Related
For example:
#dtBegin = '2012-06-29'
#input = 20
I want the output to be '2012-07-27'.
I had to tackle the same problem in my project. Gordon Lionoff's solution got me on the right track but did not always produce the right result. I also have to take in account dates that start in a weekend. I.e. adding 1 business day to a saturday or sunday should result in a monday. This is the most common approach to handling business day calculation.
I've created my own solution based on Gordon Linoff's function and Patrick McDonald's excellent C# equivalent
NOTE: My solution only works if DATEFIRST is set to the default value of 7. If you use a different DATEFIRST value, you will have to change the 1, 7 and (1,7,8,9,10) bits.
My solution consists of two functions. An "outer" function that handles edge cases and an "inner" function that performs the actual calculation. Both functions are table-valued functions so they will actually be expanded into the implementing query and fed through the query optimizer.
CREATE FUNCTION [dbo].[UTL_DateAddWorkingDays]
(
#date datetime,
#days int
)
RETURNS TABLE AS RETURN
(
SELECT
CASE
WHEN #days = 0 THEN #date
WHEN DATEPART(dw, #date) = 1 THEN (SELECT Date FROM [dbo].[UTL_DateAddWorkingDays_Inner](DATEADD(d, 1, #date), #days - 1))
WHEN DATEPART(dw, #date) = 7 THEN (SELECT Date FROM [dbo].[UTL_DateAddWorkingDays_Inner](DATEADD(d, 2, #date), #days - 1))
ELSE (SELECT Date FROM [dbo].[UTL_DateAddWorkingDays_Inner](#date, #days))
END AS Date
)
As you can see, the "outer" function handles:
When adding no days, return the original date. This will keep saturday and sunday dates intact.
When adding days to a sunday, start counting from monday. This consumes 1 day.
When adding days to a saturday, start counting from monday. This consumes 1 day.
In all other cases, perform the usual calculation
_
CREATE FUNCTION [dbo].[UTL_DateAddWorkingDays_Inner]
(
#date datetime,
#days int
)
RETURNS TABLE AS RETURN
(
SELECT
DATEADD(d
, (#days / 5) * 7
+ (#days % 5)
+ (CASE WHEN ((#days%5) + DATEPART(dw, #date)) IN (1,7,8,9,10) THEN 2 ELSE 0 END)
, #date) AS Date
)
The "inner" function is similar to Gordon Linoff's solution, except it accounts for dates crossing weekend boundaries but without crossing a full week boundary.
Finally, I created a test script to test my function. The expected values were generated using Patrick McDonald's excellent C# equivalent and I randomly cross-referenced this data with this popular calculator.
You can do this without resorting to a calendar table or user defined function:
dateadd(d,
(#input / 5) * 7 + -- get complete weeks out of the way
mod(#input, 5) + -- extra days
(case when ((#input%5) + datepart(dw, #dtbegin)%7) in (7, 1, 8) or
((#input%5) + datepart(dw, #dtbegin)%7) < (#input%5)
then 2
else 0
end),
#dtbegin
)
I'm not saying this is pretty. But sometimes arithmetic is preferable to a join or a loop.
This is what I've tried:
CREATE function [dbo].[DateAddWorkDay]
(#days int,#FromDate Date)
returns Date
as
begin
declare #result date
set #result = (
select b
from
(
SELECT
b,
(DATEDIFF(dd, a, b))
-(DATEDIFF(wk, a, b) * 2)
-(CASE WHEN DATENAME(dw, a) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, b) = 'Saturday' THEN 1 ELSE 0 END)
-COUNT(o.Holiday_Date)
as workday
from
(
select
#FromDate as a,
dateadd(DAY,num +#days,#FromDate) as b
from (select row_number() over (order by (select NULL)) as num
from Information_Schema.columns
) t
where num <= 100
) dt
left join Holiday o on o.Holiday_Date between a and b and DATENAME(dw, o.Holiday_Date) not in('Saturday','Sunday')
where DATENAME(dw, b) not in('Saturday','Sunday')
and b not in (select Holiday_Date from OP_Holiday where Holiday_Date between a and b)
group by a,b
) du
where workday =#days
)
return #result
end
Where Holiday is a table with holiday_date as a reference for holiday.
