Related
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2012/11/01'
SET #EndDate = '2012/11/05'
SELECT
(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)
Help to add # of holiday days to this solution? If there are any Holiday days between your start and end dates. Assuming I do have a table called BICalendar with a column isHoliday and value = 1 if that date is a holiday.
To directly answer the question:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2012/11/01'
SET #EndDate = '2012/11/05'
SELECT (DATEDIFF(wk, #StartDate, #EndDate) * 2) -- To calculate the number of weekend dates between the two input dates
+ (CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END) --If the Start date is Sunday, then add one to adjust for non-working days
+ (CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END) --If the End date is Saturday, then add one to adjust for non-working days
- (SELECT COUNT(*) --Minus the number of holiday dates between the two dates
FROM BICalendar
WHERE [Date] BETWEEN #STartDate AND #EndDate
AND isHoliday = 1);
I suspect what you're trying to do is calculate the number of working days between the two dates? If so, the below solution should also work:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2021-05-01'
SET #EndDate = '2021-05-15'
SELECT DATEDIFF(dd, #StartDate, #EndDate) --To calculate the number of days between the two dates
- (DATEDIFF(wk, #StartDate, #EndDate) * 2) -- To remove the weekends between the two dates
+ (CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END) --If the Start date is Sunday, then add one to adjust for non-working days
+ (CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END) --If the End date is Saturday, then add one to adjust for non-working days
- (SELECT COUNT(*) --Minus the number of holiday dates between the two dates
FROM BICalendar
WHERE [Date] BETWEEN #STartDate AND #EndDate
AND isHoliday = 1);
I'm working on SQL-Server 2012 and have the following code example to get workdays between two dates
DECLARE #StartDate AS date
DECLARE #EndDate AS date
SET #StartDate = '2019/02/18' -- this is a monday
SET #EndDate = '2019/02/23' -- this is a saturday
SELECT
DATEDIFF(DD, #StartDate, #EndDate)
- (DATEDIFF(WK, #StartDate,#EndDate) * 2)
- CASE WHEN DATEPART(DW, #StartDate) = 1 THEN 1 ELSE 0 END
- CASE WHEN DATEPART(DW, #EndDate) = 1 THEN 1 ELSE 0 END
the result is 4, which is correct...
But If I put 2019/02/24 (sunday) for the EndDate I'm getting 3... ????
I'm getting crazy here...
You're validating your Enddate to be Sunday instead of Saturday. I had a similar function available that is independent of date settings.
SELECT ISNULL((((DATEDIFF(dd,#StartDate,#EndDate)) --Start with total number of days including weekends
- (DATEDIFF(wk,#StartDate,#EndDate)*2) --Subtact 2 days for each full weekend
- (1-SIGN(DATEDIFF(dd,6,#StartDate)%7)) --If StartDate is a Sunday, Subtract 1
- (1-SIGN(DATEDIFF(dd,5,#EndDate) %7)))) , 0) --If StartDate is a Saturday, Subtract 1
WHERE #StartDate <= #EndDate
The answer by #Luis Cazares works great except that when the start date or end date fall on a Saturday or Sunday, then the number is 1 off. Also, holidays are not accounted for (if that falls within the discussion).
Here's my way to construct a function.
CREATE FUNCTION [dbo].[getWorkDays] (#startDate date, #endDate date)
RETURNS int
AS
BEGIN
DECLARE #daysct int = 0;
DECLARE #currentDate date = #startDate;
IF #endDate>#startDate
GOTO COUNTIT;
ELSE
GOTO ZERO;
COUNTIT:
BEGIN
--Pre-process: if the startDate is holiday or weekend, then step in until the next business day
WHILE datepart(weekday, #currentDate) in (1, 7) OR EXISTS (SELECT holiday_date FROM dbo.holidays WHERE holiday_date = #currentDate) --if weekend or holiday
BEGIN
SET #currentDate = dateadd(day, 1, #currentDate);
SET #daysct=1;
END
WHILE #currentDate <= #endDate
BEGIN
IF #currentDate=#endDate
BREAK;
ELSE
IF datepart(weekday, #currentDate) not in (1, 7) AND NOT EXISTS (SELECT holiday_date FROM dbo.holidays WHERE holiday_date = #currentDate) --if not weekend and not holiday
BEGIN
SET #daysct=#daysct + 1;
END
SET #currentDate=dateadd(day, 1, #currentDate);
END
--Post-process: if the end date DOES NOT fall on a weekend or holiday, then #daysct is systematically 1 below. Add 1 to straighten it.
