Related
I have a table with a birthdate field in it. I need to be able to randomize the date and month but keep the year. Is this possible in TSQL?
That is, if the given date in the field is 1/1/2012 I would like something like:
RANDBETWEEN(1, 29) / RANDBETWEEN(1, 12) / 2012
You could take this approach, it does not randomize the month and day separately, but gives you a random day you can attach to your year.
DECLARE #year INT = 2012;
SELECT DATEADD(DAY, FLOOR(RAND() * 365), CAST(#year AS CHAR(4)) + '-01-01')
If you require leap-year checking for your randomization to include 12/31/LeapYear, you can use this code instead:
DECLARE #year INT = 2012;
DECLARE #daysToAdd INT = 365;
IF #year % 400 = 0
OR
(
#year % 100 != 0
AND #year % 4 = 0
)
BEGIN
SELECT #daysToAdd = 366;
END
SELECT DATEADD(DAY,FLOOR(RAND() * #daysToAdd),CAST(#year AS CHAR(4)) + '-01-01');
Below are two methods for 2012 using NEWID() and CRYPT_GEN_RANDOM() functions.
SELECT RandomDateUsingNewId = DATEADD(DAY, ABS(CHECKSUM(NEWID())) % 366, '1/1/2012')
, RandomDateUsingCryptGenRandom = DATEADD(DAY, CONVERT(INT, CRYPT_GEN_RANDOM(2)) % 366, '1/1/2012');
If you want code that works for years with our without leap year (not just 2012), then here's a modified set of code that calculates the days in the year based on the selected year.
DECLARE #Year INT = 2012;
DECLARE #StartingDateOfYear DATE = CONVERT(DATE, CONCAT('1/1/', #Year));
DECLARE #DaysInYear INT = DATEPART(DAYOFYEAR, DATEADD(DAY, -1, DATEADD(YEAR, 1, #StartingDateOfYear)));
SELECT RandomDateUsingNewId = DATEADD(DAY, ABS(CHECKSUM(NEWID())) % #DaysInYear, #StartingDateOfYear)
, RandomDateUsingCryptGenRandom = DATEADD(DAY, CONVERT(INT, CRYPT_GEN_RANDOM(2)) % #DaysInYear, #StartingDateOfYear);
declare #t table(birthdate date)
insert #t values('20130101'), ('20120505')
update t
set birthdate = v.newbirthdate
from #t t
cross apply
(
select top 1
dateadd(year, datediff(year, 0,t.birthdate), number) newbirthdate
from master..spt_values
where type = 'P' and
number < datepart(dayofyear, dateadd(year, datediff(year, -1,t.birthdate), -1))
order by newid()
) v
select * from #t
Something like this perhaps
WITH cte AS
(
SELECT CAST('2012-01-01' AS DATE) ShortDate
UNION ALL
SELECT DATEADD(D, 1, ShortDate)
FROM cte
WHERE ShortDate < '2012-12-31'
)
SELECT TOP 10 ShortDate
FROM cte
ORDER BY NEWID()
OPTION (MAXRECURSION 0)
I have a table called FcData and the data looks like:
Op_Date
2011-02-14 11:53:40.000
2011-02-17 16:02:19.000
2010-02-14 12:53:40.000
2010-02-17 14:02:19.000
I am looking to get the Number of weeks in That Month from Op_Date. So I am looking for output like:
Op_Date Number of Weeks
2011-02-14 11:53:40.000 5
2011-02-17 16:02:19.000 5
2010-02-14 12:53:40.000 5
2010-02-17 14:02:19.000 5
This page has some good functions to figure out the last day of any given month: http://www.sql-server-helper.com/functions/get-last-day-of-month.aspx
Just wrap the output of that function with a DATEPART(wk, last_day_of_month) call. Combining it with an equivalent call for the 1st-day-of-week will let you get the number of weeks in that month.
Use this to get the number of week for ONE specific date. Replace GetDate() by your date:
declare #dt date = cast(GetDate() as date);
declare #dtstart date = DATEADD(day, -DATEPART(day, #dt) + 1, #dt);
declare #dtend date = dateadd(DAY, -1, DATEADD(MONTH, 1, #dtstart));
WITH dates AS (
SELECT #dtstart ADate
UNION ALL
SELECT DATEADD(day, 1, t.ADate)
FROM dates t
WHERE DATEADD(day, 1, t.ADate) <= #dtend
)
SELECT top 1 DatePart(WEEKDAY, ADate) weekday, COUNT(*) weeks
FROM dates d
group by DatePart(WEEKDAY, ADate)
order by 2 desc
Explained: the CTE creates a result set with all dates for the month of the given date. Then we query the result set, grouping by week day and count the number of occurrences. The max number will give us how many weeks the month overlaps (premise: if the month has 5 Mondays, it will cover five weeks of the year).
