I need to check if a given day is the last sunday of any year, if yes the return 1 using TSQL only.
I do not have much idea about TSQL.
SQL Server has a problem with weekdays, because they can be affected by internationalization settings. Assuming the defaults, you can do:
select dateadd(day,
1 - datepart(weekday, datefromparts(#year, 12, 31)),
datefromparts(#year, 12, 31)
)
Otherwise, you'll need to do a case expression to turn the day of the week into a number.
In an older version of SQL Server, you could do:
select dateadd(day,
1 - datepart(weekday, cast(#year + '0101' as date)),
cast(#year + '0101' as date)
)
I haven't worked with tsql specifically but if my sql knowledge and googling is good enough then something like this should do the trick:
... WHERE DATEPART(dw, date) = 7 and DATEDIFF (d, date, DATEFROMPARTS (DATEPART(yyyy, date), 12, 31)) <= 6
Basically we check if that day is Sunday at first and then if it's less than week away from last day of the year
Using Mr. Gordon's query, following IIF() returns 1 if given day is last Sunday of the year, returns 0 if it is not.
Using 2018 as year and 2018-12-30 as given date. You can replace values with variables.
select IIF( DATEDIFF(DAY,'2018-12-30',
DATEADD(day,
1 - datepart(weekday, datefromparts(2018, 12, 31)),
datefromparts(2018, 12, 31)
)) = 0, 1, 0)
You can use this function
Function Code :
create FUNCTION CheckIsSaturday
(
#date DATETIME
)
RETURNS int
AS
BEGIN
-- Declare the return variable here
DECLARE #result INT
DECLARE #DayOfWeek NVARCHAR(22)
DECLARE #LastDayOfYear DATETIME
select #LastDayOfYear=DATEADD(yy, DATEDIFF(yy, 0, #date) + 1, -1)
SELECT #DayOfWeek=DATENAME(dw, #date)
IF(#DayOfWeek='Saturday' AND DATEDIFF(dd,#date,#LastDayOfYear)<7)
RETURN 1;
RETURN 0;
END
GO
function Usage:
SELECT dbo.CheckIsSaturday('2017-12-23')
This becomes quite trivial if you have a Calendar Table
DECLARE #CheckDate DATE = '20181230'
;WITH cteGetDates AS
(
SELECT
[Date], WeekDayName, WeekOfMonth, [MonthName], [Year]
,LastDOWInMonth = ROW_NUMBER() OVER
(
PARTITION BY FirstDayOfMonth, [Weekday]
ORDER BY [Date] DESC
)
FROM
dbo.DateDimension
)
SELECT * FROM cteGetDates D
WHERE D.LastDOWInMonth = 1 AND D.WeekDayName = 'Sunday' and D.MonthName = 'December' AND D.[Date] = #CheckDate
You can also use this one to get every last day of the year:
;WITH getlastdaysofyear ( LastDay, DayCnt ) AS (
SELECT DATEADD(dd, -DAY(DATEADD(mm, 1, DATEADD(yy, DATEDIFF(yy, 0, GETDATE()) + 1, -1))),
DATEADD(mm, 1, DATEADD(yy, DATEDIFF(yy, 0, GETDATE()) + 1, -1))),
0 AS DayCnt
UNION ALL
SELECT LastDay,
DayCnt + 1
FROM getlastdaysofyear
)
SELECT *
FROM ( SELECT TOP 7 DATEADD(DD, -DayCnt, LastDay) LastDate,
'Last ' + DATENAME(Weekday,DATEADD(DD, -DayCnt, LastDay)) AS DayStatus
FROM getlastdaysofyear ) T
ORDER BY DATEPART(Weekday, LastDate)
Hope you like it :)
This question already has answers here:
Count work days between two dates
(24 answers)
Calculating days to excluding weekends (Monday to Friday) in SQL Server
(6 answers)
Closed 7 years ago.
I'm looking for a way to calculate the days between two dates, but on weekdays. Here is the formula, but it counts weekend days.
DATEDIFF(DAY,STARTDATE,ENDDATE)
SELECT DATEDIFF(DAY,'2015/06/01' , '2015/06/30')
Result of above query of datediff is 29 days which are weekend days. but i need week days that should be 21 by removing Saturday and Sunday(8 days).
Any suggestions?
Put it in the WHERE clause
SELECT DATEDIFF(DAY,'2015/06/01' , '2015/06/30')
FROM yourtable
WHERE DATENAME(dw, StartDate) != 'Saturday'
AND DATENAME(dw, StartDate) != 'Sunday'
Or all in a SELECT statement
SELECT (DATEDIFF(dd, StartDate, EndDate) + 1)-(DATEDIFF(wk, StartDate, EndDate) * 2)-(CASE WHEN DATENAME(dw, StartDate) = 'Sunday' THEN 1 ELSE 0 END)-(CASE WHEN DATENAME(dw, EndDate) = 'Saturday' THEN 1 ELSE 0 END)
This returns 22:
DECLARE #StartDate AS DATE = '20150601'
DECLARE #EndDate AS DATE = '20150630'
SELECT
(DATEDIFF(DAY, #StartDate, #EndDate))
-(DATEDIFF(WEEK, #StartDate, #EndDate) * 2)
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)
Read this article by Jeff Moden for more information.
Explanation:
First, (DATEDIFF(DAY, #StartDate, #EndDate)) will return the difference in number of days. In this case, it'll be 29. Now, depending on your interpretation of whole days, you may want to add 1 day to its result.
Next,(DATEDIFF(WEEK, #StartDate, #EndDate) * 2):
To quote the article:
DATEDIFF for the WEEK datepart doesn't actually calculate weeks, it
calculates the number of times a date range contains dates that
represent pairs of Saturdays and Sundays. To think of it in more
simple terms, it only counts WHOLE WEEKENDS!
So, to exclude the weekends, you must subtract twice the result of this from the first DATEDIFF. Which now will be: 29 - (2 *4) = 21.
Finally, this:
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)
removes the partial weeks, which only happens when then #StartDate occurs on a Sunday and the #EndDate occurs on a Saturday.
