Tonight someone tweeted trying to find a ColdFusion or SQL solution to counting the number of weekdays (excluding Saturdays and Sundays) between two dates. Most of the solutions online involving looping over the date range and adding to a counter if that iteration of the loop is a weekday.
That’s fine for limited, but probably not if you need to test dates that span many years or need to run the function 1,000s of times. Anyway, here’s a SQL Server UDF solution. It’s late, so hopefully the comments in the code suffice. It should be straightforward to rewrite this in ColdFusion.
This is an improved version of the solution presented here. That solution fails when the timespan is less than 7 days long and contains at least one weekend day.
create function [dbo].[dateDiffWeekdays] ( @startdaytime DATETIME; @enddaytime DATETIME; ) returns int as begin DECLARE @answer int; set @answer = 0 -- Strip Times SELECT @start = dateadd(dd,0, datediff(dd,0,@startdaytime)) SELECT @end = dateadd(dd,0, datediff(dd,0,@enddaytime)) -- handle end conditions DECLARE @firstWeekDayInRange datetime, @lastWeekDayInRange datetime; select @firstWeekDayInRange = case -- If Saturday, add two days when datepart(dw,@start) = 7 then dateadd(day,2,@start) -- If Sunday, add one day when datepart(dw,@start) = 1 then dateadd(day,1,@start) else @start end select @lastWeekDayInRange = case -- If Saturday, substract one day when datepart(dw,@end) = 7 then dateadd(day,-1,@end) -- If Sunday, substract two days when datepart(dw,@end) = 1 then dateadd(day,-2,@end) else @end end -- add one day to answer (to count Friday) if enddate was on a weekend if @end != @lastWeekDayInRange set @answer = 1 select @answer = @answer + case -- triggered if start and end date are on same weekend when dateDiff(day,@firstWeekDayInRange,@lastWeekDayInRange) < 0 then (@answer * -1) -- otherwise count the days and substract 2 days per weekend in between dates else (DateDiff(day, @firstWeekDayInRange, @lastWeekDayInRange) - DateDiff(week, @firstWeekDayInRange, @lastWeekDayInRange)*2) end return @answer end -- Call created function select dbo.dateDiffWeekdays('6/1/2009','6/13/2009')