I realize I'm about 8 years late to this party... But I took Martin's answer and updated it to:
1. a single scalar function instead of 2 nested table-valued functions
I tested using his original test script and my function passes too. Also, it seems the rebuild to a scalar function has a slight positive impact on performance. Both versions seem to benefit equally from 'buffer caching', the scalar version performing up to 25% better non-cached and up to 40% better cached. Disclaimer: I just ran both versions a bunch of times and recorded the times, I did not do any decent performance testing.
2. include support for DATEFIRST is either Monday, Saturday or Sunday
I feel a UDF should be agnostic to the datefirst setting. I'm in Europe and Monday is the default here. The original function would not work without adaptation.
According to wikipedia Monday, Saturday and Sunday are the only real world first days of the week. Support for other could easily be added, but would make the code more bulky and I have a hard time envisioning a real world use case.
CREATE FUNCTION dbo.fn_addWorkDays
(
#date datetime,
#days int
)
RETURNS DATETIME
AS
BEGIN
IF NOT ##DATEFIRST IN (1,6,7) BEGIN --UNSUPPORTED DATE FIRST
RETURN NULL
/* MONDAY = FRIST DAY */
END ELSE IF #days = 0 BEGIN
RETURN #date
END ELSE IF ##DATEFIRST = 1 AND DATEPART(dw, #date) = 7 BEGIN --SUNDAY
SET #date = DATEADD(d, 1, #date)
SET #days = #days - 1
END ELSE IF ##DATEFIRST = 1 AND DATEPART(dw, #date) = 6 BEGIN --SATURDAY
SET #date = DATEADD(d, 2, #date)
SET #days = #days - 1
/* SATURDAY = FRIST DAY */
END ELSE IF ##DATEFIRST = 7 AND DATEPART(dw, #date) = 2 BEGIN --SUNDAY
SET #date = DATEADD(d, 1, #date)
SET #days = #days - 1
END ELSE IF ##DATEFIRST = 7 AND DATEPART(dw, #date) = 1 BEGIN --SATURDAY
SET #date = DATEADD(d, 2, #date)
SET #days = #days - 1
/* SUNDAY = FRIST DAY */
END ELSE IF ##DATEFIRST = 7 AND DATEPART(dw, #date) = 1 BEGIN --SUNDAY
SET #date = DATEADD(d, 1, #date)
SET #days = #days - 1
END ELSE IF ##DATEFIRST = 7 AND DATEPART(dw, #date) = 7 BEGIN --SATURDAY
SET #date = DATEADD(d, 2, #date)
SET #days = #days - 1
END
DECLARE #return AS dateTime
SELECT #return =
DATEADD(d
, (#days / 5) * 7
+ (#days % 5)
+ (CASE
/* MONDAY = FRIST DAY */
WHEN ##DATEFIRST = 1 AND ((#days%5) + DATEPART(dw, #date)) IN (6,7,8,9) THEN 2
/* SATURDAY = FRIST DAY */
WHEN ##DATEFIRST = 7 AND ((#days%5) + DATEPART(dw, #date)) IN (1,2,8,9,10) THEN 2
/* SUNDAY = FRIST DAY */
WHEN ##DATEFIRST = 7 AND ((#days%5) + DATEPART(dw, #date)) IN (1,7,8,9,10,11) THEN 2
ELSE 0
END)
, #date)
RETURN #return
END
I hope this might benefit someone!
you can use the below mentioned code for exclude weekends
go
if object_id('UTL_DateAddWorkingDays') is not null
drop function UTL_DateAddWorkingDays
go
create FUNCTION [dbo].[UTL_DateAddWorkingDays]
(
#date datetime,
#daysToAdd int
)
RETURNS date
as
begin
declare #daysToAddCnt int=0,
#Dt datetime
declare #OutTable table
(
id int identity(1,1),
WeekDate date,
DayId int
)
while #daysToAdd>0
begin
set #Dt=dateadd(day,#daysToAddCnt,#date)
--select #daysToAddCnt cnt,#Dt date,DATEPART(weekday,#Dt) dayId,#daysToAdd daystoAdd
if(DATEPART(weekday,#Dt) <>7 and DATEPART(weekday,#Dt)<>1)
begin
insert into #outTable (WeekDate,DayId)
select #Dt,DATEPART(weekday,DATEADD(day,#daysToAddCnt,#Dt))
set #daysToAdd=#daysToAdd-1
end
set #daysToAddCnt=#daysToAddCnt+1
end
select #Dt=max(WeekDate) from #outTable
return #Dt
end
what about this?