IF datepart(weekday, #endDate) in (1, 7) OR EXISTS (SELECT holiday_date FROM dbo.holidays WHERE holiday_date = #endDate)
SET #daysct = #daysct - 1;
RETURN(#daysct);
END
ZERO:
BEGIN
SET #daysct=0;
RETURN(#daysct);
END
END;
It's not the most efficient code but it's naturally the process how I would hand-count it. The catch is you need to declare a variable or constant, otherwise it will significantly slow down your queries.
DECLARE #days_worked int = dbo.getWorkDays(dateadd(day, 1, eomonth(getdate() AT TIME ZONE 'Pacific Standard Time', -1)),getdate() AT TIME ZONE 'Pacific Standard Time');
Also, before you can use it, make sure you have a table for the holidays. My table is constructed this way.
enter image description here
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2019/02/18'
SET #EndDate = '2019/02/23'
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)
I have a table listing people along with their date of birth (currently a nvarchar(25))
How can I convert that to a date, and then calculate their age in years?
My data looks as follows
ID Name DOB
1 John 1992-01-09 00:00:00
2 Sally 1959-05-20 00:00:00
I would like to see:
ID Name AGE DOB
1 John 17 1992-01-09 00:00:00
2 Sally 50 1959-05-20 00:00:00
There are issues with leap year/days and the following method, see the update below:
try this:
DECLARE #dob datetime
SET #dob='1992-01-09 00:00:00'
SELECT DATEDIFF(hour,#dob,GETDATE())/8766.0 AS AgeYearsDecimal
,CONVERT(int,ROUND(DATEDIFF(hour,#dob,GETDATE())/8766.0,0)) AS AgeYearsIntRound
,DATEDIFF(hour,#dob,GETDATE())/8766 AS AgeYearsIntTrunc
OUTPUT:
AgeYearsDecimal AgeYearsIntRound AgeYearsIntTrunc
--------------------------------------- ---------------- ----------------
17.767054 18 17
(1 row(s) affected)
UPDATE here are some more accurate methods:
BEST METHOD FOR YEARS IN INT
DECLARE #Now datetime, #Dob datetime
SELECT #Now='1990-05-05', #Dob='1980-05-05' --results in 10
--SELECT #Now='1990-05-04', #Dob='1980-05-05' --results in 9
--SELECT #Now='1989-05-06', #Dob='1980-05-05' --results in 9
--SELECT #Now='1990-05-06', #Dob='1980-05-05' --results in 10
--SELECT #Now='1990-12-06', #Dob='1980-05-05' --results in 10
--SELECT #Now='1991-05-04', #Dob='1980-05-05' --results in 10
SELECT
(CONVERT(int,CONVERT(char(8),#Now,112))-CONVERT(char(8),#Dob,112))/10000 AS AgeIntYears
you can change the above 10000 to 10000.0 and get decimals, but it will not be as accurate as the method below.
