1) Your solution uses a non-deterministic function: datepart(dw...)
. Because of this aspect, changing DATEFIRST
setting will gives different results. For example, you should try:
SET DATEFIRST 7;
your solution;
and then
SET DATEFIRST 1;
your solution;
2) Following solution is independent of DATEFIRST
/LANGUAGE
settings:
DECLARE @NextDayID INT = 0 -- 0=Mon, 1=Tue, 2 = Wed, ..., 5=Sat, 6=Sun
SELECT DATEADD(DAY, (DATEDIFF(DAY, @NextDayID, GETDATE()) / 7) * 7 + 7, @NextDayID) AS NextDay
Result:
NextDay
-----------------------
2013-09-23 00:00:00.000
This solution is based on following property of DATETIME
type:
Day 0 = 19000101
= Mon
Day 1 = 19000102
= Tue
Day 2 = 19000103
= Wed
...
Day 5 = 19000106
= Sat
Day 6 = 19000107
= Sun
So, converting INT value 0 to DATETIME gives 19000101
.
If you want to find the next Wednesday
then you should start from day 2 (19000103
/Wed
), compute days between day 2 and current day (20130921
; 41534 days), divide by 7 (in order to get number of full weeks; 5933 weeks), multiple by 7 (41531 fays; in order to get the number of days - full weeks between the first Wednesday
/19000103
and the last Wednesday
) and then add 7 days (one week; 41538 days; in order to get following Wednesday
). Add this number (41538 days) to the starting date: 19000103
.
Note: my current date is 20130921
.
Edit #1:
DECLARE @NextDayID INT;
SET @NextDayID = 1; -- Next Sunday
SELECT DATEADD(DAY, (DATEDIFF(DAY, ((@NextDayID + 5) % 7), GETDATE()) / 7) * 7 + 7, ((@NextDayID + 5) % 7)) AS NextDay
Result:
NextDay
-----------------------
2013-09-29 00:00:00.000
Note: my current date is 20130923
.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…