOC
DNAME CITY
----------------------------
RESEARCH BOSTON
IT PARIS
SCIENCE LONDON
RESEARCH LONDON
SALES NEWYORK
RESEARCH PARIS
RESEARCH NEWYORK
MARKETING NEWYORK
So i used the following query
SELECT CITY FROM DEPTLOC
INTERSECT
(
SELECT CITY FROM DEPTLOC WHERE DNAME='SALES'
UNION
SELECT CITY FROM DEPTLOC WHERE DNAME='RESEARCH'
);
But my output is all the CITY will be displayed. My question is want find which of the DNAME='SALES' OR DNAME='RESEARCH' has its location in all cities.
So from the table above, all distinct city are
CITY
-------
BOSTON
PARIS
LONDON
NEWYORK
Since 'RESEARCH' have all the location but 'SALES' Only have some, my output should be display like this
DNAME
---------
RESEARCH
What should i change for my query in order to get the correct output

Answer 1

Here's a shorter (than my previous answer) query that checks only between DNAMES SALES and RESEARCH to see which DNAME has all CITIES. Set operators UNION ALL and MINUS are used.
SELECT DNAME
FROM
(
SELECT 'SALES' DNAME, COUNT(*) MISSING_CITIES
FROM
(
SELECT DISTINCT CITY FROM DEPTLOC
MINUS
SELECT CITY FROM DEPTLOC WHERE DNAME = 'SALES'
)
UNION ALL
SELECT 'RESEARCH', COUNT(*) FROM
(
SELECT DISTINCT CITY FROM DEPTLOC
MINUS
SELECT CITY FROM DEPTLOC WHERE DNAME = 'RESEARCH'
)
)
WHERE MISSING_CITIES = 0;

Answer 2

One way of doing this is to count the distinct locations, and join it on the departments query:
SELECT dname
FROM (SELECT dname, COUNT(*) AS dept_city_count
FROM deptloc
GROUP BY dname) d
JOIN (SELECT COUNT (DISTINCT city) AS city_count
FROM deptloc) ON city_count = dept_cirt_count

Answer 3

Here is a query using joins. First all Cities are selected. Then, a CROSS JOIN is made between all Departments and all Cities. Then, a LEFT JOIN is made to select Departments that do not have all Cities. Then, another LEFT JOIN is made to select Departments that do have all Cities.
SELECT DISTINCT
DL.DNAME
FROM
DEPTLOC DL
LEFT JOIN
(
SELECT
ALL_COMBOS.DNAME dname
FROM
(
SELECT DISTINCT
D1.DNAME DNAME,
D2.CITY CITY
FROM
DEPTLOC D1
CROSS JOIN
(
SELECT DISTINCT
CITY
FROM
DEPTLOC
)
D2 --All distinct Cities
)
ALL_COMBOS --All Departments with all Locations
LEFT JOIN DEPTLOC D2
ON
ALL_COMBOS.DNAME = D2.DNAME
AND ALL_COMBOS.CITY = D2.CITY
WHERE
D2.DNAME IS NULL
)
NOT_ALL_LOCATIONS --Departments that do not have all Locations
ON DL.DNAME = NOT_ALL_LOCATIONS.DNAME
WHERE
NOT_ALL_LOCATIONS.DNAME IS NULL; --Departments that have all Locations
As it can be observed, this method does not rely on the count of the Cities, but on the actual values of the Cities themselves.

Answer 4

Assuming that the key in deptloc is (dname, city) then the following would work:
select dname
from deptloc
group by dname
having count(*) = (select count(distinct city) from deptloc);
The query works by creating one group for every department. The count(*) would be the nr of cities that the department is located in. Each such count is compared with the nr of unique Cities cited in the same table.
Edited: Saw your comment now, that this was the solution you did not want :)
Here is another version of "relational division"
select departments.dname
from (select distinct dname from deptloc) departments
where not exists(
select 'x'
from (select distinct city from deptloc) cities
where not exists(
select 'x'
from deptloc x
where x.city = cities.city
and x.dname = departments.dname));