BEST METHOD FOR YEARS IN DECIMAL
DECLARE #Now datetime, #Dob datetime
SELECT #Now='1990-05-05', #Dob='1980-05-05' --results in 10.000000000000
--SELECT #Now='1990-05-04', #Dob='1980-05-05' --results in 9.997260273973
--SELECT #Now='1989-05-06', #Dob='1980-05-05' --results in 9.002739726027
--SELECT #Now='1990-05-06', #Dob='1980-05-05' --results in 10.002739726027
--SELECT #Now='1990-12-06', #Dob='1980-05-05' --results in 10.589041095890
--SELECT #Now='1991-05-04', #Dob='1980-05-05' --results in 10.997260273973
SELECT 1.0* DateDiff(yy,#Dob,#Now)
+CASE
WHEN #Now >= DATEFROMPARTS(DATEPART(yyyy,#Now),DATEPART(m,#Dob),DATEPART(d,#Dob)) THEN --birthday has happened for the #now year, so add some portion onto the year difference
( 1.0 --force automatic conversions from int to decimal
* DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),DATEPART(m,#Dob),DATEPART(d,#Dob)),#Now) --number of days difference between the #Now year birthday and the #Now day
/ DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),1,1),DATEFROMPARTS(DATEPART(yyyy,#Now)+1,1,1)) --number of days in the #Now year
)
ELSE --birthday has not been reached for the last year, so remove some portion of the year difference
-1 --remove this fractional difference onto the age
* ( -1.0 --force automatic conversions from int to decimal
* DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),DATEPART(m,#Dob),DATEPART(d,#Dob)),#Now) --number of days difference between the #Now year birthday and the #Now day
/ DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),1,1),DATEFROMPARTS(DATEPART(yyyy,#Now)+1,1,1)) --number of days in the #Now year
)
END AS AgeYearsDecimal
Gotta throw this one out there. If you convert the date using the 112 style (yyyymmdd) to a number you can use a calculation like this...
(yyyyMMdd - yyyyMMdd) / 10000 = difference in full years
declare #as_of datetime, #bday datetime;
select #as_of = '2009/10/15', #bday = '1980/4/20'
select
Convert(Char(8),#as_of,112),
Convert(Char(8),#bday,112),
0 + Convert(Char(8),#as_of,112) - Convert(Char(8),#bday,112),
(0 + Convert(Char(8),#as_of,112) - Convert(Char(8),#bday,112)) / 10000
output
20091015 19800420 290595 29
I have used this query in our production code for nearly 10 years:
SELECT FLOOR((CAST (GetDate() AS INTEGER) - CAST(Date_of_birth AS INTEGER)) / 365.25) AS Age
You need to consider the way the datediff command rounds.
SELECT CASE WHEN dateadd(year, datediff (year, DOB, getdate()), DOB) > getdate()
THEN datediff(year, DOB, getdate()) - 1
ELSE datediff(year, DOB, getdate())
END as Age
FROM <table>
Which I adapted from here.
Note that it will consider 28th February as the birthday of a leapling for non-leap years e.g. a person born on 29 Feb 2020 will be considered 1 year old on 28 Feb 2021 instead of 01 Mar 2021.
So many of the above solutions are wrong DateDiff(yy,#Dob, #PassedDate) will not consider the month and day of both dates. Also taking the dart parts and comparing only works if they're properly ordered.
THE FOLLOWING CODE WORKS AND IS VERY SIMPLE:
create function [dbo].[AgeAtDate](
#DOB datetime,
#PassedDate datetime
)
returns int
with SCHEMABINDING
as
begin
declare #iMonthDayDob int
declare #iMonthDayPassedDate int
select #iMonthDayDob = CAST(datepart (mm,#DOB) * 100 + datepart (dd,#DOB) AS int)
select #iMonthDayPassedDate = CAST(datepart (mm,#PassedDate) * 100 + datepart (dd,#PassedDate) AS int)
return DateDiff(yy,#DOB, #PassedDate)
- CASE WHEN #iMonthDayDob <= #iMonthDayPassedDate
THEN 0
ELSE 1
END
End
EDIT: THIS ANSWER IS INCORRECT. I leave it in here as a warning to anyone tempted to use dayofyear, with a further edit at the end.
If, like me, you do not want to divide by fractional days or risk rounding/leap year errors, I applaud #Bacon Bits comment in a post above https://stackoverflow.com/a/1572257/489865 where he says:
If we're talking about human ages, you should calculate it the way
humans calculate age. It has nothing to do with how fast the earth
moves and everything to do with the calendar. Every time the same
month and day elapses as the date of birth, you increment age by 1.
This means the following is the most accurate because it mirrors what
humans mean when they say "age".