You can try recursive cte:
WITH cte
AS ( SELECT CAST('2015/06/01' AS DATE) AS dt ,
DATEPART(WEEKDAY, '2015/06/01') AS wd
UNION ALL
SELECT DATEADD(d, 1, dt) AS dt ,
DATEPART(WEEKDAY, DATEADD(d, 1, dt))
FROM cte
WHERE dt < '2015/06/30'
)
SELECT COUNT(*)
FROM cte
WHERE wd NOT IN ( 7, 1 )
Result is 22.
You better add some useful calendar table that every database should have. Fill it with some big range and then use it to calculate business days.
I wrote a simple stored procedure - don't know if a SP is what you need but i think you can easily convert it to a function , TVF, whatever:
CREATE PROCEDURE [dbo].[BusinessdaysBetween](#DateFrom DATE, #DateTo DATE, #days INT OUTPUT)
AS
BEGIN
DECLARE #datefirst INT = ##DATEFIRST
SET DATEFIRST 1 --so week starts on monday
SET #days = 0
IF #DateFrom > #DateTo
RETURN NULL
IF #DateFrom = #DateTo
RETURN #days
WHILE #DateFrom <= #DateTo
BEGIN
IF DATEPART(WEEKDAY,#DateFrom) NOT IN (6,7) --Saturday or Sunday
BEGIN
SET #days = #days + 1
END
SET #DateFrom = DATEADD(DAY,1,#DateFrom)
END
SET DATEFIRST #datefirst --restore original setup
END
GO
in my example i call i it like this:
DECLARE #days INT = 0
DECLARE #datefrom DATETIME = GETDATE()
DECLARE #dateto DATETIME = DATEADD(DAY,25,GETDATE())
EXEC dbo.BusinessdaysBetween #datefrom, #dateto, #days OUTPUT
SELECT #days
How to get Saturday's Date. I have today's date with me.
GETDATE()
How to do this.
For eg. TODAY is 08-08-2011
I want output as 08-13-2011
This is a function that will return the next Saturday if you call it like this:
SELECT dbo.fn_Get_NextWeekDay('2011-08-08', 6)
The "6" comes from the list of possible values you can set for DATEFIRST.
You can get any other day of the week by changing the second parameter accordingly.
This is the function:
IF OBJECT_ID('dbo.fn_Get_NextWeekDay') IS NOT NULL
DROP FUNCTION dbo.fn_Get_NextWeekDay
GO
CREATE FUNCTION dbo.fn_Get_NextWeekDay(
#aDate DATETIME
, #dayofweek INT
/*
#dw - day of the week
1 - Monday
2 - Tuesday
3 - Wednesday
4 - Thursday
5 - Friday
6 - Saturday
7 - Sunday
*/
)
RETURNS DATETIME
AS
/*
SELECT dbo.fn_Get_NextWeekDay('2011-08-08', 6)
SELECT dbo.fn_Get_NextWeekDay('2011-08-08', 1)
*/
BEGIN
RETURN
DATEADD(day
, ( #dayofweek + 8 - DATEPART(dw, #aDate) - ##DATEFIRST ) % 7
, #aDate
)
END
GO
[EDIT]
This might be another solution. This should work in any language:
IF OBJECT_ID('dbo.fn_NextWeekDay') IS NOT NULL
DROP FUNCTION dbo.fn_NextWeekDay
GO
CREATE FUNCTION dbo.fn_NextWeekDay(
#aDate DATE
, #dayofweek NVARCHAR(30)
)
RETURNS DATE
AS
/*
SELECT dbo.fn_NextWeekDay('2016-12-14', 'fri')
SELECT dbo.fn_NextWeekDay('2016-03-15', 'mon')
*/
BEGIN
DECLARE #dx INT = 6
WHILE UPPER(DATENAME(weekday,#aDate)) NOT LIKE UPPER(#dayofweek) + '%'
BEGIN
SET #aDate = DATEADD(day,1,#aDate)
SET #dx=#dx-1
if #dx < 0
BEGIN
SET #aDate = NULL
BREAK
END
END
RETURN #aDate
END
GO
Use DATEPART to get the day of week of today and add the difference to the desired day of week to todays date.
DECLARE #Today date = 'TODAYS-DATE';
DECLARE #TodayNumber int = DATEPART(dw, #Today) -- Get the day number
DECLARE #Saturday date = DATEADD(DAY, (6-#TodayNumber)%7, #Today)
-- Add the number of days between today and saturday (the 6th day), modulus 7 to stop you adding negative days
Hope that helps!
Use a Calendar table (table with one row per date):
SELECT MIN(DateValue) DateValue
FROM Calendar
WHERE DateValue >= CURRENT_TIMESTAMP
AND DayOfWeek = 'Saturday';
Another approach to this takes two steps, but might be more readable (look ma, no modulus):
Go back to last saturday: DATEADD(DAY, -1 * datepart(weekday, GETDATE()), getdate())
Then, add on a week: DATEADD(WEEK, 1, #lastSaturday, getdate()))
The whole thing:
declare #today DATETIME = GETDATE()
declare #lastSaturday DATETIME = DATEADD(DAY, -1 * datepart(weekday, #today), #today)
declare #nextSaturday DATETIME = DATEADD(WEEK, 1, #lastSaturday)
Or, if you're ok with #today being GETDATE(), you can do the calculation all at once:
SELECT DATEADD(WEEK, 1, DATEADD(DAY, -1 * datepart(weekday, GETDATE()), getdate()))
Checkout the SQL DATEADD function.
DATEADD (Transact-SQL)
Which you can use this along with DATEPART function to return the correct date.
DATEPART (Transact-SQL)
Try this :
SET DATEFIRST 7
DECLARE #d DATETIME
SET #d = '2011-08-08' --GETDATE()
SELECT NEXT_SAT = DATEADD(day, (7 + ##DATEFIRST - DATEPART(dw, #d)) % 7, #d )
declare #Curdate date=( SELECT SWITCHOFFSET(SYSDATETIMEOFFSET(),'+05:30') )
declare #nextsaturdaydate date=(select dateadd(d, 7-datepart(WEEKDAY, #CurDate),#Curdate))
select #nextsaturdaydate
How will you find last sunday of a month in sql 2000?
