How to determine the number of days in a month in SQL Server? - sql

Question

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?

Answer 1

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

Answer 2

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)))

Answer 3

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

Answer 4

--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...

Answer 5

I would suggest:
SELECT DAY(EOMONTH(GETDATE()))

Answer 6

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.

Answer 7

--- 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()))

Answer 8

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]

Answer 9

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

Answer 10

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]

Answer 11

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

Answer 12

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.

Answer 13

select add_months(trunc(sysdate,'MM'),1) - trunc(sysdate,'MM') from dual;

Answer 14

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)

Answer 15

here's another one...
Select Day(DateAdd(day, -Day(DateAdd(month, 1, getdate())),
DateAdd(month, 1, getdate())))

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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]

Answer 22

RETURN day(dateadd(month, 12 * #year + #month - 22800, -1))
select day(dateadd(month, 12 * year(date) + month(date) - 22800, -1))

Answer 23

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

Answer 24

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

Answer 25

For any date
select DateDiff(Day,#date,DateAdd(month,1,#date))

Answer 26

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

Answer 27

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

Answer 28

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)))

Answer 29

simple query in SQLServer2012 :
select day(('20-05-1951 22:00:00'))
i tested for many dates and it return always a correct result