He then offers:
DATEDIFF(yy, #date, GETDATE()) -
CASE WHEN (MONTH(#date) > MONTH(GETDATE())) OR (MONTH(#date) = MONTH(GETDATE()) AND DAY(#date) > DAY(GETDATE()))
THEN 1 ELSE 0 END
There are several suggestions here involving comparing the month & day (and some get it wrong, failing to allow for the OR as correctly here!). But nobody has offered dayofyear, which seems so simple and much shorter. I offer:
DATEDIFF(year, #date, GETDATE()) -
CASE WHEN DATEPART(dayofyear, #date) > DATEPART(dayofyear, GETDATE()) THEN 1 ELSE 0 END
[Note: Nowhere in SQL BOL/MSDN is what DATEPART(dayofyear, ...) returns actually documented! I understand it to be a number in the range 1--366; most importantly, it does not change by locale as per DATEPART(weekday, ...) & SET DATEFIRST.]
EDIT: Why dayofyear goes wrong: As user #AeroX has commented, if the birth/start date is after February in a non leap year, the age is incremented one day early when the current/end date is a leap year, e.g. '2015-05-26', '2016-05-25' gives an age of 1 when it should still be 0. Comparing the dayofyear in different years is clearly dangerous. So using MONTH() and DAY() is necessary after all.
I believe this is similar to other ones posted here.... but this solution worked for the leap year examples 02/29/1976 to 03/01/2011 and also worked for the case for the first year.. like 07/04/2011 to 07/03/2012 which the last one posted about leap year solution did not work for that first year use case.
SELECT FLOOR(DATEDIFF(DAY, #date1 , #date2) / 365.25)
Found here.
Since there isn't one simple answer that always gives the correct age, here's what I came up with.
SELECT DATEDIFF(YY, DateOfBirth, GETDATE()) -
CASE WHEN RIGHT(CONVERT(VARCHAR(6), GETDATE(), 12), 4) >=
RIGHT(CONVERT(VARCHAR(6), DateOfBirth, 12), 4)
THEN 0 ELSE 1 END AS AGE
This gets the year difference between the birth date and the current date. Then it subtracts a year if the birthdate hasn't passed yet.
Accurate all the time - regardless of leap years or how close to the birthdate.
Best of all - no function.
I've done a lot of thinking and searching about this and I have 3 solutions that
calculate age correctly
are short (mostly)
are (mostly) very understandable.
Here are testing values:
DECLARE #NOW DATETIME = '2013-07-04 23:59:59'
DECLARE #DOB DATETIME = '1986-07-05'
Solution 1: I found this approach in one js library. It's my favourite.
DATEDIFF(YY, #DOB, #NOW) -
CASE WHEN DATEADD(YY, DATEDIFF(YY, #DOB, #NOW), #DOB) > #NOW THEN 1 ELSE 0 END
It's actually adding difference in years to DOB and if it is bigger than current date then subtracts one year. Simple right? The only thing is that difference in years is duplicated here.
But if you don't need to use it inline you can write it like this:
DECLARE #AGE INT = DATEDIFF(YY, #DOB, #NOW)
IF DATEADD(YY, #AGE, #DOB) > #NOW
SET #AGE = #AGE - 1
Solution 2: This one I originally copied from #bacon-bits. It's the easiest to understand but a bit long.
DATEDIFF(YY, #DOB, #NOW) -
CASE WHEN MONTH(#DOB) > MONTH(#NOW)
OR MONTH(#DOB) = MONTH(#NOW) AND DAY(#DOB) > DAY(#NOW)
THEN 1 ELSE 0 END
It's basically calculating age as we humans do.
Solution 3: My friend refactored it into this:
DATEDIFF(YY, #DOB, #NOW) -
CEILING(0.5 * SIGN((MONTH(#DOB) - MONTH(#NOW)) * 50 + DAY(#DOB) - DAY(#NOW)))
This one is the shortest but it's most difficult to understand. 50 is just a weight so the day difference is only important when months are the same. SIGN function is for transforming whatever value it gets to -1, 0 or 1. CEILING(0.5 * is the same as Math.max(0, value) but there is no such thing in SQL.