SELECT
DATEADD(day,DATEDIFF(day,'19000107',DATEADD(month,DATEDIFF(MONTH,0,GETDATE() /*YourValuehere*/),30))/7*7,'19000107')
Edit: A correct, final, working answer from my colleague.
select dateadd(day,1-datepart(dw, getdate()), getdate())
An alternative approach, borrowed from data warehousing practice. Create a date-dimension table and pre-load it for 10 years, or so.
TABLE dimDate (DateKey, FullDate, Day, Month, Year, DayOfWeek,
DayInEpoch, MonthName, LastDayInMonthIndicator, many more..)
The easiest way to fill-in the dimDate is to spend an afternoon with Excel and then import to DB from there. A half-decent dimDate table has 50+ columns -- anything you ever wanted to know about a date.
With this in place, the question becomes something like:
SELECT max(FullDate)
FROM dimDate
WHERE DayOfWeek = 'Sunday'
AND Month = 11
AND Year = 2009;
Essentially, all date related queries become simpler.
Next sunday in SQL, regardless which day is first day of week: returns 2011-01-02 23:59:59.000 on 22-dec-2010:
select DateADD(ss, -1, DATEADD(week, DATEDIFF(week, 0, getdate()), 14))
I find some of these solutions hard to understand so here's my version with variables to explain the steps.
ALTER FUNCTION dbo.fn_LastSundayInMonth
(
#StartDate DATETIME
,#RequiredDayOfWeek INT /* 1= Sunday */
)
RETURNS DATETIME
AS
/*
A detailed step by step way to get the answer...
SELECT dbo.fn_LastSundayInMonth(getdate()-31,1)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,2)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,3)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,4)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,5)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,6)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,7)
*/
BEGIN
DECLARE #MonthsSince1900 INTEGER
DECLARE #NextMonth INTEGER
DECLARE #DaysToSubtract INTEGER
DECLARE #FirstDayOfNextMonth DATETIME
DECLARE #LastDayOfMonthDayOfWeek INTEGER
DECLARE #LastDayOfMonth DATETIME
DECLARE #ReturnValue DATETIME
SET #MonthsSince1900=DateDiff(month, 0, #StartDate)
SET #NextMonth=#MonthsSince1900+1
SET #FirstDayOfNextMonth = DateAdd(month,#NextMonth, 0)
SET #LastDayOfMonth = DateAdd(day, -1, #FirstDayOfNextMonth)
SET #ReturnValue = #LastDayOfMonth
WHILE DATEPART(dw, #ReturnValue) <> #RequiredDayOfWeek
BEGIN
SET #ReturnValue = DATEADD(DAY,-1, #ReturnValue)
END
RETURN #ReturnValue
END
DECLARE #LastDateOfMonth smalldatetime
SELECT #LastDateOfMonth = DATEADD(month, DATEDIFF(month, -1, GETDATE()), 0) -1
Select DATEADD(dd,-( CASE WHEN DATEPART(weekday,#LastDateOfMonth) = 1 THEN 0 ELSE DATEPART(weekday,#LastDateOfMonth) - 1 END ),#LastDateOfMonth)
Holy cow, this is ugly, but here goes:
DECLARE #dtDate DATETIME
SET #dtDate = '2009-11-05'
SELECT DATEADD(dd, -1*(DATEPART(dw, DateAdd(day, -1, DateAdd(month, DateDiff(month, 0, #dtDate)+1, 0)))-1),
DateAdd(day, -1, DateAdd(month, DateDiff(month, 0, #dtDate)+1, 0)))
First built a tally table.
http://www.sqlservercentral.com/articles/T-SQL/62867/
then get what you want..
http://www.sqlservercentral.com/Forums/Topic515226-1291-1.aspx
DECLARE #DateStart DATETIME,
#DateEnd DATETIME
SELECT #DateStart = '20080131',
#DateEnd = '20101201'
SELECT DATEADD(wk,DATEDIFF(wk,6,DATEADD(mm,DATEDIFF(mm,-1,DATEADD(mm,t.N-1,#DateStart)),-1)),6)
FROM dbo.Tally t
WHERE t.N <= DATEDIFF(mm,#DateStart,#DateEnd)
Here's the correct way, accounting for ##DATEFIRST
IF NOT EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[fu_dtLastSundayInMonth]') AND type in (N'FN', N'IF', N'TF', N'FS', N'FT'))
BEGIN
EXECUTE(N'CREATE FUNCTION [dbo].[fu_dtLastSundayInMonth]() RETURNS int BEGIN RETURN 0 END ')
END
GO
/*
SET DATEFIRST 3; -- Monday
WITH CTE AS (
SELECT 1 AS i, CAST('20190101' AS datetime) AS mydate
UNION ALL
SELECT i+1 AS i, DATEADD(month, 1, CTE.mydate) AS mydate
FROM CTE WHERE i < 100
)
SELECT -666 AS i, dbo.fu_dtLastSundayInMonth('17530101') AS lastSundayInMonth, dbo.fu_dtLastSundayInMonth('17530101') AS Control
UNION ALL
SELECT -666 AS i, dbo.fu_dtLastSundayInMonth('99991231') AS lastSundayInMonth, dbo.fu_dtLastSundayInMonth('99991231') AS Control
UNION ALL
SELECT
mydate
,dbo.fu_dtLastSundayInMonth(mydate) AS lastSundayInMonth
,dbo.fu_dtLastSundayInMonth(mydate) AS lastSundayInMonth
,DATEADD(day,DATEDIFF(day,'19000107', DATEADD(MONTH, DATEDIFF(MONTH, 0, mydate, 30))/7*7,'19000107') AS Control
FROM CTE
*/
-- =====================================================================
-- Description: Return date of last sunday in month
-- of the same year and month as #in_DateTime
-- =====================================================================
ALTER FUNCTION [dbo].[fu_dtLastSundayInMonth](#in_DateTime datetime )
RETURNS DateTime
AS
BEGIN
-- Abrunden des Eingabedatums auf 00:00:00 Uhr
DECLARE #dtReturnValue AS DateTime
-- 26.12.9999 SO
IF #in_DateTime >= CAST('99991201' AS datetime)
RETURN CAST('99991226' AS datetime);
-- #dtReturnValue is now last day of month
SET #dtReturnValue = DATEADD
(
DAY
,-1
,DATEADD
(
MONTH
,1
,CAST(CAST(YEAR(#in_DateTime) AS varchar(4)) + RIGHT('00' + CAST(MONTH(#in_DateTime) AS varchar(2)), 2) + '01' AS datetime)
)
)
;
-- SET DATEFIRST 1 -- Monday - Super easy !
