如何找到最接近的FRI？

Question

In SQL Server 2008, given a date, how would I get a date corresponding to FRI of that week?

so for example:
6/6/2012 -> 6/8/2012
6/5/2012 -> 6/8/2012

Answer 1

Assuming you want 6/9/2012 to also return 6/8/2012 (same week), this would work. It get's the day of week of the current date and adds the difference between that and Friday which is hardcoded to a value of 6.

SET DATEFIRST 7;    
declare @date date = '6/5/2012'

select dateadd(dd,6-datepart(dw,@date),@date) as Friday

If you want 6/9/2012 to return the next Friday, you just have to make a small modification:

SET DATEFIRST 7;
declare @date date = '6/9/2012'
set @date = dateadd(dd,1,@date) -- this adds a day to the date you inputted but doesn't matter since the function will always return to you a Friday
-- Sunday resets the week with datepart so adding a day to Saturday resets the week resulting in the next week being returned.

select dateadd(dd,6-datepart(dw,@date),@date) as Friday

Answer 2

Here's a function I created that seems to work. It does not change DATEFIRST and will give you the next date for the DOW. The function will return the date you passed in if it is on the DOW you are looking for.

CREATE FUNCTION [dbo].[func_NextDate]
(
    @dt DATE,
    @dow INT -- Use the day-of-week as defined by SQL Server (1=Sun, 7=Sat)
)
RETURNS DATE
AS 
BEGIN

DECLARE @dtDiff INT = 7-((DATEPART(dw, @dt)+(7-@dow))%7)
IF @dtDiff = 7
    SET @dtDiff = 0 -- Return the date if it is on the dow requested

RETURN DATEADD(dd, @dtDiff, @dt)

END