How to access columns from correlated subquery - sql

Question

With below data i am trying to get employees of same location and department with type as manager whose salary is equals to sum of other employees sal along with emp ids as empids are unequal query not returning any result also grouping query should have more than 1 record
so result should not include Raghu record and also i want emp ids of all the records that are matched :
EMP_ID EMP_Name EMP_Loc EMP_Dept EMP_Sal Emp_type
1 Arjun Hyd Comp 1000 Manager
2 Ramesh Hyd Comp 500 Interim
3 Ragav Hyd Comp 300 Interim
4 Rajesh Hyd Comp 200 Interim
5 Raghu Hyd Comp 1000 Interim
select a.emp_dept , a.emp_loc ,a.emp_dept,b.emp_dept,a.emp_id,b.emp_id
from
(select sum(emp_sal) as sett,emp_loc,emp_dept,emp_id
from employee
where emp_type = 'Interim'
group by emp_loc,emp_dept,emp_id having count(emp_sal)>1
) a
inner join
(select emp_sal ,emp_loc,emp_dept,emp_id
from employee
where emp_type = 'Manager'
) b
on a.sett=b.emp_sal and a.emp_loc=b.emp_loc and a.emp_dept=b.emp_dept;

Answer 1

Sample data
solution:
SELECT a.EMP_ID, a.EMP_Name, a.EMP_Loc, a.EMP_Dept, a.EMP_Sal, a.Emp_type
FROM employee AS a INNER JOIN
(SELECT SUM(EMP_Sal) AS sal, EMP_Loc,EMP_Dept
FROM employee AS employee_1
WHERE (Emp_type = 'Interim')
GROUP BY EMP_Loc, EMP_Dept having Count(EMP_Sal)>1) AS b
ON a.EMP_Dept = b.EMP_Dept AND a.EMP_Loc = b.EMP_Loc AND a.EMP_Sal = b.sal
WHERE a.Emp_type= 'Manager'
Results:

Answer 2

Your problem is that you are including the emp_id in your first sub-select and grouping on it. This means that you get three rows, and the sum becomes pointless. There is never a case where the manager has 500, 200 or 300 salary, so the join doesn't work and there are therefore no records returned. If you do this instead it should work:
select a.emp_dept, a.emp_loc, b.emp_id
from
(select sum(emp_sal) as sett,emp_loc,emp_dept
from employee
where emp_type = 'Interim' and emp_sal > 1
group by emp_loc,emp_dept
) a
inner join
(select emp_sal ,emp_loc,emp_dept,emp_id
from employee
where emp_type = 'Manager'
) b
on a.sett=b.emp_sal and a.emp_loc=b.emp_loc and a.emp_dept=b.emp_dept;
Note that I also removed duplicate columns in the resultset. There is no point showing b.emp_dept as well as a.emp_dept, because the join condition guarantees that they will be the same.
EDIT
Following your revision to the question, I understand now what you are trying to do. This is really difficult to achieve, since what you need to do, is to take a cartesian sum of all the possible Interim salaries for a location and department to check if any possible combination of salary sums matches that of the Manager.
One possible way to do this would be the following:
WITH cte as
(SELECT a.emp_id as aempid, b.emp_id as bempid, 0 as cempid, 0 as dempid, a.emp_sal + b.emp_sal AS SumSal, a.emp_loc, a.emp_dept
FROM employee a
CROSS JOIN employee b WHERE a.emp_loc = b.emp_loc AND a.emp_dept = b.emp_dept
AND a.emp_type = b.emp_type AND a.emp_type = 'Interim' AND a.emp_id <> b.emp_id
AND b.emp_id > a.emp_id
UNION
SELECT a.emp_id, b.emp_id, c.emp_id, 0, a.emp_sal + b.emp_sal + c.emp_sal AS SumSal, a.emp_loc, a.emp_dept
FROM employee a
CROSS JOIN employee b
CROSS JOIN employee c
WHERE a.emp_loc = b.emp_loc AND a.emp_dept = b.emp_dept
AND a.emp_type = b.emp_type AND a.emp_type = 'Interim' AND a.emp_id <> b.emp_id
AND a.emp_loc = c.emp_loc AND a.emp_dept = c.emp_dept
AND a.emp_type = c.emp_type AND a.emp_id <> c.emp_id AND b.emp_id <> c.emp_id
AND b.emp_id > a.emp_id AND c.emp_id > b.emp_id
UNION
SELECT a.emp_id, b.emp_id, c.emp_id, d.emp_id, a.emp_sal + b.emp_sal + c.emp_sal + d.emp_sal AS SumSal, a.emp_loc, a.emp_dept
FROM employee a
CROSS JOIN employee b
CROSS JOIN employee c
CROSS JOIN employee d
WHERE a.emp_loc = b.emp_loc AND a.emp_dept = b.emp_dept
AND a.emp_type = b.emp_type AND a.emp_type = 'Interim' AND a.emp_id <> b.emp_id
AND a.emp_loc = c.emp_loc AND a.emp_dept = c.emp_dept
AND a.emp_type = c.emp_type AND a.emp_id <> c.emp_id AND b.emp_id <> c.emp_id
AND a.emp_loc = d.emp_loc AND a.emp_dept = d.emp_dept
AND a.emp_type = d.emp_type AND a.emp_id <> d.emp_id AND b.emp_id <> d.emp_id AND c.emp_id <> d.emp_id
AND b.emp_id > a.emp_id AND c.emp_id > b.emp_id AND d.emp_id > c.emp_id
)
SELECT m.emp_id as managerid, cte.aempid, cte.bempid, cte.cempid, cte.dempid, m.emp_sal, m.emp_loc, m.emp_dept
FROM employee m
INNER JOIN cte ON cte.emp_loc = m.emp_loc AND cte.emp_dept = m.emp_dept AND m.emp_sal = cte.sumsal
WHERE m.emp_type = 'Manager';
This is really ugly, but I could think of nothing better, and at least it returns the desired result. In practice of course, you would have to create it dynamically to match the maximum number of Interim employees in any one location/department.
A couple of points in case it is not clear. With this approach there is no need for a HAVING COUNT(*) > 1 since the requirement a.emp_id <> b.emp_id etc. guarantees that there are at least two different Interim employees. Also the requirement that b.emp_id > a.emp_id (and c and d) is there to ensure that each possible combination of a, b, c and d only occurs once (so in this case we get just 2, 3, 4 (in that order) plus 0 for d, not all possible combinations of 2, 3, and 4.

