SQL to get an daily average from month total - sql

Question

I have a table that lists month totals (targets)
person total month
----------- --------------------- -----------
1001 114.00 201005
1001 120.00 201006
1001 120.00 201007
1001 120.00 201008
.
1002 114.00 201005
1002 222.00 201006
1002 333.00 201007
1002 111.00 201008
.
.
but month is an integer(!)
I also have another table that has a list of working days (calendar)
tran_date day_type
----------------------- ---------------------------------
1999-05-01 00:00:00.000 WEEKEND
1999-05-02 00:00:00.000 WEEKEND
1999-05-03 00:00:00.000 WORKING_DAY
1999-05-04 00:00:00.000 WORKING_DAY
1999-06-01 00:00:00.000 .....
.
.
.
What I want to do is get a list of dates with the average for that day based on the number of days in the month where day_type is 'WORKING_DAY' / the month's total.
so if I had say 20 working days in 201005 then I'd get an average of 114/20 on each working day, while the other days would be 0.
somthing like
person tran_date day_avg
------- ----------------------- ---------------------------------
1001 2010-05-01 00:00:00.000 0
1001 2010-05-02 00:00:00.000 0
1001 2010-05-03 00:00:00.000 114/2 (as there are two working days)
1001 2010-05-04 00:00:00.000 114/2 (as there are two working days)
.
.
.
It has to be done as a CTE as this is a limitation of the target system (I can only do one statement)
I can start off with (Dates to
WITH
Dates AS
(
SELECT CAST('19990501' as datetime) TRAN_DATE
UNION ALL
SELECT TRAN_DATE + 1
FROM Dates
WHERE TRAN_DATE + 1 <= CAST('20120430' as datetime)
),
Targets as
(
select CAST(cast(month as nvarchar) + '01' as dateTime) mon_start,
DATEADD(MONTH, 1, CAST(cast(month as nvarchar) + '01' as dateTime)) mon_end,
total
from targets
)
select ????

Answer 1

Sample data (may vary):
select * into #totals from (
select '1001' as person, 114.00 as total, 199905 as month union
select '1001', 120.00, 199906 union
select '1001', 120.00, 199907 union
select '1001', 120.00, 199908
) t
select * into #calendar from (
select cast('19990501' as datetime) as tran_date, 'WEEKEND' as day_type union
select '19990502', 'WEEKEND' union
select '19990503', 'WORKING_DAY' union
select '19990504', 'WORKING_DAY' union
select '19990505', 'WORKING_DAY' union
select '19990601', 'WEEKEND' union
select '19990602', 'WORKING_DAY' union
select '19990603', 'WORKING_DAY' union
select '19990604', 'WORKING_DAY' union
select '19990605', 'WORKING_DAY' union
select '19990606', 'WORKING_DAY' union
select '19990701', 'WORKING_DAY' union
select '19990702', 'WEEKEND' union
select '19990703', 'WEEKEND' union
select '19990704', 'WORKING_DAY' union
select '19990801', 'WORKING_DAY' union
select '19990802', 'WORKING_DAY' union
select '19990803', 'WEEKEND' union
select '19990804', 'WEEKEND' union
select '19990805', 'WORKING_DAY' union
select '19990901', 'WORKING_DAY'
) t
Select statement, it returns 0 if the day is 'weekend' or not exists in calendar table. Please keep in mind that MAXRECURSION is a value between 0 and 32,767.
;with dates as (
select cast('19990501' as datetime) as tran_date
union all
select dateadd(dd, 1, tran_date)
from dates where dateadd(dd, 1, tran_date) <= cast('20010101' as datetime)
)
select t.person , d.tran_date, (case when wd.tran_date is not null then t.total / w_days else 0 end) as day_avg
from dates d
left join #totals t on
datepart(yy, d.tran_date) * 100 + datepart(mm, d.tran_date) = t.month
left join (
select datepart(yy, tran_date) * 100 + datepart(mm, tran_date) as month, count(*) as w_days
from #calendar
where day_type = 'WORKING_DAY'
group by datepart(yy, tran_date) * 100 + datepart(mm, tran_date)
) c on t.month = c.month
left join #calendar wd on d.tran_date = wd.tran_date and wd.day_type = 'WORKING_DAY'
where t.person is not null
option(maxrecursion 20000)

Answer 2

You could calculate the number of working days per month in a subquery. Only the subquery would have to use group by. For example:
select t.person
, wd.tran_date
, t.total / m.WorkingDays as day_avg
from #Targets t
join #WorkingDays wd
on t.month = convert(varchar(6), wd.tran_date, 112)
left join
(
select convert(varchar(6), tran_date, 112) as Month
, sum(case when day_type = 'WORKING_DAY' then 1 end) as WorkingDays
from #WorkingDays
group by
convert(varchar(6), tran_date, 112)
) as m
on m.Month = t.month
Working example at SE Data.
For the "magic number" 112 in convert, see the MSDN page.

Answer 3

If I understood your question correctly, the following query should do it:
SELECT
*,
ISNULL(
(
SELECT total
FROM targets
WHERE
MONTH(tran_date) = month - ROUND(month, -2)
AND c1.day_type = 'WORKING_DAY'
) /
(
SELECT COUNT(*)
FROM calendar c2
WHERE
MONTH(c1.tran_date) = MONTH(c2.tran_date)
AND c2.day_type = 'WORKING_DAY'
),
0
) day_avg
FROM
calendar c1
In plain English:
For each row in calendar,
get the total of the corresponding month if this row is a working day (otherwise get NULL),
get the number of working days in the same month
and divide them.
Finally, convert the NULL (of non-working days) into 0.