Related
I have searched and have yet to find this little helpful snippet.
I want to input...
Year (2014)
Weeknumber (2)
Weekday (2 = Tuesday, in my case)
Expected result: 2014-01-07 (seventh of January)
And get the full date in return, anyone?
EDIT: My server is SQL 2008
The finished code thanks to all!.
declare #year int = 2014
declare #weeknr int = 2
declare #daynroffset int = 2
SELECT
DATEADD(DAY,+ (#daynroffset-1),
DATEADD(DAY,-DATEPART(DW,CAST('1/1/' + cast(#year as varchar) AS Date))+2,DATEADD(WK,#weeknr- 1,CAST('1/1/' + cast(#year as varchar) AS Date)))
)
The other answers (so far) use SQL Server default mechanisms to determine week and day of week. In this case the current language setting determines the day of the week (through the ##DATEFIRST setting) and the DATEPART(wk uses Jan 1st as the fixed date contained in week 1.
To get a deterministic answer independent of the language setting one can use the ISO 8601 week standard which starts a week on Mondays and where the first week always contains Jan 4th.
This code determines the date based on ISO weeks:
declare #year int = 2016
declare #isoweek int = 22
declare #isoday int = 2
-- ISO-WEEK 1 always contains 4th Jan, so let's use this as a base
declare #date datetime = cast(cast(#year as varchar(4)) + '-01-04T12:00:00' as datetime)
-- Offset the wanted DayOfWeek versus our base date
-- We also set DATEFIRST temporarily because it affects DayOfWeek
-- ISO-Weeks always start on Monday
declare #datefirst int = ##DATEFIRST
SET DATEFIRST 1
declare #offset int = datepart(dw, #date) - 1
SET DATEFIRST #datefirst
-- Add given day and week to basedate
set #date = dateadd(day, #isoday - 1 - #offset, dateadd(wk, #isoweek - 1, #date))
print #date
CODE:
2012+:
DATEADD(DAY,-DATEPART(DW,DATEFROMPARTS("YEAR",1,1))+1+"DAY OF WEEK",DATEADD(WK,"WEEK NUMBER"-1,DATEFROMPARTS(2014,1,1)))
2008+:
SELECT DATEADD(DAY,-DATEPART(DW,CAST(CONCAT('1/1/',"YEAR") AS Date))+1+"DAY OF WEEK",DATEADD(WK,"WEEK NUMBER"-1,CAST(CONCAT('1/1/',"YEAR") AS Date)))
Simply substitue the values where necessary.
This will work for any date.
declare #year int = 2014
declare #week int = 2
declare #day int = 2
declare #date datetime = cast(cast(#year as varchar(20)) + '-01-01' as datetime)
declare #offset int = datepart(dw, #date) - 1
set #date = dateadd(day, #day - #offset, dateadd(ww, #week - 1, #date))
print #date
This may have issues near the year boundary, but it works for the example data given. You may want to add further validations. I've broken down each step of the datetime manipulation into a new field, so you can see it being constructed
2008
DECLARE #Year INT = 2014
DECLARE #WeekNum INT = 2
DECLARE #WeekDay INT = 2
SELECT
BaseDate = CAST( #year AS VARCHAR(4) )
, RoundToWeekStart = DATEADD(WEEK, DATEDIFF(WEEK, 0, CAST( #year AS VARCHAR(4) )), 0) -- Will be a Monday
, AddWeeksToRoundedDate = DATEADD(WEEK, #WeekNum - 1, DATEADD(WEEK, DATEDIFF(WEEK, 0, CAST( #year AS VARCHAR(4) )), 0) )
, AddWeekDay = DATEADD( DAY, #WeekDay - 1, DATEADD(WEEK, #WeekNum - 1, DATEADD(WEEK, DATEDIFF(WEEK, 0, CAST( #year AS VARCHAR(4) )), 0) ) )
2012+
DECLARE #Year INT = 2014
DECLARE #WeekNum INT = 2
DECLARE #WeekDay INT = 2
SELECT
BaseDate = DATEFROMPARTS(#Year, 1, 1)
, RoundToWeekStart = DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEFROMPARTS(#Year, 1, 1)), 0) -- Will be a Monday
, AddWeeksToRoundedDate = DATEADD(WEEK, #WeekNum - 1, DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEFROMPARTS(#Year, 1, 1)), 0) )
, AddWeekDay = DATEADD( DAY, #WeekDay - 1, DATEADD(WEEK, #WeekNum - 1, DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEFROMPARTS(#Year, 1, 1)), 0) ) )
I have the date value like this - 12/2011 or 11/2011 (MM/yyyy)
I want to find the sundays on the particular month....
For Example, If i select the month 01/2012, The query should give the result like this
01
08
15
22
29
The above date are sunday.
Expected Output for 01/2012 Month
01
08
15
22
29
How to make a query
Need Query Help
With a little help of a number table (master..spt_values)
declare #M varchar(7)
set #M = '01/2012'
declare #D datetime
set #D = convert(datetime, '01/'+#M, 103)
set datefirst 7
select dateadd(day, N.Number, #D)
from master..spt_values as N
where N.type = 'P' and
dateadd(day, N.Number, #D) >= #D and
dateadd(day, N.Number, #D) < dateadd(month, 1, #D) and
datepart(weekday, dateadd(day, N.Number, #D)) = 1
Here it comes:
SET DATEFIRST 1
DECLARE #givenMonth CHAR(7)
SET #givenMonth = '12/2011'
DECLARE #month INT
SET #month = CAST(SUBSTRING(#givenMonth, 1, 2) AS INT)
DECLARE #year INT
SET #year = CAST(SUBSTRING(#givenMonth, 4, 4) AS INT)
DECLARE #Date DATETIME
SET #Date = DATEADD(month, #month - 1, CAST(CAST(#year AS CHAR(4)) AS DATETIME))
DECLARE #nextMonth DATETIME
SET #nextMonth = DATEADD(MONTH, 1, #date)
DECLARE #firstSunday DATETIME
SET #firstSunday = DATEADD(day, 7 - DATEPART(weekday, #date), #date)
CREATE TABLE #Days(Sunday INT)
WHILE #firstSunday < #nextMonth
BEGIN
INSERT #Days
SELECT DATEPART(DAY, #firstSunday) Sunday
SET #firstSunday = DATEADD(day, 7, #firstSunday)
END
SELECT Sunday
FROM #Days
ORDER BY 1
DROP TABLE #Days
Edit: use a numbers table in place of the CTE because you are in SQL Server 2000. C'mon, upgrade and do yourself a favour
DECLARE #monthyear varchar(10) = '11/2012';
DECLARE #start smalldatetime, #end smalldatetime;
-- use yyyymmdd format
SET #start = CAST(RIGHT(#monthyear, 4)+ LEFT(#monthyear, 2) + '01' AS smalldatetime);
-- work backwards from start of next month
SET #end = DATEADD(day, -1, DATEADD(month, 1, #start));
-- recursive CTE. Would be easier with a numbers table
;WITH daycte AS
(
SELECT #start AS TheDay
UNION ALL
SELECT DATEADD(day, 1, TheDay)
FROM daycte
WHERE TheDay < #end
)
SELECT DATENAME(day, TheDay)
FROM daycte
-- One of many ways.
-- This is independent of ##datefirst but fails with Chinese and Japanese language settings
WHERE DATENAME(weekday, TheDay) = 'Sunday';
You can change date format and use DateName function.
SELECT DateName(dw, GETDATE())
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