Reputation: 43
Please help me with below problem.
table 1 employee details
emp name empno.
---------------------------------
John 1234
Joe 6789
table 2 employee assignment
empno assignmentstartdate assignmentenddate assignmentID empassignmentID
-----------------------------------------------------------------------------
1234 01JAN2017 02JAN2017 A1 X1
6789 01jan2017 02JAN2017 B1 Z1
table 3 employee assignment property
empassignmentID assignmentID propertyname propertyvalue
-------------------------------------------------------------------
X1 A1 COMPLETED true
X1 A1 STARTED true
Z1 B1 STARTED true
Z1 B1 COMPLETED false
Result wanted: (count of completed and started for each employee)
emp name emp no. COMPLETED STARTED
------------------------------------------
John 1234 1 1
Joe 6789 0 1
Currently with my query it is not putting count correctly for propertyvalue if I run for one employee it works correctly but not for multiple employees. Please help.
SELECT empno ,
empname ,
(SELECT COUNT(A.propertyvalue)
FROM employeedetails C ,
employees_ASSIGNMENT RCA,
employee_assignment_property A
WHERE TRUNC(startdate) >= '14jun2017'
AND TRUNC(endate) <= '20jun2017'
AND RCA.empno = C.empno
AND RCA.empassignmetid = A.empassignmetid
AND rca.EMPNO IN ('1234','6789')
AND RCA.assignmentid = A.assignmentid
AND A.Name = 'COMPLETED'
AND A.propertyvalue = 'true') ,
(SELECT COUNT(A.propertyvalue)
FROM employeedetails C ,
employees_ASSIGNMENT RCA,
employee_assignment_property A
WHERE TRUNC(startdate) >= '14jun2017'
AND TRUNC(endate) <= '20jun2017'
AND RCA.empno = C.empno
AND RCA.empassignmetid = A.empassignmetid
AND rca.EMPNO IN ('1234','6789')
AND RCA.assignmentid = A.assignmentid
AND A.Name = 'STARTED'
AND A.propertyvalue = 'true')FROM employeedetails WHERE EMPNO IN
('1234','6789') GROUP BY C.empno ,
C.EMPNAME
Upvotes: 4
Views: 1125
Reputation: 15150
I think you are simply looking for this:
SELECT DET.empname
, COUNT(CASE WHEN PROP.propertyname = 'COMPLETED' THEN 1 END) COMP_COUNT
, COUNT(CASE WHEN PROP.propertyname = 'STARTED' THEN 1 END) START_COUNT
FROM employeedetails DET
INNER JOIN employees_ASSIGNMENT ASS
ON ASS.empno = DET.empno
INNER JOIN employee_assignment_property PROP
ON PROP.empassignmentID = ASS.empassignmentID
AND PROP.assignmentID = ASS.assignmentID
GROUP BY DET.empname
Just add a WHERE
clause if you need one.
Upvotes: 2
Reputation: 639
if you want you result as a query without CTEs this should work:
select empName,
empNo,
(select employee_details.empNo, count(employee_assignment.assId)
from employee_details as t1
join employee_assignment on (t1.empno = employee_assignment.empno)
join employee_assignment_property on (employee_assignment.assId = employee_assignment_property.assId)
where employee_assignment.ptop = 'COMPLETED'
and t.empNo = t1.empNo
group by t1.empNo ) as [COMPLETED],
(select employee_details.empNo, count(employee_assignment.assId)
from employee_details as t1
join employee_assignment on (t1.empno = employee_assignment.empno)
join employee_assignment_property on (employee_assignment.assId = employee_assignment_property.assId)
where employee_assignment.ptop = 'STARTED'
and t.empNo = t1.empNo
group by t1.empNo ) as [STARTED],
from employee_details as t
Upvotes: 1
Reputation: 639
this should work using CTEs: Using Common Table Expressions
with numComplet()
as
(
select tbl1.empNo, count(tbl2.assId)
from tbl1
join tbl2 on (tbl1.empno = tbl2.empno)
join tbl3 on (tbl2.assId = tbl3.assId)
where tbl2.ptop = 'COMPLETED'
group by tbl1.empNo
),
with numStarted()
as
(
select tbl1.empNo, count(tbl2.assId)
from tbl1
join tbl2 on (tbl1.empno = tbl2.empno)
join tbl3 on (tbl2.assId = tbl3.assId)
where tbl2.ptop = 'STARTED'
group by tbl1.empNo
)
select *
from tbl1
join numComplet on (tbl1.empNo = numComplet.empNo)
join numStarted on (tbl1.empNo = numStarted.empNo)
I put down table names as tbl[1|2|3]
Upvotes: 0
Reputation: 342
If you don't want to do a dirty query composed of subqueries, you can try creating a view
(if your database permits it).
What does it mean : I'll be useless in front of this. In summary, a view is a temporary table.
Hope this helps
Upvotes: 0