How to get daily budget based on monthly budget and workings days - sql

Question

Have have 2 tables.
One table with month budget, and one table with workings days.
What I want, is find out daily budget based on the monthly budget and working days.
Example:
August have a budget on 1000 and have 21 workings day.
September have a budget on 2000 and 23 workings days
I want to figure out what the total budget betweens two dates.
Ex: between 2020-08-02 and 2020-09-15
But must be sure that, days in august takes budget from august, days from september takes budget from september etc.
tbBudget:
Date | Amount
2020-08-01 | 1000
2020-09-01 | 2000
2020-10-01 | 3000
tbWorkingDays
Date | WorkingDay
2020-08-01 | 0
2020-08-02 | 0
2020-08-03 | 1
2020-08-04 | 1
2020-08-05 | 1
2020-08-06 | 1
2020-08-07 | 1
2020-08-08 | 1
...
2020-09-01 | 1
2020-09-02 | 1
2020-09-03 | 0
2020-09-04 | 1
...
2020-10-01 | 1
2020-10-02 | 0
2020-10-03 | 1
2020-10-04 | 1
I have no idea how to solve this issue. Can you help me?
My result should be like:
Date | WorkingDay | BudgetAmount
2020-08-02 | 0 | 0.0
2020-08-03 | 1 | 47.6
2020-08-04 | 1 | 47.6
2020-08-05 | 1 | 47.6
..
2020-09-13 | 1 | 86.9
2020-09-14 | 1 | 86.9
2020-09-15 | 1 | 86.9

Answer 1

Using CTE and group by:
with CTE1 AS(
SELECT FORMAT(A.DATE, 'MMyyyy') DATE, B.AMOUNT, SUM(CASE WHEN [WorkingDay] = 1 THEN 1 ELSE 0 END) AS TOTAL_WORKING_DAYS
FROM tbWorkingDays A INNER JOIN tbBudget B
ON (FORMAT(A.DATE, 'MMyyyy') = FORMAT(B.DATE, 'MMyyyy')) GROUP BY FORMAT(A.[DATE], 'MMyyyy'), B.AMOUNT
)
SELECT A.DATE,
A.WORKINGDAY,
CASE WHEN A.WORKINGDAY = 1 THEN B.AMOUNT/B.TOTAL_WORKING_DAYS
ELSE 0 END AS BudgetAmount
FROM CTE1 B
INNER JOIN
tbWorkingDays A
ON (FORMAT(A.DATE, 'MMyyyy') = B.DATE);

Answer 2

Assuming that the budgets are by month:
select wd.*,
(case when workingday = 0 then 0
else wd.budget * 1.0 / sum(wd.workingday) over (partition by wd.date)
end) as daily_amount
from tbWorkingDays wd join
tblBudget b
on wd.date >= b.date and wd.date < dateadd(month, 1, wd.date);
If the budget dates are not per month, then use apply instead:
select wd.*,
(case when workingday = 0 then 0
else wd.budget * 1.0 / sum(wd.workingday) over (partition by wd.date)
end) as daily_amount
from tbWorkingDays wd cross apply
(select top (1) b.*
from tblBudget b
where wd.date >= b.date
order by b.date desc
) b

Answer 3

Use sum as an analytical function to get the number of workingdays pr month, then divide out
Here is a functioning solution
with tally as
(
SELECT
row_number() over (order by (select null))-1 n
from (values (null),(null),(null),(null),(null),(null),(null),(null),(null),(null),(null)) a(a)
cross join (values (null),(null),(null),(null),(null),(null),(null),(null),(null),(null),(null)) b(b)
cross join (values (null),(null),(null),(null),(null),(null),(null),(null),(null),(null),(null)) c(c)
)
, tbWorkingDays as
(
select
cast(dateadd(day,n,'2020-01-01') as date) [Date],
iif(DATEPART(WEEKDAY,cast(dateadd(day,n,'2020-01-01') as date)) in (1,7),0,1) WorkingDay
from tally
where n<365
)
, tbBudget AS
(
select * from
(values
(cast('2020-08-01' as date), cast(1000 as decimal(19,2)))
,(cast('2020-09-01' as date), cast(2000as decimal(19,2)))
,(cast('2020-10-01' as date), cast(3000as decimal(19,2)))
) a([Date],[Amount])
)
select
a.[Date]
,a.WorkingDay*
(b.Amount/
sum(a.WorkingDay) over (partition by year(a.Date)*100+month(a.Date)))
from tbWorkingDays a
inner join tbBudget b
on a.Date between b.Date and dateadd(day,-1,dateadd(month,1,b.date))
The work is done here:
select
a.[Date]
,a.WorkingDay*
(b.Amount/
sum(a.WorkingDay) over (partition by year(a.Date)*100+month(a.Date)))
from tbWorkingDays a
inner join tbBudget b
on a.Date between b.Date and dateadd(day,-1,dateadd(month,1,b.date))
The expression
sum(a.WorkingDay) over (partition by year(a.Date)*100+month(a.Date))
Sums the number of workingdays for the current month. I then join against the budget and take the sum for the month and divide by the expression above.
To make sure there only is budget on workingdays, I simply multiply by "workingday", since 0 is a non workingday, the sum will be 0 for all non workingdays.