declare #d1 date='2012-06-29'
declare #days int=20
select dateadd(dd,#days,#d1)
select dateadd(dd,case DATEPART(dw,t1.d) when 6 then +2 when 7 then +1 else +0 end,t1.d)
from
(
select dateadd(dd,CEILING((convert(float,#days)/5)*2)+#days,#d1)d
)t1
i found how many we in the range by CEILING((convert(float,#days)/5)*2)
then i added them to date and at the end i check if is a saturday or sunday and i add 1 or 2 days.
i have an ssis Package which runs on business days (mon-Fri). if i receive file on tuesday , background(DB), it takes previous business day date and does some transactions. If i run the job on friday, it has to fetch mondays date and process the transactions.
i have used the below query to get previous business date
Select Convert(varchar(50), Position_ID) as Position_ID,
TransAmount_Base,
Insert_Date as InsertDate
from tblsample
Where AsOfdate = Dateadd(dd, -1, Convert(datetime, Convert(varchar(10), '03/28/2012', 101), 120))
Order By Position_ID
if i execute this query i'll get the results of yesterdays Transactios. if i ran the same query on monday, it has to fetch the Fridays transactions instead of Sundays.
SELECT DATEADD(DAY, CASE DATENAME(WEEKDAY, GETDATE())
WHEN 'Sunday' THEN -2
WHEN 'Monday' THEN -3
ELSE -1 END, DATEDIFF(DAY, 0, GETDATE()))
I prefer to use DATENAME for things like this over DATEPART as it removes the need for Setting DATEFIRST And ensures that variations on time/date settings on local machines do not affect the results. Finally DATEDIFF(DAY, 0, GETDATE()) will remove the time part of GETDATE() removing the need to convert to varchar (much slower).
EDIT (almost 2 years on)
This answer was very early in my SO career and it annoys me everytime it gets upvoted because I no longer agree with the sentiment of using DATENAME.
A much more rubust solution would be:
SELECT DATEADD(DAY, CASE (DATEPART(WEEKDAY, GETDATE()) + ##DATEFIRST) % 7
WHEN 1 THEN -2
WHEN 2 THEN -3
ELSE -1
END, DATEDIFF(DAY, 0, GETDATE()));
This will work for all language and DATEFIRST settings.
This function returns last working day and takes into account holidays and weekends. You will need to create a simple holiday table.
-- =============================================
-- Author: Dale Kilian
-- Create date: 2019-04-29
-- Description: recursive function returns last work day for weekends and
-- holidays
-- =============================================
ALTER FUNCTION dbo.fnGetWorkWeekday
(
#theDate DATE
)
RETURNS DATE
AS
BEGIN
DECLARE #importDate DATE = #theDate
DECLARE #returnDate DATE
--Holidays
IF EXISTS(SELECT 1 FROM dbo.Holidays WHERE isDeleted = 0 AND #theDate = Holiday_Date)
BEGIN
SET #importDate = DATEADD(DAY,-1,#theDate);
SET #importDate = (SELECT dbo.fnGetWorkWeekday(#importDate))
END
--Satruday
IF(DATEPART(WEEKDAY,#theDate) = 7)
BEGIN
SET #importDate = DATEADD(DAY,-1,#theDate);
SET #importDate = (SELECT dbo.fnGetWorkWeekday(#importDate))
END
--Sunday
IF(DATEPART(WEEKDAY,#theDate) = 1)
BEGIN
SET #importDate = DATEADD(DAY,-2,#theDate);
SET #importDate = (SELECT dbo.fnGetWorkWeekday(#importDate))
END
RETURN #importDate;
END
GO
Then how about:
declare #dt datetime='1 dec 2012'
select case when 8-##DATEFIRST=DATEPART(dw,#dt)
then DATEADD(d,-2,#dt)
when (9-##DATEFIRST)%7=DATEPART(dw,#dt)%7
then DATEADD(d,-3,#dt)
else DATEADD(d,-1,#dt)
end
The simplest solution to find the previous business day is to use a calendar table with a column called IsBusinessDay or something similar. The your query is something like this:
select max(BaseDate)
from dbo.Calendar c
where c.IsBusinessDay = 0x1 and c.BaseDate < #InputDate
The problem with using functions is that when (not if) you have to create exceptions for any reason (national holidays etc.) the code quickly becomes unmaintainable; with the table, you just UPDATE a single value. A table also makes it much easier to answer questions like "how many business days are there between dates X and Y", which are quite common in reporting tasks.