Answer 3

You are looking for all groups of employees that have a total salary that equals their manager's salary. In order to find all these groups you need a recursive query in SQL. In PostgreSQL you can use an array to collect the employees in that query.
Sample data:
emp_id
emp_type
salary
1
Manager
1000
2
Interim
500
3
Interim
500
4
Interim
300
5
Interim
200
Result:
manager_id
employee_ids
1
{2,3}
1
{2,4,5}
1
{3,4,5}
The query:
with recursive cte(manager_id, manager_salary, employee_ids, total, loc, dept) as
(
select emp_id, emp_sal, array[]::integer[], 0, emp_loc, emp_dept
from employee
where emp_type = 'Manager'
union
select cte.manager_id, cte.manager_salary, cte.employee_ids || e.emp_id,
cte.total + e.emp_sal, cte.loc, cte.dept
from cte
join employee e
on e.emp_loc = cte.loc
and e.emp_dept = cte.dept
and e.emp_type <> 'Manager'
and e.emp_id > all(cte.employee_ids)
and cte.total + e.emp_sal <= cte.manager_salary
)
select manager_id, manager_salary, employee_ids
from cte
where total = manager_salary
and array_length(employee_ids, 1) > 1
order by manager_id, employee_ids;
Demo: https://dbfiddle.uk/?rdbms=postgres_13&fiddle=5311cdaebdc875440d9e226ace8dfefc
As a bonus:
At last the query with the employee IDs unnested, so you see the groups in multiple line with one line per employee:
with recursive cte(manager_id, manager_salary, employee_ids, total, loc, dept) as
(
select emp_id, emp_sal, array[]::integer[], 0, emp_loc, emp_dept
from employee
where emp_type = 'Manager'
union
select cte.manager_id, cte.manager_salary, cte.employee_ids || e.emp_id,
cte.total + e.emp_sal, cte.loc, cte.dept
from cte
join employee e
on e.emp_loc = cte.loc
and e.emp_dept = cte.dept
and e.emp_type <> 'Manager'
and e.emp_id > all(cte.employee_ids)
and cte.total + e.emp_sal <= cte.manager_salary
)
select cte.manager_id, cte.manager_salary,
dense_rank() over (partition by cte.manager_id order by cte.employee_ids) as grp,
e.*
from cte
cross join lateral unnest(employee_ids) link(employee_id)
join employee e on e.emp_id = link.employee_id
where cte.total = cte.manager_salary
and array_length(cte.employee_ids, 1) > 1
order by cte.manager_id, grp, e.emp_id;
Demo: https://dbfiddle.uk/?rdbms=postgres_13&fiddle=7b6218f28cbf1e9d4c1a0c76c46b9601