-- SET DATEFIRST != 1 - PHUK THIS !
SET #dtReturnValue = DATEADD
(
day
,
-
(
(
-- DATEPART(WEEKDAY, #lastDayofMonth) -- with SET DATEFIRST 1
DATEPART(WEEKDAY, #dtReturnValue) + ##DATEFIRST - 2 % 7 + 1
)
%7
)
, #dtReturnValue
);
RETURN #dtReturnValue;
END
GO
select next_day(last_day(sysdate)-7, 'Sunday') from dual
How can I calculate the number of work days between two dates in SQL Server?
Monday to Friday and it must be T-SQL.
For workdays, Monday to Friday, you can do it with a single SELECT, like this:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2008/10/01'
SET #EndDate = '2008/10/31'
SELECT
(DATEDIFF(dd, #StartDate, #EndDate) + 1)
-(DATEDIFF(wk, #StartDate, #EndDate) * 2)
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)
If you want to include holidays, you have to work it out a bit...
In Calculating Work Days you can find a good article about this subject, but as you can see it is not that advanced.
--Changing current database to the Master database allows function to be shared by everyone.
USE MASTER
GO
--If the function already exists, drop it.
IF EXISTS
(
SELECT *
FROM dbo.SYSOBJECTS
WHERE ID = OBJECT_ID(N'[dbo].[fn_WorkDays]')
AND XType IN (N'FN', N'IF', N'TF')
)
DROP FUNCTION [dbo].[fn_WorkDays]
GO
CREATE FUNCTION dbo.fn_WorkDays
--Presets
--Define the input parameters (OK if reversed by mistake).
(
#StartDate DATETIME,
#EndDate DATETIME = NULL --#EndDate replaced by #StartDate when DEFAULTed
)
--Define the output data type.
RETURNS INT
AS
--Calculate the RETURN of the function.
BEGIN
--Declare local variables
--Temporarily holds #EndDate during date reversal.
DECLARE #Swap DATETIME
--If the Start Date is null, return a NULL and exit.
IF #StartDate IS NULL
RETURN NULL
--If the End Date is null, populate with Start Date value so will have two dates (required by DATEDIFF below).
IF #EndDate IS NULL
SELECT #EndDate = #StartDate
--Strip the time element from both dates (just to be safe) by converting to whole days and back to a date.
--Usually faster than CONVERT.
--0 is a date (01/01/1900 00:00:00.000)
SELECT #StartDate = DATEADD(dd,DATEDIFF(dd,0,#StartDate), 0),
#EndDate = DATEADD(dd,DATEDIFF(dd,0,#EndDate) , 0)
--If the inputs are in the wrong order, reverse them.
IF #StartDate > #EndDate
SELECT #Swap = #EndDate,
#EndDate = #StartDate,
#StartDate = #Swap
--Calculate and return the number of workdays using the input parameters.
--This is the meat of the function.
--This is really just one formula with a couple of parts that are listed on separate lines for documentation purposes.
RETURN (
SELECT
--Start with total number of days including weekends
(DATEDIFF(dd,#StartDate, #EndDate)+1)
--Subtact 2 days for each full weekend
-(DATEDIFF(wk,#StartDate, #EndDate)*2)
--If StartDate is a Sunday, Subtract 1
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday'
THEN 1
ELSE 0
END)
--If EndDate is a Saturday, Subtract 1
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday'
THEN 1
ELSE 0
END)
)
END
GO
If you need to use a custom calendar, you might need to add some checks and some parameters. Hopefully it will provide a good starting point.
All Credit to Bogdan Maxim & Peter Mortensen. This is their post, I just added holidays to the function (This assumes you have a table "tblHolidays" with a datetime field "HolDate".
--Changing current database to the Master database allows function to be shared by everyone.
USE MASTER
GO
--If the function already exists, drop it.
IF EXISTS
(
SELECT *
FROM dbo.SYSOBJECTS
WHERE ID = OBJECT_ID(N'[dbo].[fn_WorkDays]')
AND XType IN (N'FN', N'IF', N'TF')
)
DROP FUNCTION [dbo].[fn_WorkDays]
GO
CREATE FUNCTION dbo.fn_WorkDays
--Presets
--Define the input parameters (OK if reversed by mistake).
(
#StartDate DATETIME,
#EndDate DATETIME = NULL --#EndDate replaced by #StartDate when DEFAULTed
)
--Define the output data type.
RETURNS INT
AS
--Calculate the RETURN of the function.
BEGIN
--Declare local variables
--Temporarily holds #EndDate during date reversal.
DECLARE #Swap DATETIME
--If the Start Date is null, return a NULL and exit.
IF #StartDate IS NULL
RETURN NULL
--If the End Date is null, populate with Start Date value so will have two dates (required by DATEDIFF below).
IF #EndDate IS NULL
SELECT #EndDate = #StartDate
--Strip the time element from both dates (just to be safe) by converting to whole days and back to a date.
--Usually faster than CONVERT.
--0 is a date (01/01/1900 00:00:00.000)
SELECT #StartDate = DATEADD(dd,DATEDIFF(dd,0,#StartDate), 0),
#EndDate = DATEADD(dd,DATEDIFF(dd,0,#EndDate) , 0)
--If the inputs are in the wrong order, reverse them.
IF #StartDate > #EndDate
SELECT #Swap = #EndDate,
#EndDate = #StartDate,
#StartDate = #Swap
--Calculate and return the number of workdays using the input parameters.
--This is the meat of the function.
--This is really just one formula with a couple of parts that are listed on separate lines for documentation purposes.