You can easily make this a function call, adding a second param to replace GetDate() with whatever date you wanted.
It will work for any day of the week, at any date range, if you change GetDate().
It will not change the date if the day of week is the input date (GetDate())
Declare #DayOfWeek As Integer = 2 -- Monday
Select DateAdd(Day, ((DatePart(dw,GetDate()) + (7 - #DayOfWeek)) * -1) % 7, Convert(Date,GetDate()))
More elegant:
select DATEADD(DAY,
CASE when datepart (dw,Getdate()) < 3 then datepart (dw,Getdate()) * -1 + -1 ELSE -1 END,
cast(GETDATE() as date))
select
dateadd(dd,
case DATEPART(dw, getdate())
when 1
then -2
when 2
then -3
else -1
end, GETDATE())
thanks for the tips above, I had a slight variant on the query in that my user needed all values for the previous business date. For example, today is a Monday so he needs everything between last Friday at midnight through to Saturday at Midnight. I did this using a combo of the above, and "between", just if anyone is interested. I'm not a massive techie.
-- Declare a variable for the start and end dates.
declare #StartDate as datetime
declare #EndDate as datetime
SELECT #StartDate = DATEADD(DAY, CASE DATENAME(WEEKDAY, GETDATE())
WHEN 'Sunday' THEN -2
WHEN 'Monday' THEN -3
ELSE -1 END, DATEDIFF(DAY, 0, GETDATE()))
select #EndDate = #StartDate + 1
select #StartDate , #EndDate
-- Later on in the query use "between"
and mydate between #StartDate and #EndDate
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
Looking for a way to get the date in the format "11/1/2009", which would be the first sunday of next month. I want to run this query after the first sunday in october to get the first sunday of the upcoming month. What is the best method to accomplish this with a T-SQL query?
Thanks
try this:
Declare #D Datetime
Set #D = [Some date for which you want the following months' first sunday]
Select DateAdd(day, (8-DatePart(weekday,
DateAdd(Month, 1+DateDiff(Month, 0, #D), 0)))%7,
DateAdd(Month, 1+DateDiff(Month, 0, #D), 0))
EDIT Notes:
The first of next Month is given by the expression:
DateAdd(Month, 1+DateDiff(Month, 0, #D), 0)
or by:
which can be modified to give the first of the month two months from now by changing the 1 to a 2:
DateAdd(Month, 2+DateDiff(Month, 0, #D), 0)
EDIT: In response to #NissanFan, and #Anthony: to modify this to return the first Monday Tuesday Wednesday, etc, change the value 8 to a 9, 10, 11, etc....
Declare #Sun TinyInt Set #Sun = 8
Declare #Mon TinyInt Set #Mon = 9
Declare #Tue TinyInt Set #Tue = 10
Declare #Wed TinyInt Set #Wed = 11
Declare #Thu TinyInt Set #Thu = 12
Declare #Fri TinyInt Set #Fri = 13
Declare #Sat TinyInt Set #Sat = 14
Declare #D Datetime, #FONM DateTime -- FirstofNextMonth
Set #D = [Some date for which you want the following months' first sunday]
Set #FONM = DateAdd(Month, 1+DateDiff(Month, 0, #D),0)
Select
DateAdd(day, (#Sun -DatePart(weekday, #FONM))%7, #FONM) firstSunInNextMonth,
DateAdd(day, (#Mon -DatePart(weekday, #FONM))%7, #FONM) firstMonInNextMonth,
DateAdd(day, (#Tue -DatePart(weekday, #FONM))%7, #FONM) firstTueInNextMonth,
DateAdd(day, (#Wed -DatePart(weekday, #FONM))%7, #FONM) firstWedInNextMonth,
DateAdd(day, (#Thu -DatePart(weekday, #FONM))%7, #FONM) firstThuInNextMonth,
DateAdd(day, (#Fri -DatePart(weekday, #FONM))%7, #FONM) firstFriInNextMonth,
DateAdd(day, (#Sat -DatePart(weekday, #FONM))%7, #FONM) firstSatInNextMonth
Just an FYI rather then coming up with some code to do this how about using a calendar table.