What about:
DECLARE #DOB datetime
SET #DOB='19851125'
SELECT Datepart(yy,convert(date,GETDATE())-#DOB)-1900
Wouldn't that avoid all those rounding, truncating and ofsetting issues?
Just check whether the below answer is feasible.
DECLARE #BirthDate DATE = '09/06/1979'
SELECT
(
YEAR(GETDATE()) - YEAR(#BirthDate) -
CASE WHEN (MONTH(GETDATE()) * 100) + DATEPART(dd, GETDATE()) >
(MONTH(#BirthDate) * 100) + DATEPART(dd, #BirthDate)
THEN 1
ELSE 0
END
)
select floor((datediff(day,0,#today) - datediff(day,0,#birthdate)) / 365.2425) as age
There are a lot of 365.25 answers here. Remember how leap years are defined:
Every four years
except every 100 years
except every 400 years
There are many answers to this question, but I think this one is close to the truth.
The datediff(year,…,…) function, as we all know, only counts the boundaries crossed by the date part, in this case the year. As a result it ignores the rest of the year.
This will only give the age in completed years if the year were to start on the birthday. It probably doesn’t, but we can fake it by adjusting the asking date back by the same amount.
In pseudopseudo code, it’s something like this:
adjusted_today = today - month(dob) + 1 - day(dob) + 1
age = year(adjusted_today - dob)
The + 1 is to allow for the fact that the month and day numbers start from 1 and not 0.
The reason we subtract the month and the day separately rather than the day of the year is because February has the annoying tendency to change its length.
The calculation in SQL is:
datediff(year,dob,dateadd(month,-month(dob)+1,dateadd(day,-day(dob)+1,today)))
where dob and today are presumed to be the date of birth and the asking date.
You can test this as follows:
WITH dates AS (
SELECT
cast('2022-03-01' as date) AS today,
cast('1943-02-25' as date) AS dob
)
select
datediff(year,dob,dateadd(month,-month(dob)+1,dateadd(day,-day(dob)+1,today))) AS age
from dates;
which gives you George Harrison’s age in completed years.
This is much cleaner than fiddling about with quarter days which will generally give you misleading values on the edges.
If you have the luxury of creating a scalar function, you can use something like this:
DROP FUNCTION IF EXISTS age;
GO
CREATE FUNCTION age(#dob date, #today date) RETURNS INT AS
BEGIN
SET #today = dateadd(month,-month(#dob)+1,#today);
SET #today = dateadd(day,-day(#dob)+1,#today);
RETURN datediff(year,#dob,#today);
END;
GO
Remember, you need to call dbo.age() because, well, Microsoft.
DECLARE #DOB datetime
set #DOB ='11/25/1985'
select floor(
( cast(convert(varchar(8),getdate(),112) as int)-
cast(convert(varchar(8),#DOB,112) as int) ) / 10000
)
source: http://beginsql.wordpress.com/2012/04/26/how-to-calculate-age-in-sql-server/
Try This
DECLARE #date datetime, #tmpdate datetime, #years int, #months int, #days int
SELECT #date = '08/16/84'
SELECT #tmpdate = #date
SELECT #years = DATEDIFF(yy, #tmpdate, GETDATE()) - CASE WHEN (MONTH(#date) > MONTH(GETDATE())) OR (MONTH(#date) = MONTH(GETDATE()) AND DAY(#date) > DAY(GETDATE())) THEN 1 ELSE 0 END
SELECT #tmpdate = DATEADD(yy, #years, #tmpdate)
SELECT #months = DATEDIFF(m, #tmpdate, GETDATE()) - CASE WHEN DAY(#date) > DAY(GETDATE()) THEN 1 ELSE 0 END
SELECT #tmpdate = DATEADD(m, #months, #tmpdate)
SELECT #days = DATEDIFF(d, #tmpdate, GETDATE())
SELECT Convert(Varchar(Max),#years)+' Years '+ Convert(Varchar(max),#months) + ' Months '+Convert(Varchar(Max), #days)+'days'
After trying MANY methods, this works 100% of the time using the modern MS SQL FORMAT function instead of convert to style 112. Either would work but this is the least code.
