Is there a way to get the number of days in a month in SQL Server, if we input the month number or month name, or even a date?
You can use:
select day(eomonth ('2018-02-01')) as NoOfDays
and the result will be:
NoOfDays
-----------
28
If you have a date, then simply do:
select day(eomonth(date))
If you have a month number, then:
select day(eomonth(datefromparts(2020, month_number, 1)))
If you have a date and are on 2012 or later :
SELECT day(eomonth(yourdate))
Month name / number is automatically prone to an error when dealing with February - do you consider it 28 or 29, which year are you referring to when making that calculation etc.
In case you are using sql-server 2008 or earlier:
Date as input
DECLARE #date DATETIME = getdate()
SELECT day(dateadd(m, datediff(m, -1, #date), -1))
Month and year as input
DECLARE #year INT = 2018
DECLARE #month INT = 2
SELECT day(dateadd(m, #month + datediff(m, 0, cast(#year as char(4))), -1))
try using
SELECT day(eomonth(yourdate))
for pre 2012 where eomonth() is not available
if you have a date
select day(dateadd(month, datediff(month, 0, #date) + 1, -1))
if you have the year & month
declare #year int = 2018,
#month int = 8
select dateadd(month, #month, dateadd(year, #year - 1900, -1))
I created a function "ufngetFirstDateOfMonth" and "ufngetLastDateOfMonth" stored in Microsoft SQL Server 2008. My purpose is to send some date into the function and it will return the first date of month with '00:00:00' or the last date of month with '23:59:59'.
I call the function like this:
exec ufngetLastDateOfMonth('2014-10-15')
and normally it returns '2014-10-31 23:59:59'
but when I send the last date of months that have 31 days (august, january,...):
exec ufngetLastDateOfMonth('2014-10-31')
it return '2014-10-30 23:59:59' whick is not correct Actally, it should be '2014-10-31 23:59:59'
Something goes wrong here...
This is my function:
CREATE FUNCTION [dbo].[ufnLastDateOfMonth](#Date date)
RETURNS varchar(50)
AS
BEGIN
DECLARE #New_Date varchar(50)
select #New_date = cast(dateadd(dd,-(DAY(#Date )),DATEADD(mm,1,#Date ))as varchar(50)) + ' 23:59:59'
RETURN #New_Date
END
To get the last day you can do this:
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,'2014-08-12')+1,0))
Adding to your function:
select #New_date = DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#date)+1,0))
Source:
SQL SERVER – Find Last Day of Any Month – Current Previous Next
For those who are using SQL Server 2012,
EOMONTH function could be an alternative.
DECLARE #date DATETIME = '12/1/2011';
SELECT EOMONTH ( #date ) AS Result;
GO
Source: https://msdn.microsoft.com/en-us/library/hh213020.aspx
Go to the first day of the month. Add one month. Then subtract one day. Or, in your case, one second:
select #New_date = dateadd(second, -1, dateadd(month, 1, dateadd(day, -(DAY(#Date) + 1)) ) )
You should use convert() if you want this as a string. I would instead suggest that the function return a datetime.
Use this
DECLARE #curdate datetime;
SET #curdate = GETDATE(); -- you can pass your date here
--SET #curdate = '2014-10-15'; Example
SELECT DATEADD(MONTH,1+DATEDIFF(MONTH,0,#curdate),-1);
OR
SELECT DATEADD(MONTH,1+DATEDIFF(MONTH,0,GETDATE()),-1);
you can try this
select DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, GETDATE()) + 1, 0)) As EndDateOfCurrentMonth
or this
select DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0) As StartDateOfCurrentMonth
Just try this:
SELECT EOMONTH('2000/2/1') Last_Date
Note: EOMONTH function is available in SQL Server version >= 2012
SELECT
#begindate = dateadd(mm,-1,CAST(datepart(mm,getdate()) AS VARCHAR(2)) + '/15/' + CAST(datepart(YYYY,getdate() -1) AS varchar(4))),
#enddate = CAST(datepart(mm,getdate()) AS VARCHAR(2)) + '/14/' + CAST(datepart(YYYY,getdate() -1) AS varchar(4))
Right now this code returns the dates May 15th - June 14th. Those are the required dates that I need, but what I also need is for it to be in the year 2013, and this returns the year 2014. I've tried doing a dateadd(yyyy, -1) and it didn't work. So I was wondering if anyone knew how I would be able to get last years date.
If DateAdd didn't work, there must have been an error in your implementation. Try it like this after your existing code:
SET #begindate = dateadd(year, -1, #begindate);
SET #enddate = dateadd(year, -1, #enddate)
declare #begindate datetime
declare #enddate datetime
set #begindate = dateadd(mm,-1,CAST(datepart(mm,getdate()) AS VARCHAR(2)) + '/15/' + CAST(datepart(YYYY, getdate()) - 1 AS varchar(4)))
set #enddate = CAST(datepart(mm,getdate() ) AS VARCHAR(2)) + '/14/' + CAST(datepart(YYYY, getdate()) - 1 AS varchar(4))
select #begindate
union all
select #enddate
Instead of working with strings you can use date variables and date arithmetic:
declare #lastyear date=dateadd(yy,-1,cast(getdate() as date))
declare #currentMonth date=dateadd(d,-day(#lastyear),#lastyear)
declare #prevMonth date=dateadd(m,-1,#currentMonth)
select #lastyear,DATEADD(d,14,#currentMonth),DATEADD(d,15,#prevMonth)
In SQL Server 2012 it's even easier because you can use the DATEFROMPARTS function to construct a new date from its parts.
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