Take a look at this: http://web.archive.org/web/20070611150639/http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-calendar-table.html
This also may help:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=99696
Here is a query to get first working day of next month
DECLARE #DAYOFWEEK INT,#ReminderDate DateTime
SET #DAYOFWEEK = DATEPART( WEEKDAY,DateAdd(D,- Day(GetDate())+1, DATEADD(M,1,GetDate())) )
Print #DAYOFWEEK
If #DAYOFWEEK = 1
Set #ReminderDate = DateAdd(D,- Day(GetDate())+2, DATEADD(M,1,GetDate()))
Else If #DAYOFWEEK =7
Set #ReminderDate = DateAdd(D,- Day(GetDate())+3, DATEADD(M,1,GetDate()))
Else
Set #ReminderDate = DateAdd(D,- Day(GetDate())+1, DATEADD(M,1,GetDate()))
Print #ReminderDate
Reference taken from this blog:
SQL Server 2012 introduced one new TSQL EOMONTH to return the last day of the month that contains the specified date with an optional offset.
CREATE TABLE tbl_Test_EOMONTH
(
SampleDate DATETIME
)
GO
INSERT INTO tbl_Test_EOMONTH VALUES ('2015-12-20')
INSERT INTO tbl_Test_EOMONTH VALUES ('2015-11-08')
INSERT INTO tbl_Test_EOMONTH VALUES ('2015-10-16')
INSERT INTO tbl_Test_EOMONTH VALUES ('2015-09-26')
INSERT INTO tbl_Test_EOMONTH VALUES ('2016-01-31')
GO
SELECT
DATEADD(DAY,8-DATEPART(WEEKDAY,DATEADD(DAY,0,EOMONTH([SampleDate])))
,EOMONTH([SampleDate])) AS FirstSunday_ofTheNextMonth
FROM tbl_Test_EOMONTH
GO
You can use DATENAME to determine the day you want, I might recommend a loop to move the date from the 01 of the month in question to get to the first sunday.
So lets try:
DECLARE #DateTime DATETIME
Set to the date to start off with, then add 1 day until you find what you are looking for. Use datename with dw...
We have used this to determine weekends, but holidays will be a problem, where we use a table to store that.
Try this code as a function:
-- Variables
DECLARE #DATE DATETIME
DECLARE #DAY INT
DECLARE #DAYOFWEEK INT
DECLARE #TESTDATE DATETIME
-- Set
SET #TESTDATE = GETDATE()
SET #DATE = DATEADD( MONTH, 1, #TESTDATE )
SET #DAY = DATEPART( DAY, #TESTDATE )
SET #DATE = DATEADD( DAY, -#DAY + 1, #DATE )
SET #DAYOFWEEK = DATEPART( WEEKDAY, #DATE )
IF #DAYOFWEEK > 1
BEGIN
SET #DAYOFWEEK = 8 - #DAYOFWEEK
END
ELSE
BEGIN
SET #DAYOFWEEK = 0
END
SET #DATE = DATEADD( DAY, #DAYOFWEEK, #DATE )
-- Display
PRINT #TESTDATE
PRINT #DAY
PRINT #DAYOFWEEK
PRINT #DATE
Here is the non-system specific way to determine the first Sunday of the following month:
First, get the current month and add one month.
Next, set the date of that variable to be on the first.
Next, find the day value of that date (let's assume Mondays are 1 and Sundays are 7).
Next, subtract the day value of the 1st of the month from the day value of Sunday (7).
You now have the number of days between the first of the month and the first Sunday. You could then add that to the date variable to get the first Sunday, or, since we know the first of the month is 1, you could just add one to the difference (found in that last step above) and that is the date of the first Sunday. (You have to add one because it's subtracting and thus if the first of the given month IS Sunday, you'd end up with 0).
I have been looking through the T-SQL documentation and it is not at all intuitive as to how how you would use my method, but you will need the concept of "day of week number" to make it work no matter what.
This would be simplest with an auxiliary calendar table. See this link, for example.