RETURN (
SELECT
--Start with total number of days including weekends
(DATEDIFF(dd,#StartDate, #EndDate)+1)
--Subtact 2 days for each full weekend
-(DATEDIFF(wk,#StartDate, #EndDate)*2)
--If StartDate is a Sunday, Subtract 1
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday'
THEN 1
ELSE 0
END)
--If EndDate is a Saturday, Subtract 1
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday'
THEN 1
ELSE 0
END)
--Subtract all holidays
-(Select Count(*) from [DB04\DB04].[Gateway].[dbo].[tblHolidays]
where [HolDate] between #StartDate and #EndDate )
)
END
GO
-- Test Script
/*
declare #EndDate datetime= dateadd(m,2,getdate())
print #EndDate
select [Master].[dbo].[fn_WorkDays] (getdate(), #EndDate)
*/
My version of the accepted answer as a function using DATEPART, so I don't have to do a string comparison on the line with
DATENAME(dw, #StartDate) = 'Sunday'
Anyway, here's my business datediff function
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION BDATEDIFF
(
#startdate as DATETIME,
#enddate as DATETIME
)
RETURNS INT
AS
BEGIN
DECLARE #res int
SET #res = (DATEDIFF(dd, #startdate, #enddate) + 1)
-(DATEDIFF(wk, #startdate, #enddate) * 2)
-(CASE WHEN DATEPART(dw, #startdate) = 1 THEN 1 ELSE 0 END)
-(CASE WHEN DATEPART(dw, #enddate) = 7 THEN 1 ELSE 0 END)
RETURN #res
END
GO
Another approach to calculating working days is to use a WHILE loop which basically iterates through a date range and increment it by 1 whenever days are found to be within Monday – Friday. The complete script for calculating working days using the WHILE loop is shown below:
CREATE FUNCTION [dbo].[fn_GetTotalWorkingDaysUsingLoop]
(#DateFrom DATE,
#DateTo DATE
)
RETURNS INT
AS
BEGIN
DECLARE #TotWorkingDays INT= 0;
WHILE #DateFrom <= #DateTo
BEGIN
IF DATENAME(WEEKDAY, #DateFrom) IN('Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday')
BEGIN
SET #TotWorkingDays = #TotWorkingDays + 1;
END;
SET #DateFrom = DATEADD(DAY, 1, #DateFrom);
END;
RETURN #TotWorkingDays;
END;
GO
Although the WHILE loop option is cleaner and uses less lines of code, it has the potential of being a performance bottleneck in your environment particularly when your date range spans across several years.
You can see more methods on how to calculate work days and hours in this article:
https://www.sqlshack.com/how-to-calculate-work-days-and-hours-in-sql-server/
DECLARE #TotalDays INT,#WorkDays INT
DECLARE #ReducedDayswithEndDate INT
DECLARE #WeekPart INT
DECLARE #DatePart INT
SET #TotalDays= DATEDIFF(day, #StartDate, #EndDate) +1
SELECT #ReducedDayswithEndDate = CASE DATENAME(weekday, #EndDate)
WHEN 'Saturday' THEN 1
WHEN 'Sunday' THEN 2
ELSE 0 END
SET #TotalDays=#TotalDays-#ReducedDayswithEndDate
SET #WeekPart=#TotalDays/7;
SET #DatePart=#TotalDays%7;
SET #WorkDays=(#WeekPart*5)+#DatePart
RETURN #WorkDays
(I'm a few points shy of commenting privileges)
If you decide to forgo the +1 day in CMS's elegant solution, note that if your start date and end date are in the same weekend, you get a negative answer. Ie., 2008/10/26 to 2008/10/26 returns -1.
my rather simplistic solution:
select #Result = (..CMS's answer..)
if (#Result < 0)
select #Result = 0
RETURN #Result
.. which also sets all erroneous posts with start date after end date to zero. Something you may or may not be looking for.
For difference between dates including holidays I went this way:
1) Table with Holidays:
CREATE TABLE [dbo].[Holiday](
[Id] [int] IDENTITY(1,1) NOT NULL,
[Name] [nvarchar](50) NULL,
[Date] [datetime] NOT NULL)
2) I had my plannings Table like this and wanted to fill column Work_Days which was empty:
CREATE TABLE [dbo].[Plan_Phase](
[Id] [int] IDENTITY(1,1) NOT NULL,
[Id_Plan] [int] NOT NULL,
[Id_Phase] [int] NOT NULL,
[Start_Date] [datetime] NULL,
[End_Date] [datetime] NULL,
[Work_Days] [int] NULL)
3) So in order to get "Work_Days" to later fill in my column just had to:
SELECT Start_Date, End_Date,
(DATEDIFF(dd, Start_Date, End_Date) + 1)
-(DATEDIFF(wk, Start_Date, End_Date) * 2)
-(SELECT COUNT(*) From Holiday Where Date >= Start_Date AND Date <= End_Date)
-(CASE WHEN DATENAME(dw, Start_Date) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, End_Date) = 'Saturday' THEN 1 ELSE 0 END)
-(CASE WHEN (SELECT COUNT(*) From Holiday Where Start_Date = Date) > 0 THEN 1 ELSE 0 END)
-(CASE WHEN (SELECT COUNT(*) From Holiday Where End_Date = Date) > 0 THEN 1 ELSE 0 END) AS Work_Days
from Plan_Phase
Hope that I could help.
Cheers
Here is a version that works well (I think). Holiday table contains Holiday_date columns that contains holidays your company observe.
DECLARE #RAWDAYS INT
SELECT #RAWDAYS = DATEDIFF(day, #StartDate, #EndDate )--+1
-( 2 * DATEDIFF( week, #StartDate, #EndDate ) )
+ CASE WHEN DATENAME(dw, #StartDate) = 'Saturday' THEN 1 ELSE 0 END
- CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END
SELECT #RAWDAYS - COUNT(*)
FROM HOLIDAY NumberOfBusinessDays
WHERE [Holiday_Date] BETWEEN #StartDate+1 AND #EndDate
I know this is an old question but I needed a formula for workdays excluding the start date since I have several items and need the days to accumulate correctly.
None of the non-iterative answers worked for me.