Can anyone find a date combination which does not work? I don't think there is one :)
--Set parameters, or choose from table.column instead:
DECLARE #DOB DATE = '2000/02/29' -- If #DOB is a leap day...
,#ToDate DATE = '2018/03/01' --...there birthday in this calculation will be
--0+ part tells SQL to calc the char(8) as numbers:
SELECT [Age] = (0+ FORMAT(#ToDate,'yyyyMMdd') - FORMAT(#DOB,'yyyyMMdd') ) /10000
CASE WHEN datepart(MM, getdate()) < datepart(MM, BIRTHDATE) THEN ((datepart(YYYY, getdate()) - datepart(YYYY, BIRTH_DATE)) -1 )
ELSE
CASE WHEN datepart(MM, getdate()) = datepart(MM, BIRTHDATE)
THEN
CASE WHEN datepart(DD, getdate()) < datepart(DD, BIRTHDATE) THEN ((datepart(YYYY, getdate()) - datepart(YYYY, BIRTHDATE)) -1 )
ELSE (datepart(YYYY, getdate()) - datepart(YYYY, BIRTHDATE))
END
ELSE (datepart(YYYY, getdate()) - datepart(YYYY, BIRTHDATE)) END
END
SELECT ID,
Name,
DATEDIFF(yy,CONVERT(DATETIME, DOB),GETDATE()) AS AGE,
DOB
FROM MyTable
How about this:
SET #Age = CAST(DATEDIFF(Year, #DOB, #Stamp) as int)
IF (CAST(DATEDIFF(DAY, DATEADD(Year, #Age, #DOB), #Stamp) as int) < 0)
SET #Age = #Age - 1
Try this solution:
declare #BirthDate datetime
declare #ToDate datetime
set #BirthDate = '1/3/1990'
set #ToDate = '1/2/2008'
select #BirthDate [Date of Birth], #ToDate [ToDate],(case when (DatePart(mm,#ToDate) < Datepart(mm,#BirthDate))
OR (DatePart(m,#ToDate) = Datepart(m,#BirthDate) AND DatePart(dd,#ToDate) < Datepart(dd,#BirthDate))
then (Datepart(yy, #ToDate) - Datepart(yy, #BirthDate) - 1)
else (Datepart(yy, #ToDate) - Datepart(yy, #BirthDate))end) Age
This will correctly handle the issues with the birthday and rounding:
DECLARE #dob datetime
SET #dob='1992-01-09 00:00:00'
SELECT DATEDIFF(YEAR, '0:0', getdate()-#dob)
Ed Harper's solution is the simplest I have found which never returns the wrong answer when the month and day of the two dates are 1 or less days apart. I made a slight modification to handle negative ages.
DECLARE #D1 AS DATETIME, #D2 AS DATETIME
SET #D2 = '2012-03-01 10:00:02'
SET #D1 = '2013-03-01 10:00:01'
SELECT
DATEDIFF(YEAR, #D1,#D2)
+
CASE
WHEN #D1<#D2 AND DATEADD(YEAR, DATEDIFF(YEAR,#D1, #D2), #D1) > #D2
THEN - 1
WHEN #D1>#D2 AND DATEADD(YEAR, DATEDIFF(YEAR,#D1, #D2), #D1) < #D2
THEN 1
ELSE 0
END AS AGE
The answer marked as correct is nearer to accuracy but, it fails in following scenario - where Year of birth is Leap year and day are after February month
declare #ReportStartDate datetime = CONVERT(datetime, '1/1/2014'),
#DateofBirth datetime = CONVERT(datetime, '2/29/1948')
FLOOR(DATEDIFF(HOUR,#DateofBirth,#ReportStartDate )/8766)
OR
FLOOR(DATEDIFF(HOUR,#DateofBirth,#ReportStartDate )/8765.82) -- Divisor is more accurate than 8766
-- Following solution is giving me more accurate results.
