Number of days of a week inside a month in SQL - sql

Question

Using SQL Server 2019, therefore also using T-SQL, I'm trying to get the number of days of a particular week, identified by the ISO 8601 week definition, that fall inside a particular month.
The table that contains the data has the following fields:
Year | Month | IsoWeek
-----------------------
2020 | 12 | 50
2020 | 12 | 51
2020 | 12 | 52
2020 | 12 | 53
2021 | 01 | 01
2021 | 01 | 02
2021 | 01 | 03
For example the last week of 2020, week 53, has started on Monday 28th Dec 2020 and has finished on Sunday 3rd Jan 2021.
So I want to get 4 for Dec2020 and 3 for Jan2021 in order to create a new table like the following:
Year | Month | IsoWeek | WeekDays
------------------------------------
2020 | 12 | 52 | 7
2020 | 12 | 53 | 4
2021 | 01 | 53 | 3
2021 | 01 | 01 | 7
2021 | 01 | 02 | 7
Is there a way to do it entirely in SQL?

Answer 1

This is sort of a SQL Server setting (or trick) which works because the 1st of January, 1900 was a Monday. Since that's where SQL Server starts counting from it makes it easier to locate the first Thursday of any month. Thanks to Jeff Moden btw. I got this from something he wrote. Maybe there's a better way to this now, idk
with iso_dts_cte(yr, mo, wk) as (
select * from (values ('2020', '12', '50'),
('2020', '12', '51'),
('2020', '12', '52'),
('2020', '12', '53'),
('2021', '01', '01'),
('2021', '01', '02'),
('2021', '01', '03')) v(yr, mo, wk))
select iso.*, v.*
from iso_dts_cte iso
cross apply (values (cast(dateadd(wk,datediff(wk,0,'01/04/'+iso.yr),0)+((iso.wk-1)*7) as date),
cast(dateadd(wk,datediff(wk,0,'01/04/'+iso.yr),0)+((iso.wk)*7)-1 as date))) v(start_dt, end_dt);
yr mo wk start_dt end_dt
2020 12 50 2020-12-07 2020-12-13
2020 12 51 2020-12-14 2020-12-20
2020 12 52 2020-12-21 2020-12-27
2020 12 53 2020-12-28 2021-01-03
2021 01 01 2021-01-04 2021-01-10
2021 01 02 2021-01-11 2021-01-17
2021 01 03 2021-01-18 2021-01-24
To expand the week ranges into days and then count by calendar year and calendar month you could try something like this.
[Edit] It's my understanding the date hierarchy you're looking for is 1) calendar year, 2) calendar month, 3) iso week. The output seems to match the example now. However, there's not a way to ORDER BY to display like the example.
with
iso_dts_cte(yr, mo, wk) as (
select * from (values ('2020', '12', '50'),
('2020', '12', '51'),
('2020', '12', '52'),
('2020', '12', '53'),
('2021', '01', '01'),
('2021', '01', '02'),
('2021', '01', '03')) v(yr, mo, wk)),
days_cte(n) as (
select * from (values (1),(2),(3),(4),(5),(6),(7)) v(n))
select year(dt.calc_dt) cal_yr, month(dt.calc_dt) cal_mo, iso.wk, count(*) day_count
from iso_dts_cte iso
cross apply (values (cast(dateadd(wk,datediff(wk,0,'01/04/'+iso.yr),0)+((iso.wk-1)*7) as date),
cast(dateadd(wk,datediff(wk,0,'01/04/'+iso.yr),0)+((iso.wk)*7)-1 as date))) v(start_dt, end_dt)
cross join days_cte d
cross apply (values (dateadd(day, d.n-1, v.start_dt))) dt(calc_dt)
group by year(dt.calc_dt), month(dt.calc_dt), iso.wk;
cal_yr cal_mo wk day_count
2020 12 50 7
2020 12 51 7
2020 12 52 7
2020 12 53 4
2021 1 01 7
2021 1 02 7
2021 1 03 7
2021 1 53 3

Answer 2

I would assume you could use:
Select DatePart(weekday, '2021/01/01')
and loop for each months, the value here is 5 as the day of the week is Friday, thus there is 3 days for that first week and 4 days for the previous month last week since there is 7 days per week.
If you have the week number 1 to 53 and the year, if it's the first or last week of the month, you need to calculate using the weekday logic written above. Otherwise it's 7.
You could create a scalar function with a similar logic and create your table using the function in it. I believe you are expecting something simpler which nothing come to mind right now.
You can also use "SET DATEFIRST 1;" to change what is considered the first day of the week (Monday / Sunday).
Ended up creating a proof of concept:
CREATE FUNCTION calculateWeekDays(#y int,#m int,#w int)
RETURNS INT
AS BEGIN
DECLARE #numberOfDays INT = 7;
DECLARE #firstDayOfMonth datetime = DATEFROMPARTS (#y, #m, 1);
DECLARE #nextMonth datetime = (SELECT DATEADD(MONTH, 1, #firstDayOfMonth));
DECLARE #lastDayOfMonth datetime = (SELECT DATEADD(DAY, -1, #nextMonth));
DECLARE #weekOfYearStartOfMonth INT = (select DatePart(week, #firstDayOfMonth));
DECLARE #weekOfYearEndOfMonth INT = (select DatePart(week, #lastDayOfMonth));
DECLARE #firstWeekOfMonth INT = (select datediff(week, dateadd(week, datediff(day,0,dateadd(month, datediff(month,0,#firstDayOfMonth),0))/7, 0),#firstDayOfMonth-1) + 1);
DECLARE #lastWeekOfMonth INT = (select datediff(week, dateadd(week, datediff(day,0,dateadd(month, datediff(month,0,#lastDayOfMonth),0))/7, 0),#lastDayOfMonth-1) + 1);
IF #w - #weekOfYearStartOfMonth % 52 = 0
SET #numberOfDays = (select 8-DatePart(weekday, #firstDayOfMonth));
ELSE IF #w = #weekOfYearEndOfMonth
SET #numberOfDays = (select DatePart(weekday, #nextMonth)-1);
RETURN #numberOfDays;
END
GO
from that scalar function you can get the result using:
select yr, mo, wk, dbo.calculateWeekDays(yr,mo,wk) as wd from (
values (2020, 12, 50),
(2020, 12, 51),
(2020, 12, 52),
(2020, 12, 53),
(2021, 01, 01),
(2021, 01, 02),
(2021, 01, 03),
(2021, 03, 13),
(2021, 03, 14),
(2021, 04, 14),
(2021, 04, 15)) v(yr, mo, wk)
it give the following result:
year
mo
wk
wd
2020
12
50
7
2020
12
51
7
2020
12
52
7
2020
12
53
4
2021
1
1
3
2021
1
2
7
2021
1
3
7
----
--
--
--
2021
3
13
7
2021
3
14
3
2021
4
14
4
2021
4
15
7