I used a defintion like
Number of times midnight to monday, tuesday, wednesday, thursday and friday is passed
(others might count midnight to saturday instead of monday)
I ended up with this formula
SELECT DATEDIFF(day, #StartDate, #EndDate) /* all midnights passed */
- DATEDIFF(week, #StartDate, #EndDate) /* remove sunday midnights */
- DATEDIFF(week, DATEADD(day, 1, #StartDate), DATEADD(day, 1, #EndDate)) /* remove saturday midnights */
This is basically CMS's answer without the reliance on a particular language setting. And since we're shooting for generic, that means it should work for all ##datefirst settings as well.
datediff(day, <start>, <end>) + 1 - datediff(week, <start>, <end>) * 2
/* if start is a Sunday, adjust by -1 */
+ case when datepart(weekday, <start>) = 8 - ##datefirst then -1 else 0 end
/* if end is a Saturday, adjust by -1 */
+ case when datepart(weekday, <end>) = (13 - ##datefirst) % 7 + 1 then -1 else 0 end
datediff(week, ...) always uses a Saturday-to-Sunday boundary for weeks, so that expression is deterministic and doesn't need to be modified (as long as our definition of weekdays is consistently Monday through Friday.) Day numbering does vary according to the ##datefirst setting and the modified calculations handle this correction with the small complication of some modular arithmetic.
A cleaner way to deal with the Saturday/Sunday thing is to translate the dates prior to extracting a day of week value. After shifting, the values will be back in line with a fixed (and probably more familiar) numbering that starts with 1 on Sunday and ends with 7 on Saturday.
datediff(day, <start>, <end>) + 1 - datediff(week, <start>, <end>) * 2
+ case when datepart(weekday, dateadd(day, ##datefirst, <start>)) = 1 then -1 else 0 end
+ case when datepart(weekday, dateadd(day, ##datefirst, <end>)) = 7 then -1 else 0 end
I've tracked this form of the solution back at least as far as 2002 and an Itzik Ben-Gan article. (https://technet.microsoft.com/en-us/library/aa175781(v=sql.80).aspx) Though it needed a small tweak since newer date types don't allow date arithmetic, it is otherwise identical.
EDIT:
I added back the +1 that had somehow been left off. It's also worth noting that this method always counts the start and end days. It also assumes that the end date is on or after the start date.
None of the functions above work for the same week or deal with holidays. I wrote this:
create FUNCTION [dbo].[ShiftHolidayToWorkday](#date date)
RETURNS date
AS
BEGIN
IF DATENAME( dw, #Date ) = 'Saturday'
SET #Date = DATEADD(day, - 1, #Date)
ELSE IF DATENAME( dw, #Date ) = 'Sunday'
SET #Date = DATEADD(day, 1, #Date)
RETURN #date
END
GO
create FUNCTION [dbo].[GetHoliday](#date date)
RETURNS varchar(50)
AS
BEGIN
declare #s varchar(50)
SELECT #s = CASE
WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year] ) + '-01-01') = #date THEN 'New Year'
WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]+1) + '-01-01') = #date THEN 'New Year'
WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year] ) + '-07-04') = #date THEN 'Independence Day'
WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year] ) + '-12-25') = #date THEN 'Christmas Day'
--WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]) + '-12-31') = #date THEN 'New Years Eve'
--WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]) + '-11-11') = #date THEN 'Veteran''s Day'
WHEN [Month] = 1 AND [DayOfMonth] BETWEEN 15 AND 21 AND [DayName] = 'Monday' THEN 'Martin Luther King Day'
WHEN [Month] = 5 AND [DayOfMonth] >= 25 AND [DayName] = 'Monday' THEN 'Memorial Day'
WHEN [Month] = 9 AND [DayOfMonth] <= 7 AND [DayName] = 'Monday' THEN 'Labor Day'
WHEN [Month] = 11 AND [DayOfMonth] BETWEEN 22 AND 28 AND [DayName] = 'Thursday' THEN 'Thanksgiving Day'
WHEN [Month] = 11 AND [DayOfMonth] BETWEEN 23 AND 29 AND [DayName] = 'Friday' THEN 'Day After Thanksgiving'
ELSE NULL END
FROM (
SELECT
[Year] = YEAR(#date),
[Month] = MONTH(#date),
[DayOfMonth] = DAY(#date),
[DayName] = DATENAME(weekday,#date)
) c
RETURN #s
END
GO
create FUNCTION [dbo].GetHolidays(#year int)
RETURNS TABLE
AS
RETURN (
select dt, dbo.GetHoliday(dt) as Holiday
from (
select dateadd(day, number, convert(varchar,#year) + '-01-01') dt
from master..spt_values
where type='p'
) d
where year(dt) = #year and dbo.GetHoliday(dt) is not null
)
create proc UpdateHolidaysTable
as
if not exists(select TABLE_NAME from INFORMATION_SCHEMA.TABLES where TABLE_NAME = 'Holidays')
create table Holidays(dt date primary key clustered, Holiday varchar(50))
declare #year int
set #year = 1990
while #year < year(GetDate()) + 20
begin
insert into Holidays(dt, Holiday)
select a.dt, a.Holiday
from dbo.GetHolidays(#year) a
left join Holidays b on b.dt = a.dt
where b.dt is null
set #year = #year + 1
end
create FUNCTION [dbo].[GetWorkDays](#StartDate DATE = NULL, #EndDate DATE = NULL)
RETURNS INT
AS
BEGIN
IF #StartDate IS NULL OR #EndDate IS NULL
RETURN 0
IF #StartDate >= #EndDate
RETURN 0
DECLARE #Days int
SET #Days = 0
IF year(#StartDate) * 100 + datepart(week, #StartDate) = year(#EndDate) * 100 + datepart(week, #EndDate)
--same week
select #Days = (DATEDIFF(dd, #StartDate, #EndDate))
- (CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
- (CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)
- (select count(*) from Holidays where dt between #StartDate and #EndDate)
ELSE
--diff weeks
select #Days = (DATEDIFF(dd, #StartDate, #EndDate) + 1)
- (DATEDIFF(wk, #StartDate, #EndDate) * 2)
- (CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
- (CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)
- (select count(*) from Holidays where dt between #StartDate and #EndDate)
RETURN #Days
END
Using a date table:
DECLARE
#StartDate date = '2014-01-01',
#EndDate date = '2014-01-31';
SELECT
COUNT(*) As NumberOfWeekDays
FROM dbo.Calendar
WHERE CalendarDate BETWEEN #StartDate AND #EndDate
AND IsWorkDay = 1;
If you don't have that, you can use a numbers table:
DECLARE
#StartDate datetime = '2014-01-01',
#EndDate datetime = '2014-01-31';
SELECT
SUM(CASE WHEN DATEPART(dw, DATEADD(dd, Number-1, #StartDate)) BETWEEN 2 AND 6 THEN 1 ELSE 0 END) As NumberOfWeekDays
FROM dbo.Numbers
WHERE Number <= DATEDIFF(dd, #StartDate, #EndDate) + 1 -- Number table starts at 1, we want a 0 base
They should both be fast and it takes out the ambiguity/complexity. The first option is the best but if you don't have a calendar table you can allways create a numbers table with a CTE.
