Reputation: 87
The database:
EMPLOYEE (fname, minit, lname, ssn, birthdate, address, sex, salary, superssn, dno) KEY: ssn
DEPARTMENT (dname, dnumber, mgrssn, mgrstartdate) KEY: dnumber.
PROJECT (pname, pnumber, plocation, dnum) KEY: pnumber.
WORKS_ON (essn, pno, hours) KEY: (essn, pno)
DEPENDENT (essn, dependent-name, sex, bdate, relationship) KEY: (essn, dependent-name)
I want to use left outer join
and group by
to...
Find the last name and SSN of those managers who work on 3 or more projects and who are not located in Cleveland.
Here is what I have so far:
select Lname
from Employee e outer join Department d
where (e.ssn = d.mgrssn)
and ssn NOT in (
select w.essn
from works_on w outer join Project p
where w.pno = p.pnumber
and p.plocation = 'Cleveland'
group by w.essn
having count(*) >= 3
)
Did I do it right using left outer join
and group by
? Should I divide this code into two parts, like loops?
Upvotes: 1
Views: 83
Reputation: 48197
Select JOIN
find all project for the employee
First HAVING
tell you this user doesnt have project in 'Cleveland'
Second HAVING
tell you this user has 3 project or more
.
SELECT e.Lname, e.ssn
FROM Employee e
JOIN works_on w
ON e.ssn = w.essn
JOIN Project p
ON w.pno = p.pnumber
GROUP BY e.ssn
HAVING
SUM(CASE WHEN p.plocation = 'Cleveland' THEN 1 ELSE 0 END) = 0
AND COUNT(*) >= 3
Upvotes: 1