SQL Count Of Open Orders Each Day Between Two Dates

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 d

Answer 1

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