DECLARE #StartDate datetime,#EndDate datetime
select #StartDate='3/2/2010', #EndDate='3/7/2010'
DECLARE #TotalDays INT,#WorkDays INT
DECLARE #ReducedDayswithEndDate INT
DECLARE #WeekPart INT
DECLARE #DatePart INT
SET #TotalDays= DATEDIFF(day, #StartDate, #EndDate) +1
SELECT #ReducedDayswithEndDate = CASE DATENAME(weekday, #EndDate)
WHEN 'Saturday' THEN 1
WHEN 'Sunday' THEN 2
ELSE 0 END
SET #TotalDays=#TotalDays-#ReducedDayswithEndDate
SET #WeekPart=#TotalDays/7;
SET #DatePart=#TotalDays%7;
SET #WorkDays=(#WeekPart*5)+#DatePart
SELECT #WorkDays
CREATE FUNCTION x
(
#StartDate DATETIME,
#EndDate DATETIME
)
RETURNS INT
AS
BEGIN
DECLARE #Teller INT
SET #StartDate = DATEADD(dd,1,#StartDate)
SET #Teller = 0
IF DATEDIFF(dd,#StartDate,#EndDate) <= 0
BEGIN
SET #Teller = 0
END
ELSE
BEGIN
WHILE
DATEDIFF(dd,#StartDate,#EndDate) >= 0
BEGIN
IF DATEPART(dw,#StartDate) < 6
BEGIN
SET #Teller = #Teller + 1
END
SET #StartDate = DATEADD(dd,1,#StartDate)
END
END
RETURN #Teller
END
I took the various examples here, but in my particular situation we have a #PromisedDate for delivery and a #ReceivedDate for the actual receipt of the item. When an item was received before the "PromisedDate" the calculations were not totaling correctly unless I ordered the dates passed into the function by calendar order. Not wanting to check the dates every time, I changed the function to handle this for me.
Create FUNCTION [dbo].[fnGetBusinessDays]
(
#PromiseDate date,
#ReceivedDate date
)
RETURNS integer
AS
BEGIN
DECLARE #days integer
SELECT #days =
Case when #PromiseDate > #ReceivedDate Then
DATEDIFF(d,#PromiseDate,#ReceivedDate) +
ABS(DATEDIFF(wk,#PromiseDate,#ReceivedDate)) * 2 +
CASE
WHEN DATENAME(dw, #PromiseDate) <> 'Saturday' AND DATENAME(dw, #ReceivedDate) = 'Saturday' THEN 1
WHEN DATENAME(dw, #PromiseDate) = 'Saturday' AND DATENAME(dw, #ReceivedDate) <> 'Saturday' THEN -1
ELSE 0
END +
(Select COUNT(*) FROM CompanyHolidays
WHERE HolidayDate BETWEEN #ReceivedDate AND #PromiseDate
AND DATENAME(dw, HolidayDate) <> 'Saturday' AND DATENAME(dw, HolidayDate) <> 'Sunday')
Else
DATEDIFF(d,#PromiseDate,#ReceivedDate) -
ABS(DATEDIFF(wk,#PromiseDate,#ReceivedDate)) * 2 -
CASE
WHEN DATENAME(dw, #PromiseDate) <> 'Saturday' AND DATENAME(dw, #ReceivedDate) = 'Saturday' THEN 1
WHEN DATENAME(dw, #PromiseDate) = 'Saturday' AND DATENAME(dw, #ReceivedDate) <> 'Saturday' THEN -1
ELSE 0
END -
(Select COUNT(*) FROM CompanyHolidays
WHERE HolidayDate BETWEEN #PromiseDate and #ReceivedDate
AND DATENAME(dw, HolidayDate) <> 'Saturday' AND DATENAME(dw, HolidayDate) <> 'Sunday')
End
RETURN (#days)
END
If you need to add work days to a given date, you can create a function that depends on a calendar table, described below:
CREATE TABLE Calendar
(
dt SMALLDATETIME PRIMARY KEY,
IsWorkDay BIT
);
--fill the rows with normal days, weekends and holidays.
create function AddWorkingDays (#initialDate smalldatetime, #numberOfDays int)
returns smalldatetime as
begin
declare #result smalldatetime
set #result =
(
select t.dt from
(
select dt, ROW_NUMBER() over (order by dt) as daysAhead from calendar
where dt > #initialDate
and IsWorkDay = 1
) t
where t.daysAhead = #numberOfDays
)
return #result
end
As with DATEDIFF, I do not consider the end date to be part of the interval.
The number of (for example) Sundays between #StartDate and #EndDate is the number of Sundays between an "initial" Monday and the #EndDate minus the number of Sundays between this "initial" Monday and the #StartDate. Knowing this, we can calculate the number of workdays as follows:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2018/01/01'
SET #EndDate = '2019/01/01'
SELECT DATEDIFF(Day, #StartDate, #EndDate) -- Total Days
- (DATEDIFF(Day, 0, #EndDate)/7 - DATEDIFF(Day, 0, #StartDate)/7) -- Sundays
- (DATEDIFF(Day, -1, #EndDate)/7 - DATEDIFF(Day, -1, #StartDate)/7) -- Saturdays
Best regards!
I borrowed some ideas from others to create my solution. I use inline code to ignore weekends and U.S. federal holidays. In my environment, EndDate may be null, but it will never precede StartDate.
CREATE FUNCTION dbo.ufn_CalculateBusinessDays(
#StartDate DATE,
#EndDate DATE = NULL)
RETURNS INT
AS
BEGIN
DECLARE #TotalBusinessDays INT = 0;
DECLARE #TestDate DATE = #StartDate;
IF #EndDate IS NULL
RETURN NULL;
WHILE #TestDate < #EndDate
BEGIN
DECLARE #Month INT = DATEPART(MM, #TestDate);
DECLARE #Day INT = DATEPART(DD, #TestDate);
DECLARE #DayOfWeek INT = DATEPART(WEEKDAY, #TestDate) - 1; --Monday = 1, Tuesday = 2, etc.
