Return number of weekdays between 2 dates in T-SQL - sql

Question

I have 2 dates and want to know how many weekdays (mon-fri) there are
e.g.
thu jan 1 20xx
fri jan 2 20xx
sat jan 3 20xx
sun jan 4 20xx
mon jan 5 20xx
jan 1, jan 5 would return 3
(can ignore public holidays)

Answer 1

Try
DateDiff(day, #DtA, #DtB) - 2 * DateDiff(Week, #DtA, #DtB)
this may not work exactly, but you can see the idea. Some slight modification will work.

Answer 2

Assuming the dates can't be more than five and a half years from each other (or use your own tally table instead of master..spt_values):
DECLARE #date1 datetime, #date2 datetime;
SET #date1 = '20110901';
SET #date2 = '20110905';
SELECT COUNT(*)
FROM (
SELECT
Date = DATEADD(day, number, #date1)
FROM master..spt_values
WHERE type = 'P'
AND number between 0 AND DATEDIFF(day, #date1, #date2)
) s
WHERE DATENAME(DW, Date) NOT IN ('Saturday', 'Sunday')

Answer 3

try this:
SET DATEFIRST 1
DECLARE #StartDate datetime
,#EndDate datetime
SELECT #StartDate='6/21/2011'
,#EndDate='6/28/2011'
;with AllDates AS
(
SELECT #StartDate AS DateOf, datepart(weekday,getdate()) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datepart(weekday,DateOf+1)
FROM AllDates
WHERE DateOf<#EndDate
)
SELECT COUNT(*) AS WeekDayCount FROM AllDates WHERE WeekDayNumber<=5
OUTPUT:
WeekDayCount
------------
6
(1 row(s) affected)
If you have a holiday table, you can join it in and remove those as well.
EDIT based on #Ross Watson comment:
SET DATEFIRST 1
DECLARE #StartDate datetime
,#EndDate datetime
SELECT #StartDate='6/21/2011'
,#EndDate='6/28/2011'
;with AllDates AS
(
SELECT #StartDate AS DateOf, datepart(weekday,getdate()) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, (WeekDayNumber+1) % 7
FROM AllDates
WHERE DateOf<#EndDate
)
SELECT COUNT(*) AS WeekDayCount FROM AllDates WHERE WeekDayNumber>0 AND WeekDayNumber<6
--I don't like using "BETWEEN", ">", ">=", "<", and "<=" are more explicit in defining end points
produces same output as original query.

Answer 4

Try the following:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2011/06/01'
SET #EndDate = '2011/06/31'
SELECT
(DATEDIFF(dd, #StartDate, #EndDate) + 1) -
(DATEDIFF(wk, #StartDate, #EndDate) * 5) -
(
CASE
WHEN DATENAME(dw, #StartDate) in
('Sunday', 'Tuesday', 'Wednesday', 'Turesday', 'Saturday')
THEN 1
ELSE 0
END
) -
(
CASE
WHEN DATENAME(dw, #EndDate) in
('Sunday', 'Tuesday', 'Wednesday', 'Turesday', 'Saturday')
THEN 1
ELSE 0
END
)

Answer 5

Also see if this is helpful
http://blog.sqlauthority.com/2007/06/08/sql-server-udf-function-to-display-current-week-date-and-day-weekly-calendar/

Answer 6

This approach is limited to ~100 days due to recursion. This works for the date ranges i've tested. Same idea above, removed the math and simplified:
BEGIN
SET DATEFIRST 1
DECLARE #StartDate datetime
,#EndDate datetime
SELECT #StartDate='12/16/2015'
,#EndDate='1/8/2016'
;with AllDates AS
(
SELECT #StartDate AS DateOf
UNION ALL
SELECT DateOf+1
FROM AllDates
WHERE DateOf<#EndDate
)
SELECT COUNT(*) AS WeekDayCount
FROM
AllDates
WHERE
datepart(weekday,DateOf) between 1 AND 5
--SELECT DateOf [date], datepart(weekday,DateOf) [day]
--FROM
-- AllDates
--WHERE
-- datepart(weekday,DateOf) between 1 AND 5
END