However, it can be done without one, though it's rather tricky. Here I assume you want the first future date that is the first Sunday of a month. I've written this with a variable #now for - well - now, so it's easier to test. It might be possible to write more simply, but I think this has a better chance of being correct than quickly-written simpler solutions. [Corrected by adding DAY(d) < 8]
SET #now = GETDATE();
WITH Seven(i) AS (
SELECT -1 UNION ALL SELECT 0 UNION ALL SELECT 1
UNION ALL SELECT 3 UNION ALL SELECT 4
UNION ALL SELECT 5 UNION ALL SELECT 6
), Candidates(d) AS (
SELECT DATEADD(WEEK,i+DATEDIFF(WEEK,'19000107',#now),'19000107')
FROM Seven
)
SELECT TOP (1) d AS SoonestFutureFirstSunday
FROM Candidates
WHERE DAY(d) < 8 AND MONTH(d) >= MONTH(#now)
AND (MONTH(d) > MONTH(#now) OR DAY(d) > DAY(#now))
ORDER BY d; ORDER BY d;
I reckon that the answer is this
SELECT DATEADD(Month, DATEDIFF(Month, 0, GETDATE()) + 1, 0) + 6 - (DATEPART(Weekday,
DATEADD(Month,
DATEDIFF(Month,0, GETDATE()) + 1, 0))
+ ##DateFirst + 5) % 7 --FIRST sunday of following month
I need to determine the number of days in a month for a given date in SQL Server.
Is there a built-in function? If not, what should I use as the user-defined function?
In SQL Server 2012 you can use EOMONTH (Transact-SQL) to get the last day of the month and then you can use DAY (Transact-SQL) to get the number of days in the month.
DECLARE #ADate DATETIME
SET #ADate = GETDATE()
SELECT DAY(EOMONTH(#ADate)) AS DaysInMonth
You can use the following with the first day of the specified month:
datediff(day, #date, dateadd(month, 1, #date))
To make it work for every date:
datediff(day, dateadd(day, 1-day(#date), #date),
dateadd(month, 1, dateadd(day, 1-day(#date), #date)))
Most elegant solution: works for any #DATE
DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,#DATE),0)))
Throw it in a function or just use it inline. This answers the original question without all the extra junk in the other answers.
examples for dates from other answers:
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'1/31/2009'),0))) Returns 31
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'2404-feb-15'),0))) Returns 29
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'2011-12-22'),0))) Returns 31
--Last Day of Previous Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE()),0)))
--Last Day of Current Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+1,0)))
--Last Day of Next Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+2,0)))
Personally though, I would make a UDF for it if there is not a built in function...
I would suggest:
SELECT DAY(EOMONTH(GETDATE()))
This code gets you the number of days in current month:
SELECT datediff(dd,getdate(),dateadd(mm,1,getdate())) as datas
Change getdate() to the date you need to count days for.
--- sql server below 2012---
select day( dateadd(day,-1,dateadd(month, 1, convert(date,'2019-03-01'))))
-- this for sql server 2012--
select day(EOMONTH(getdate()))
Solution 1: Find the number of days in whatever month we're currently in
DECLARE #dt datetime
SET #dt = getdate()
SELECT #dt AS [DateTime],
DAY(DATEADD(mm, DATEDIFF(mm, -1, #dt), -1)) AS [Days in Month]
Solution 2: Find the number of days in a given month-year combo
DECLARE #y int, #m int
SET #y = 2012
SET #m = 2
SELECT #y AS [Year],
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m - 1, 0)),
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m, 0))
) AS [Days in Month]
You do need to add a function, but it's a simple one. I use this:
CREATE FUNCTION [dbo].[ufn_GetDaysInMonth] ( #pDate DATETIME )
RETURNS INT
AS
BEGIN
SET #pDate = CONVERT(VARCHAR(10), #pDate, 101)
SET #pDate = #pDate - DAY(#pDate) + 1
RETURN DATEDIFF(DD, #pDate, DATEADD(MM, 1, #pDate))
END
GO
SELECT Datediff(day,
(Convert(DateTime,Convert(varchar(2),Month(getdate()))+'/01/'+Convert(varchar(4),Year(getdate())))),
(Convert(DateTime,Convert(varchar(2),Month(getdate())+1)+'/01/'+Convert(varchar(4),Year(getdate()))))) as [No.of Days in a Month]
select datediff(day,
dateadd(day, 0, dateadd(month, ((2013 - 1900) * 12) + 3 - 1, 0)),
dateadd(day, 0, dateadd(month, ((2013 - 1900) * 12) + 3, 0))
)
Nice Simple and does not require creating any functions Work Fine
You need to create a function, but it is for your own convenience. It works perfect and I never encountered any faulty computations using this function.
