SQL Count Of Open Orders Each Day Between Two Dates - sql

Question

I've tried searching but it's likely I'm using the wrong keywords as I can't find an answer.
I'm trying to find the number of orders that are open between two dates and by employee. I have one table that shows a list of employees, another that shows a list of orders that contains an open and close date and also a dates table if that helps.
The employee and order tables joined will return something like:
employee order ref opened closed
a 123 01/01/2012 04/01/2012
b 124 02/01/2012 03/01/2012
a 125 02/01/2012 03/01/2012
And I need to transform this data into:
Date employee Count
01/01/2012 a 1
02/01/2012 a 2
02/01/2012 b 1
03/01/2012 a 2
03/01/2012 b 1
04/01/2012 a 1
I'm pulling the data from SQL server.
Any ideas?
Thanks
Nick

Answer 1

Join Dates to the result of the join between Employees and Orders, then group by dates and employees to obtain the counts, something like this:
SELECT
d.Date,
o.Employee,
COUNT(*) AS count
FROM Employees e
INNER JOIN Orders o ON e.ID = o.Employee
INNER JOIN Dates d ON d.Date BETWEEN o.Opened AND o.Closed
GROUP BY
d.Date,
o.Employee

Answer 2

My favorite way to do this counts the number of cumulative opens and the number of cumulative closes over time.
with cumopens as
(select employee, opened as thedate,
row_number() over (partition by employee order by opened) as cumopens,
0 as cumcloses
from eo
),
cumcloses as
(select employee, closed as thedate, 0 as cumopens,
row_number() over (partition by employee order by closed ) as cumcloses
from eo
)
select employee, c.thedate, max(cumopens), max(cumcloses),
max(cumopens) - max(cumcloses) as stillopened
from ((select *
from cumopens
) union all
(select *
from cumcloses
)
) c
group by employee, thedate
The only problem with this approach is that only dates where there is employee activity get reported. This works in your case.
The more general solution requires a sequence numbers to generate dates. For this, I often create one from some existing table with enough rows:
with nums as
(select row_number() over (partition by null order by null) as seqnum
from employees
)
select employee, dateadd(day, opened, seqnum) as thedate, count(*)
from eo join
nums
on datediff(day, opened, closed) < seqnum
group by employee, dateadd(day, opened, seqnum)
order by 1, 2

Answer 3

SELECT opened,employee,count(*)
FROM employee LEFT JOIN orders
WHERE opened < firstDate and opened > secondDate
GROUP BY opened,employee
or you can change the first condition in
WHERE opened BETWEEN firstDate and secondDate

Answer 4

Calling the result column count was a bit odd because it seems to be in fact a row number.
You can do that by using ROW_NUMBER.
The other interesting part is that you also want open date and close date as separate rows. Using a simple UNION will solve that.
WITH cte
AS (SELECT Row_number() OVER ( PARTITION BY employee
ORDER BY order_ref) count,
employee,
opened,
closed
FROM orders)
SELECT employee, opened date, count
FROM cte
UNION ALL
SELECT employee, closed date, count
FROM cte
ORDER BY Date,
employee
DEMO