Reputation: 445
SELECT dep.depName, SUM(worker.salary) as total
FROM team, worker
WHERE worker.depID = dep.depID
GROUP BY dep.depID
SELECT dep.depName, SUM(manager.salary) as total
FROM manager, dep
WHERE manager.depID = dep.depID
GROUP BY team.depID
I tried something along the line of: SELECT dep.depName, SUM(manager.salary)+SUM(worker.salary) as total but it gave me some strange output, it adds N elements N number of time.
Upvotes: 0
Views: 127
Reputation: 5050
What about something like this :
SELECT result.depName, SUM(worker.salary) as total, manager_total
FROM team, worker
INNER JOIN
(SELECT dep.depID as depID, dep.depName as depName, SUM(manager.salary) as manager_total
FROM dep
LEFT OUTER JOIN manager ON manager.depID = dep.depID
GROUP BY dep.depID, dep.depName) as result ON result.depID = team.depID
WHERE worker.depID = team.depID
GROUP BY result.depName, manager_total
Upvotes: 0
Reputation: 1271131
The more efficient way is to sum the results by department without doing a join. Then join in the department name.
This method also allows you to separate the department salaries into the two groups:
select depId, sum(wsalary) as wsalary, sum(msalary) as msalary,
sum(wsalary) + sum(msalary) as total
from ((select depId, sum(w.salary) as wsalary, null as msalary
from worker w
group by depId
)
union all
(select depId, NULL as wsalary, sum(m.salary) as msalary
from manager m
group by depId
)
) t join
dep d
on t.depId = d.depId
group by t.depId
I am assuming the team
in the first from
clause really refers to dep
, as used in the rest of the query.
Upvotes: 0
Reputation: 10841
You need to do a subquery of a UNION statement
SELECT dep.depName, sum(salary) as total, dep.depID
FROM
(
(SELECT dep.depName, worker.salary as salary, dep.depID
FROM team, worker
WHERE worker.depID = dep.depID
)
UNION
(SELECT dep.depName, manager.salary as salary, dep.depID
FROM manager, dep
WHERE manager.depID = dep.depID
)
) dep GROUP BY dep.depID
Upvotes: 1