CREATE FUNCTION [dbo].[get_days](#date datetime)
RETURNS int
AS
BEGIN
SET #date = DATEADD(MONTH, 1, #date)
DECLARE #result int = (select DAY(DATEADD(DAY, -DAY(#date), #date)))
RETURN #result
END
How it works: subtracting the date's day number from the date itself gives you the last day of previous month. So, you need to add one month to the given date, subtract the day number and get the day component of the result.
select add_months(trunc(sysdate,'MM'),1) - trunc(sysdate,'MM') from dual;
I upvoted Mehrdad, but this works as well. :)
CREATE function dbo.IsLeapYear
(
#TestYear int
)
RETURNS bit
AS
BEGIN
declare #Result bit
set #Result =
cast(
case when ((#TestYear % 4 = 0) and (#testYear % 100 != 0)) or (#TestYear % 400 = 0)
then 1
else 0
end
as bit )
return #Result
END
GO
CREATE FUNCTION dbo.GetDaysInMonth
(
#TestDT datetime
)
RETURNS INT
AS
BEGIN
DECLARE #Result int
DECLARE #MonthNo int
Set #MonthNo = datepart(m,#TestDT)
Set #Result =
case #MonthNo
when 1 then 31
when 2 then
case
when dbo.IsLeapYear(datepart(yyyy,#TestDT)) = 0
then 28
else 29
end
when 3 then 31
when 4 then 30
when 5 then 31
when 6 then 30
when 7 then 31
when 8 then 31
when 9 then 30
when 10 then 31
when 11 then 30
when 12 then 31
end
RETURN #Result
END
GO
To Test
declare #testDT datetime;
set #testDT = '2404-feb-15';
select dbo.GetDaysInMonth(#testDT)
here's another one...
Select Day(DateAdd(day, -Day(DateAdd(month, 1, getdate())),
DateAdd(month, 1, getdate())))
I know this question is old but I thought I would share what I'm using.
DECLARE #date date = '2011-12-22'
/* FindFirstDayOfMonth - Find the first date of any month */
-- Replace the day part with -01
DECLARE #firstDayOfMonth date = CAST( CAST(YEAR(#date) AS varchar(4)) + '-' +
CAST(MONTH(#date) AS varchar(2)) + '-01' AS date)
SELECT #firstDayOfMonth
and
DECLARE #date date = '2011-12-22'
/* FindLastDayOfMonth - Find what is the last day of a month - Leap year is handled by DATEADD */
-- Get the first day of next month and remove a day from it using DATEADD
DECLARE #lastDayOfMonth date = CAST( DATEADD(dd, -1, DATEADD(mm, 1, FindFirstDayOfMonth(#date))) AS date)
SELECT #lastDayOfMonth
Those could be combine to create a single function to retrieve the number of days in a month if needed.
SELECT DAY(SUBDATE(ADDDATE(CONCAT(YEAR(NOW()), '-', MONTH(NOW()), '-1'), INTERVAL 1 MONTH), INTERVAL 1 DAY))
Nice 'n' Simple and does not require creating any functions
Mehrdad Afshari reply is most accurate one, apart from usual this answer is based on formal mathematical approach given by Curtis McEnroe in his blog https://cmcenroe.me/2014/12/05/days-in-month-formula.html
DECLARE #date DATE= '2015-02-01'
DECLARE #monthNumber TINYINT
DECLARE #dayCount TINYINT
SET #monthNumber = DATEPART(MONTH,#date )
SET #dayCount = 28 + (#monthNumber + floor(#monthNumber/8)) % 2 + 2 % #monthNumber + 2 * floor(1/#monthNumber)
SELECT #dayCount + CASE WHEN #dayCount = 28 AND DATEPART(YEAR,#date)%4 =0 THEN 1 ELSE 0 END -- leap year adjustment
To get the no. of days in a month we can directly use Day() available in SQL.
Follow the link posted at the end of my answer for SQL Server 2005 / 2008.
The following example and the result are from SQL 2012
alter function dbo.[daysinm]
(
#dates nvarchar(12)
)
returns int
as
begin
Declare #dates2 nvarchar(12)
Declare #days int
begin
select #dates2 = (select DAY(EOMONTH(convert(datetime,#dates,103))))
set #days = convert(int,#dates2)
end
return #days
end
--select dbo.daysinm('08/12/2016')
Result in SQL Server SSMS
(no column name)
1 31
Process:
When EOMONTH is used, whichever the date format we use it is converted into DateTime format of SQL-server. Then the date output of EOMONTH() will be 2016-12-31 having 2016 as Year, 12 as Month and 31 as Days.
