Convert iso_week to calendar date in SQL - sql

Question

I've been searching though the archives without finding what I am looking for- I'd be happy for some guidance.
I have a data set where I want to report aggregated number of appointments by provider (STAFFID) and work week, the latter defined by the week's Monday date. I've played with datepart(iso_week, appointment_date) as week_of_yr which gets me part of the way there- I can group by week to get the right numbers. However, I can't figure out if there's a simple way to display the date of the week's Monday given the iso_week integer (and year).
I found ISO8601 Convert Week Date to Calendar Date helpful, though I do not know whether (or how) I can automate that process for many values at once.
Here's the tidbit of code I have. Ideally I could add another expression to the select statement which would display the desired date.
select STAFFID
, count(*) as appointment_ct
, datepart(iso_week, appointment_date) as iso_wk --this returns the week # of the year as an int
from [dbo].[view_APPT_DATA]
where program_code in ('99999')
and appointment_date >= '1/1/2016' and appointment_date <='3/31/2016'
group by iso_wk, STAFFID

Answer 1

I would find the first Monday of that year and then use DATEADD to add the number of weeks to that day
select STAFFID
, count(*) as appointment_ct
, datepart(iso_week, appointment_date) as iso_wk --this returns the week # of the year as an int
, dateadd(week, datepart(week, DATEADD(DAY, (##DATEFIRST - DATEPART(WEEKDAY, dateadd(year, datepart(year, appointment_date) - 1900, 0)) + (8 - ##DATEFIRST) * 2) % 7, dateadd(year, datepart(year, appointment_date) - 1900, 0))) as monday_wk
from [dbo].[view_APPT_DATA]
where program_code in ('99999')
and appointment_date >= '1/1/2016' and appointment_date <='3/31/2016'
group by iso_wk, STAFFID, monday_wk

Answer 2

I didn't quite get the cha's query to return correct results for "special" years that have overlapping weeks.
This is the (inline) function i ended up with to calculate iso week to first day of that week:
CREATE OR ALTER FUNCTION dbo.FN_ISOWEEK_TO_DAY (
-- ISOWeek in format YYYYWW
#pISOWeek INT
)
RETURNS TABLE AS RETURN
SELECT DATEADD(week, CAST(RIGHT(#pISOWeek, 2) AS INT) - 1
- CASE WHEN (##DATEFIRST - DATEPART(WEEKDAY, DATEADD(YEAR, cast(LEFT(#pISOWeek, 4) AS INT) - 1900, 0)) + (8 - ##DATEFIRST) * 2) % 7 >= 4 -- means first monday is one week ahead
THEN 1 ELSE 0 END
, DATEADD(DAY, (##DATEFIRST - DATEPART(WEEKDAY, DATEADD(YEAR, cast(LEFT(#pISOWeek, 4) AS INT) - 1900, 0)) + (8 - ##DATEFIRST) * 2) % 7, DATEADD(YEAR, cast(LEFT(#pISOWeek, 4) AS INT) - 1900, 0)))
AS firstDay
Some test code:
SELECT *
FROM (
SELECT 202001, 20191230
UNION ALL
SELECT 202053, 20201228
) x (dt, expectedValue)
CROSS APPLY dbo.FN_ISOWEEK_TO_DAY(x.dt) y