DECLARE #DayOccurrence INT = (#Day - 1) / 7 + 1; --Nth day of month (3rd Monday, for example)
--Increment business day counter if not a weekend or holiday
SELECT #TotalBusinessDays += (
SELECT CASE
--Saturday OR Sunday
WHEN #DayOfWeek IN (6,7) THEN 0
--New Year's Day
WHEN #Month = 1 AND #Day = 1 THEN 0
--MLK Jr. Day
WHEN #Month = 1 AND #DayOfWeek = 1 AND #DayOccurrence = 3 THEN 0
--G. Washington's Birthday
WHEN #Month = 2 AND #DayOfWeek = 1 AND #DayOccurrence = 3 THEN 0
--Memorial Day
WHEN #Month = 5 AND #DayOfWeek = 1 AND #Day BETWEEN 25 AND 31 THEN 0
--Independence Day
WHEN #Month = 7 AND #Day = 4 THEN 0
--Labor Day
WHEN #Month = 9 AND #DayOfWeek = 1 AND #DayOccurrence = 1 THEN 0
--Columbus Day
WHEN #Month = 10 AND #DayOfWeek = 1 AND #DayOccurrence = 2 THEN 0
--Veterans Day
WHEN #Month = 11 AND #Day = 11 THEN 0
--Thanksgiving
WHEN #Month = 11 AND #DayOfWeek = 4 AND #DayOccurrence = 4 THEN 0
--Christmas
WHEN #Month = 12 AND #Day = 25 THEN 0
ELSE 1
END AS Result);
SET #TestDate = DATEADD(dd, 1, #TestDate);
END
RETURN #TotalBusinessDays;
END
That's working for me, in my country on Saturday and Sunday are non-working days.
For me is important the time of #StartDate and #EndDate.
CREATE FUNCTION [dbo].[fnGetCountWorkingBusinessDays]
(
#StartDate as DATETIME,
#EndDate as DATETIME
)
RETURNS INT
AS
BEGIN
DECLARE #res int
SET #StartDate = CASE
WHEN DATENAME(dw, #StartDate) = 'Saturday' THEN DATEADD(dd, 2, DATEDIFF(dd, 0, #StartDate))
WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN DATEADD(dd, 1, DATEDIFF(dd, 0, #StartDate))
ELSE #StartDate END
SET #EndDate = CASE
WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN DATEADD(dd, 0, DATEDIFF(dd, 0, #EndDate))
WHEN DATENAME(dw, #EndDate) = 'Sunday' THEN DATEADD(dd, -1, DATEDIFF(dd, 0, #EndDate))
ELSE #EndDate END
SET #res =
(DATEDIFF(hour, #StartDate, #EndDate) / 24)
- (DATEDIFF(wk, #StartDate, #EndDate) * 2)
SET #res = CASE WHEN #res < 0 THEN 0 ELSE #res END
RETURN #res
END
GO
Create function like:
CREATE FUNCTION dbo.fn_WorkDays(#StartDate DATETIME, #EndDate DATETIME= NULL )
RETURNS INT
AS
BEGIN
DECLARE #Days int
SET #Days = 0
IF #EndDate = NULL
SET #EndDate = EOMONTH(#StartDate) --last date of the month
WHILE DATEDIFF(dd,#StartDate,#EndDate) >= 0
BEGIN
IF DATENAME(dw, #StartDate) <> 'Saturday'
and DATENAME(dw, #StartDate) <> 'Sunday'
and Not ((Day(#StartDate) = 1 And Month(#StartDate) = 1)) --New Year's Day.
and Not ((Day(#StartDate) = 4 And Month(#StartDate) = 7)) --Independence Day.
BEGIN
SET #Days = #Days + 1
END
SET #StartDate = DATEADD(dd,1,#StartDate)
END
RETURN #Days
END
You can call the function like:
select dbo.fn_WorkDays('1/1/2016', '9/25/2016')
Or like:
select dbo.fn_WorkDays(StartDate, EndDate)
from table1
Create Function dbo.DateDiff_WeekDays
(
#StartDate DateTime,
#EndDate DateTime
)
Returns Int
As
Begin
Declare #Result Int = 0
While #StartDate <= #EndDate
Begin
If DateName(DW, #StartDate) not in ('Saturday','Sunday')
Begin
Set #Result = #Result +1
End
Set #StartDate = DateAdd(Day, +1, #StartDate)
End
Return #Result
End
I found the below TSQL a fairly elegant solution (I don't have permissions to run functions). I found the DATEDIFF ignores DATEFIRST and I wanted my first day of the week to be a Monday. I also wanted the first working day to be set a zero and if it falls on a weekend Monday will be a zero. This may help someone who has a slightly different requirement :)
It does not handle bank holidays
SET DATEFIRST 1
SELECT
,(DATEDIFF(DD, [StartDate], [EndDate]))
-(DATEDIFF(wk, [StartDate], [EndDate]))
-(DATEDIFF(wk, DATEADD(dd,-##DATEFIRST,[StartDate]), DATEADD(dd,-##DATEFIRST,[EndDate]))) AS [WorkingDays]
FROM /*Your Table*/
One approach is to 'walk the dates' from start to finish in conjunction with a case expression which checks if the day is not a Saturday or a Sunday and flagging it(1 for weekday, 0 for weekend). And in the end just sum flags(it would be equal to the count of 1-flags as the other flag is 0) to give you the number of weekdays.
You can use a GetNums(startNumber,endNumber) type of utility function which generates a series of numbers for 'looping' from start date to end date. Refer http://tsql.solidq.com/SourceCodes/GetNums.txt for an implementation. The logic can also be extended to cater for holidays(say if you have a holidays table)
declare #date1 as datetime = '19900101'
declare #date2 as datetime = '19900120'
select sum(case when DATENAME(DW,currentDate) not in ('Saturday', 'Sunday') then 1 else 0 end) as noOfWorkDays
from dbo.GetNums(0,DATEDIFF(day,#date1, #date2)-1) as Num
cross apply (select DATEADD(day,n,#date1)) as Dates(currentDate)