FLOOR(DATEDIFF(YEAR,#DateofBirth,#ReportStartDate) - (CASE WHEN DATEADD(YY,DATEDIFF(YEAR,#DateofBirth,#ReportStartDate),#DateofBirth) > #ReportStartDate THEN 1 ELSE 0 END ))
It worked in almost all scenarios, considering leap year, date as 29 feb, etc.
Please correct me if this formula have any loophole.
Declare #dob datetime
Declare #today datetime
Set #dob = '05/20/2000'
set #today = getdate()
select CASE
WHEN dateadd(year, datediff (year, #dob, #today), #dob) > #today
THEN datediff (year, #dob, #today) - 1
ELSE datediff (year, #dob, #today)
END as Age
Here is how i calculate age given a birth date and current date.
select case
when cast(getdate() as date) = cast(dateadd(year, (datediff(year, '1996-09-09', getdate())), '1996-09-09') as date)
then dateDiff(yyyy,'1996-09-09',dateadd(year, 0, getdate()))
else dateDiff(yyyy,'1996-09-09',dateadd(year, -1, getdate()))
end as MemberAge
go
CREATE function dbo.AgeAtDate(
#DOB datetime,
#CompareDate datetime
)
returns INT
as
begin
return CASE WHEN #DOB is null
THEN
null
ELSE
DateDiff(yy,#DOB, #CompareDate)
- CASE WHEN datepart(mm,#CompareDate) > datepart(mm,#DOB) OR (datepart(mm,#CompareDate) = datepart(mm,#DOB) AND datepart(dd,#CompareDate) >= datepart(dd,#DOB))
THEN 0
ELSE 1
END
END
End
GO
DECLARE #FromDate DATETIME = '1992-01-2623:59:59.000',
#ToDate DATETIME = '2016-08-10 00:00:00.000',
#Years INT, #Months INT, #Days INT, #tmpFromDate DATETIME
SET #Years = DATEDIFF(YEAR, #FromDate, #ToDate)
- (CASE WHEN DATEADD(YEAR, DATEDIFF(YEAR, #FromDate, #ToDate),
#FromDate) > #ToDate THEN 1 ELSE 0 END)
SET #tmpFromDate = DATEADD(YEAR, #Years , #FromDate)
SET #Months = DATEDIFF(MONTH, #tmpFromDate, #ToDate)
- (CASE WHEN DATEADD(MONTH,DATEDIFF(MONTH, #tmpFromDate, #ToDate),
#tmpFromDate) > #ToDate THEN 1 ELSE 0 END)
SET #tmpFromDate = DATEADD(MONTH, #Months , #tmpFromDate)
SET #Days = DATEDIFF(DAY, #tmpFromDate, #ToDate)
- (CASE WHEN DATEADD(DAY, DATEDIFF(DAY, #tmpFromDate, #ToDate),
#tmpFromDate) > #ToDate THEN 1 ELSE 0 END)
SELECT #FromDate FromDate, #ToDate ToDate,
#Years Years, #Months Months, #Days Days
What about a solution with only date functions, not math, not worries about leap year
CREATE FUNCTION dbo.getAge(#dt datetime)
RETURNS int
AS
BEGIN
RETURN
DATEDIFF(yy, #dt, getdate())
- CASE
WHEN
MONTH(#dt) > MONTH(GETDATE()) OR
(MONTH(#dt) = MONTH(GETDATE()) AND DAY(#dt) > DAY(GETDATE()))
THEN 1
ELSE 0
END
END
declare #birthday as datetime
set #birthday = '2000-01-01'
declare #today as datetime
set #today = GetDate()
select
case when ( substring(convert(varchar, #today, 112), 5,4) >= substring(convert(varchar, #birthday, 112), 5,4) ) then
(datepart(year,#today) - datepart(year,#birthday))
else
(datepart(year,#today) - datepart(year,#birthday)) - 1
end
The following script checks the difference in years between now and the given date of birth; the second part checks whether the birthday is already past in the current year; if not, it subtracts it:
SELECT year(NOW()) - year(date_of_birth) - (CONCAT(year(NOW()), '-', month(date_of_birth), '-', day(date_of_birth)) > NOW()) AS Age
FROM tableName;
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 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)