Това е Релационно разделение без остатък (RDNR) проблем. Вижте това статия от Dwain Camps, който предоставя много решения на този вид проблеми.
Първо решение
SELECT empId
FROM (
SELECT
empID, cc = COUNT(DISTINCT department)
FROM employe
WHERE department IN('Y', 'Z')
GROUP BY empID
)t
WHERE
t.cc = 2
AND t.cc = (
SELECT COUNT(*)
FROM employe
WHERE empID = t.empID
)
Второ решение
SELECT e.empId
FROM employe e
WHERE e.department IN('Y', 'Z')
GROUP BY e.empID
HAVING
COUNT(e.department) = 2
AND COUNT(e.department) = (SELECT COUNT(*) FROM employe WHERE empID = e.empId)
Без да използвате GROUP BY
и HAVING
:
SELECT DISTINCT e.empID
FROM employe e
WHERE
EXISTS(
SELECT 1 FROM employe WHERE department = 'Z' AND empID = e.empID
)
AND EXISTS(
SELECT 1 FROM employe WHERE department = 'Y' AND empID = e.empID
)
AND NOT EXISTS(
SELECT 1 FROM employe WHERE department NOT IN('Y', 'Z') AND empID = e.empID
)