Update
Now, if you have multiple dates, you should tweak accordingly, joining your query with the dates CTE.
Here is my take on it, might have missed something.
In Linq:
from u in TblUsers
let date = u.CreateDate.Value
let firstDay = new DateTime(date.Year, date.Month, 1)
let lastDay = firstDay.AddMonths(1)
where u.CreateDate.HasValue
select Math.Ceiling((lastDay - firstDay).TotalDays / 7)
And generated SQL:
-- Region Parameters
DECLARE #p0 Int = 1
DECLARE #p1 Int = 1
DECLARE #p2 Float = 7
-- EndRegion
SELECT CEILING(((CONVERT(Float,CONVERT(BigInt,(((CONVERT(BigInt,DATEDIFF(DAY, [t3].[value], [t3].[value2]))) * 86400000) + DATEDIFF(MILLISECOND, DATEADD(DAY, DATEDIFF(DAY, [t3].[value], [t3].[value2]), [t3].[value]), [t3].[value2])) * 10000))) / 864000000000) / #p2) AS [value]
FROM (
SELECT [t2].[createDate], [t2].[value], DATEADD(MONTH, #p1, [t2].[value]) AS [value2]
FROM (
SELECT [t1].[createDate], CONVERT(DATETIME, CONVERT(NCHAR(2), DATEPART(Month, [t1].[value])) + ('/' + (CONVERT(NCHAR(2), #p0) + ('/' + CONVERT(NCHAR(4), DATEPART(Year, [t1].[value]))))), 101) AS [value]
FROM (
SELECT [t0].[createDate], [t0].[createDate] AS [value]
FROM [tblUser] AS [t0]
) AS [t1]
) AS [t2]
) AS [t3]
WHERE [t3].[createDate] IS NOT NULL
According to this MSDN article: http://msdn.microsoft.com/en-us/library/ms174420.aspx you can only get the current week in the year, not what that month returns.
There may be various approaches to implementing the idea suggested by #Marc B. Here's one, where no UDFs are used but the first and the last days of month are calculated directly:
WITH SampleData AS (
SELECT CAST('20110214' AS datetime) AS Op_Date
UNION ALL SELECT '20110217'
UNION ALL SELECT '20100214'
UNION ALL SELECT '20100217'
UNION ALL SELECT '20090214'
UNION ALL SELECT '20090217'
),
MonthStarts AS (
SELECT
Op_Date,
MonthStart = DATEADD(DAY, 1 - DAY(Op_Date), Op_Date)
/* alternatively: DATEADD(MONTH, DATEDIFF(MONTH, 0, Op_Date), 0) */
FROM FcData
),
Months AS (
SELECT
Op_Date,
MonthStart,
MonthEnd = DATEADD(DAY, -1, DATEADD(MONTH, 1, MonthStart))
FROM FcData
)
Weeks AS (
SELECT
Op_Date,
StartWeek = DATEPART(WEEK, MonthStart),
EndWeek = DATEPART(WEEK, MonthEnd)
FROM MonthStarts
)
SELECT
Op_Date,
NumberOfWeeks = EndWeek - StartWeek + 1
FROM Weeks
All calculations could be done in one SELECT, but I chose to split them into steps and place every step in a separate CTE so it could be seen better how the end result was obtained.
You can get number of weeks per month using the following method.
Datepart(WEEK,
DATEADD(DAY,
-1,
DATEADD(MONTH,
1,
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())))
-
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())
+1
)
Here how you can get accurate amount of weeks:
DECLARE #date DATETIME
SET #date = GETDATE()
SELECT ROUND(cast(datediff(day, dateadd(day, 1-day(#date), #date), dateadd(month, 1, dateadd(day, 1-day(#date), #date))) AS FLOAT) / 7, 2)
With this code for Sep 2014 you'll get 4.29 which is actually true since there're 4 full weeks and 2 more days.
I need a select to return Month and year Within a specified date range where I would input the start year and month and the select would return month and year from the date I input till today.
I know I can do this in a loop but I was wondering if it is possible to do this in a series selects?
Year Month
---- -----
2010 1
2010 2
2010 3
2010 4
2010 5
2010 6
2010 7
and so on.
Gosh folks... using a "counting recursive CTE" or "rCTE" is as bad or worse than using a loop. Please see the following article for why I say that.
http://www.sqlservercentral.com/articles/T-SQL/74118/
Here's one way to do it without any RBAR including the "hidden RBAR" of a counting rCTE.
--===== Declare and preset some obviously named variables
DECLARE #StartDate DATETIME,
#EndDate DATETIME
;
SELECT #StartDate = '2010-01-14', --We'll get the month for both of these
#EndDate = '2020-12-05' --dates and everything in between
;
WITH
cteDates AS
(--==== Creates a "Tally Table" structure for months to add to start date
-- calulated by the difference in months between the start and end date.