This output when passed into Day() it gives you the total days count in the month.
If we want to get the instant result for checking we can directly run the below code,
select DAY(EOMONTH(convert(datetime,'08/12/2016',103)))
or
select DAY(EOMONTH(convert(datetime,getdate(),103)))
for reference to work in SQL Server 2005/2008/2012, please follow the following external link ...
Find No. of Days in a Month in SQL
DECLARE #date DATETIME = GETDATE(); --or '12/1/2018' (month/day/year)
SELECT DAY(EOMONTH ( #date )) AS 'This Month';
SELECT DAY(EOMONTH ( #date, 1 )) AS 'Next Month';
result:
This Month
31
Next Month
30
DECLARE #m int
SET #m = 2
SELECT
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, +#m -1, 0)),
DATEADD(DAY, 0, DATEADD(m,+ #m, 0))
) AS [Days in Month]
RETURN day(dateadd(month, 12 * #year + #month - 22800, -1))
select day(dateadd(month, 12 * year(date) + month(date) - 22800, -1))
A cleaner way of implementing this is using the datefromparts function to construct the first day of the month, and calculate the days from there.
CREATE FUNCTION [dbo].[fn_DaysInMonth]
(
#year INT,
#month INT
)
RETURNS INT
AS
BEGIN
IF #month < 1 OR #month > 12 RETURN NULL;
IF #year < 1753 OR #year > 9998 RETURN NULL;
DECLARE #firstDay DATE = datefromparts(#year, #month, 1);
DECLARE #lastDay DATE = dateadd(month, 1, #firstDay);
RETURN datediff(day, #firstDay, #lastDay);
END
GO
Similarily, you can calculate the days in a year:
CREATE FUNCTION [dbo].[fn_DaysInYear]
(
#year INT
)
RETURNS INT
AS
BEGIN
IF #year < 1753 OR #year > 9998 RETURN NULL;
DECLARE #firstDay DATE = datefromparts(#year, 1, 1);
DECLARE #lastDay DATE = dateadd(year, 1, #firstDay);
RETURN datediff(day, #firstDay, #lastDay);
END
GO
use SQL Server EOMONTH Function nested with day to get last day of month
select Day(EOMONTH('2020-02-1')) -- Leap Year returns 29
select Day(EOMONTH('2021-02-1')) -- returns 28
select Day(EOMONTH('2021-03-1')) -- returns 31
For any date
select DateDiff(Day,#date,DateAdd(month,1,#date))
select first_day=dateadd(dd,-1*datepart(dd,getdate())+1,getdate()),
last_day=dateadd(dd,-1*datepart(dd,dateadd(mm,1,getdate())),dateadd(mm,1,getdate())),
no_of_days = 1+datediff(dd,dateadd(dd,-1*datepart(dd,getdate())+1,getdate()),dateadd(dd,-1*datepart(dd,dateadd(mm,1,getdate())),dateadd(mm,1,getdate())))
replace any date with getdate to get the no of months in that particular date
DECLARE #Month INT=2,
#Year INT=1989
DECLARE #date DateTime=null
SET #date=CAST(CAST(#Year AS nvarchar) + '-' + CAST(#Month AS nvarchar) + '-' + '1' AS DATETIME);
DECLARE #noofDays TINYINT
DECLARE #CountForDate TINYINT
SET #noofDays = DATEPART(MONTH,#date )
SET #CountForDate = 28 + (#noofDays + floor(#noofDays/8)) % 2 + 2 % #noofDays + 2 * floor(1/#noofDays)
SET #noofDays= #CountForDate + CASE WHEN #CountForDate = 28 AND DATEPART(YEAR,#date)%4 =0 THEN 1 ELSE 0 END
PRINT #noofDays
DECLARE #date nvarchar(20)
SET #date ='2012-02-09 00:00:00'
SELECT DATEDIFF(day,cast(replace(cast(YEAR(#date) as char)+'-'+cast(MONTH(#date) as char)+'-01',' ','')+' 00:00:00' as datetime),dateadd(month,1,cast(replace(cast(YEAR(#date) as char)+'-'+cast(MONTH(#date) as char)+'-01',' ','')+' 00:00:00' as datetime)))
simple query in SQLServer2012 :
select day(('20-05-1951 22:00:00'))
i tested for many dates and it return always a correct result