Get the date of Sunday and Saturday 2 weeks ago - sql

Question

I have a stored procedure that produces a report each night. This SP references a view that takes a snapshot of data from 2 weeks ago (Sunday to Saturday).
I need to get the dates of the Saturday and Sunday from two weeks ago (so for today that would be 23/24 of December.
I have searched through many solutions but I am unable to find a decent response to this - any one able to help with an answer as well as explaining a little bit of the reason for doing it that way (it seems like there are a thousand ways to do it).
Thanks

Answer 1

Here is the answer in case any one needs it in future.
SELECT DATEADD(wk, -2, DATEADD(wk, DATEDIFF(wk, 6, GETDATE()), 6))

Answer 2

Try this (for Saturday) is the same for Sunday:
DECLARE #today datetime
SET #today = GETDATE()
SELECT
CASE
WHEN DATENAME(weekday, #today) = 'Monday' THEN DATEADD(day, -15, #today)
WHEN DATENAME(weekday, #today) = 'Tuesday' THEN DATEADD(day, -16, #today)
WHEN DATENAME(weekday, #today) = 'Wednesday' THEN DATEADD(day, -17, #today)
WHEN DATENAME(weekday, #today) = 'Thursday' THEN DATEADD(day, -18, #today)
WHEN DATENAME(weekday, #today) = 'Friday' THEN DATEADD(day, -19, #today)
WHEN DATENAME(weekday, #today) = 'Saturday' THEN DATEADD(day, -13, #today)
WHEN DATENAME(weekday, #today) = 'Sunday' THEN DATEADD(day, -14, #today)
END
DATENAME with weekday parameter returns the name of the day, so if you are on Monday you must subtract 15 days to reach Saturday and so on.
Pay attention: If you have another language set you can change it for your query using:
SET LANGUAGE us_english
You can see SqlFiddle

Answer 3

Since the day of week value can differ on different machines you either need to set it using the set datefirst 1 command as suggeste in Podiluska's answer,
or you can do it by first determining the actual value for sunday and add that to the days you are subtracting.
To determine the actual value for sunday simply ask the value for a know sunday, like 2018-01-07 for example
Below example should return 2017-12-23 / 2017-12-24
It will also return this if you are running this on friday, or thursday, or whatever day...
-- this will contain are test date
declare #date date = getdate() - 2
-- determine the number for sunday
declare #Sunday int = (select datepart(dw, '20180107'))
select convert(date,dateadd(d, -15 - datepart(dw, #date) + (datepart(dw, #date) - #Sunday), getdate())),
convert(date,dateadd(d, -14 - datepart(dw, #date) + (datepart(dw, #date) - #Sunday), getdate()))

Answer 4

That depends on how you define when your week starts, but the basic principle is to find the start of this week, and take away 14 days.
eg:
set datefirst 1
select convert(date,dateadd(d, -15-datepart(dw, getdate()), getdate())),
convert(date,dateadd(d, -14-datepart(dw, getdate()), getdate()))

Answer 5

Sorry, I'll work on a general form for these; this is the quick and dirty.
I like the NEXT_DAY function. Simple, and obvious as to what it does.
declare
vFromDate date := trunc( sysdate ) - 9 ; -- '27 Feb 2019'
vSatDate date ;
vSunDate date ;
begin
-- test a range of dates
while vFromDate <= sysdate -- '3 Mar 2019'
loop
select NEXT_DAY ( TRUNC ( vFromDate) - 21, 'SAT') -- 7: general form for these?
-- keep or discard the " - 1" depending on if the 2 wk Sunday is the desired Sunday
, NEXT_DAY ( TRUNC ( vFromDate) - 20 - 1, 'SUN') -- 1:
into vSatDate, vSunDate
from dual ;
dbms_output.put_line( to_char( vFromDate, 'dd Mon": "' )
|| to_char( vSatDate, 'dd Mon' )
|| to_char( vSunDate, '", "dd Mon' )
);
vFromDate := vFromDate + 1 ;
end loop ;
dbms_output.put_line( rpad('*', 21, '*' ) );
end ;