-- Then adds those numbers to the start of the month of the start date.
SELECT TOP (DATEDIFF(mm,#StartDate,#EndDate) + 1)
MonthDate = DATEADD(mm,DATEDIFF(mm,0,#StartDate)
+ (ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1),0)
FROM sys.all_columns ac1
CROSS JOIN sys.all_columns ac2
)
--===== Slice each "whole month" date into the desired display values.
SELECT [Year] = YEAR(MonthDate),
[Month] = MONTH(MonthDate)
FROM cteDates
;
I know this is an old question, but I'm mildly horrified at the complexity of some of the answers. Using a CTE is definitely the simplest way to go for selecting these values:
WITH months(dt) AS
(SELECT getdate() dt
UNION ALL
SELECT dateadd(month, -1, dt)
FROM months)
SELECT
top (datediff(month, '2017-07-01' /* start date */, getdate()) + 1)
YEAR(months.dt) yr, MONTH(months.dt) mnth
FROM months
OPTION (maxrecursion 0);
Just slap in whichever start date you'd like in place of the '2017-07-01' above and you're good to go with an efficient and easily-integrated solution.
Edit: Jeff Moden's answer quite effectively advocates against using rCTEs. However, in this case it appears to be a case of premature optimization - we're talking about 10's of records in all likelihood, and even if you span back to 1900 from today, it's still a minuscule hit. Using rCTEs to achieve code maintainability seems to be worth the trade if the expected result set is small.
You can use something like this: Link
To generate the equivalent of a numbers table using date ranges.
But could you please clarify your inputs and outputs?
Do you want to input a start date, for example, '2010-5-1' and end date, for example, '2010-8-1' and have it return every month between the two? Do you want to include the start month and end month, or exclude them?
Here's some code that I wrote that will quickly generate an inclusive result of every month between two dates.
--Inputs here:
DECLARE #StartDate datetime;
DECLARE #EndDate datetime;
SET #StartDate = '2010-1-5 5:00PM';
SET #EndDate = GETDATE();
--Procedure here:
WITH RecursiveRowGenerator (Row#, Iteration) AS (
SELECT 1, 1
UNION ALL
SELECT Row# + Iteration, Iteration * 2
FROM RecursiveRowGenerator
WHERE Iteration * 2 < CEILING(SQRT(DATEDIFF(MONTH, #StartDate, #EndDate)+1))
UNION ALL
SELECT Row# + (Iteration * 2), Iteration * 2
FROM RecursiveRowGenerator
WHERE Iteration * 2 < CEILING(SQRT(DATEDIFF(MONTH, #StartDate, #EndDate)+1))
)
, SqrtNRows AS (
SELECT *
FROM RecursiveRowGenerator
UNION ALL
SELECT 0, 0
)
SELECT TOP(DATEDIFF(MONTH, #StartDate, #EndDate)+1)
DATEADD(month, DATEDIFF(month, 0, #StartDate) + A.Row# * POWER(2,CEILING(LOG(SQRT(DATEDIFF(MONTH, #StartDate, #EndDate)+1))/LOG(2))) + B.Row#, 0) Row#
FROM SqrtNRows A, SqrtNRows B
ORDER BY A.Row#, B.Row#;
Code below generates the values for the range between 21 Jul 2013 and 15 Jan 2014.
I usually use it in SSRS reports for generating lookup values for the Month parameter.
declare
#from date = '20130721',
#to date = '20140115';
with m as (
select * from (values ('Jan', '01'), ('Feb', '02'),('Mar', '03'),('Apr', '04'),('May', '05'),('Jun', '06'),('Jul', '07'),('Aug', '08'),('Sep', '09'),('Oct', '10'),('Nov', '11'),('Dec', '12')) as t(v, c)),
y as (select cast(YEAR(getdate()) as nvarchar(4)) [v] union all select cast(YEAR(getdate())-1 as nvarchar(4)))
select m.v + ' ' + y.v [value_field], y.v + m.c [label_field]
from m
cross join y
where y.v + m.c between left(convert(nvarchar, #from, 112),6) and left(convert(nvarchar, #to, 112),6)
order by y.v + m.c desc
Results:
value_field label_field
---------------------------
Jan 2014 201401
Dec 2013 201312
Nov 2013 201311
Oct 2013 201310
Sep 2013 201309
Aug 2013 201308
Jul 2013 201307
you can do the following
SELECT DISTINCT YEAR(myDate) as [Year], MONTH(myDate) as [Month]
FROM myTable
WHERE <<appropriate criteria>>
ORDER BY [Year], [Month]
---Here is a version that gets the month end dates typically used for accounting purposes
DECLARE #StartDate datetime;
DECLARE #EndDate datetime;
SET #StartDate = '2010-1-1';
SET #EndDate = '2020-12-31';
--Procedure here:
WITH RecursiveRowGenerator (Row#, Iteration)
AS ( SELECT 1, 1
UNION ALL
SELECT Row# + Iteration, Iteration * 2
FROM RecursiveRowGenerator
WHERE Iteration * 2 < CEILING(SQRT(DATEDIFF(MONTH, #StartDate, #EndDate)+1))
UNION ALL SELECT Row# + (Iteration * 2), Iteration * 2
FROM RecursiveRowGenerator
WHERE Iteration * 2 < CEILING(SQRT(DATEDIFF(MONTH, #StartDate, #EndDate)+1)) )
, SqrtNRows AS ( SELECT * FROM RecursiveRowGenerator
UNION ALL SELECT 0, 0 )
SELECT TOP(DATEDIFF(MONTH, #StartDate, #EndDate)+1)
DateAdd(d,-1,DateAdd(m,1, DATEADD(month, DATEDIFF(month, 0, #StartDate) + A.Row# * POWER(2,CEILING(LOG(SQRT(DATEDIFF(MONTH, #StartDate, #EndDate)+1))/LOG(2))) + B.Row#, 0) ))
Row# FROM SqrtNRows A, SqrtNRows B ORDER BY A.Row#, B.Row#;
DECLARE #Date1 DATE
DECLARE #Date2 DATE
SET #Date1 = '20130401'
SET #Date2 = DATEADD(MONTH, 83, #Date1)
SELECT DATENAME(MONTH, #Date1) "Month", MONTH(#Date1) "Month Number", YEAR(#Date1) "Year"
INTO #Month
WHILE (#Date1 < #Date2)
BEGIN
SET #Date1 = DATEADD(MONTH, 1, #Date1)
INSERT INTO #Month
SELECT DATENAME(MONTH, #Date1) "Month", MONTH(#Date1) "Month Number", YEAR(#Date1) "Year"
END
SELECT * FROM #Month
ORDER BY [Year], [Month Number]
DROP TABLE #Month
declare #date1 datetime,
#date2 datetime,
#date datetime,
#month integer,
#nm_bulan varchar(20)
create table #month_tmp
( bulan integer null, keterangan varchar(20) null )
select #date1 = '2000-01-01',
#date2 = '2000-12-31'
select #month = month(#date1)
while (#month < 13)
Begin
IF #month = 1
Begin
SELECT #date = CAST( CONVERT(VARCHAR(25),DATEADD(dd,-(DAY(DATEADD(mm,0,#date1))-1),DATEADD(mm,0,#date1)),111) + ' 00:00:00' as DATETIME )
End
ELSE
Begin
SELECT #date = CAST( CONVERT(VARCHAR(25),DATEADD(dd,-(DAY(DATEADD(mm,#month -1,#date1))-1),DATEADD(mm,#month -1,#date1)),111) + ' 00:00:00' as DATETIME )
End
select #nm_bulan = DATENAME(MM, #date)
insert into #month_tmp
select #month as nilai, #nm_bulan as nama
select #month = #month + 1
End
select * from #month_tmp
drop table #month_tmp
go
How will you find last sunday of a month in sql 2000?
SELECT
DATEADD(day,DATEDIFF(day,'19000107',DATEADD(month,DATEDIFF(MONTH,0,GETDATE() /*YourValuehere*/),30))/7*7,'19000107')
Edit: A correct, final, working answer from my colleague.
select dateadd(day,1-datepart(dw, getdate()), getdate())
An alternative approach, borrowed from data warehousing practice. Create a date-dimension table and pre-load it for 10 years, or so.
TABLE dimDate (DateKey, FullDate, Day, Month, Year, DayOfWeek,
DayInEpoch, MonthName, LastDayInMonthIndicator, many more..)
The easiest way to fill-in the dimDate is to spend an afternoon with Excel and then import to DB from there. A half-decent dimDate table has 50+ columns -- anything you ever wanted to know about a date.
With this in place, the question becomes something like:
SELECT max(FullDate)
FROM dimDate
WHERE DayOfWeek = 'Sunday'
AND Month = 11
AND Year = 2009;
Essentially, all date related queries become simpler.
Next sunday in SQL, regardless which day is first day of week: returns 2011-01-02 23:59:59.000 on 22-dec-2010:
select DateADD(ss, -1, DATEADD(week, DATEDIFF(week, 0, getdate()), 14))
I find some of these solutions hard to understand so here's my version with variables to explain the steps.
ALTER FUNCTION dbo.fn_LastSundayInMonth
(
#StartDate DATETIME
,#RequiredDayOfWeek INT /* 1= Sunday */
)
RETURNS DATETIME
AS
/*
A detailed step by step way to get the answer...
SELECT dbo.fn_LastSundayInMonth(getdate()-31,1)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,2)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,3)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,4)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,5)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,6)
SELECT dbo.fn_LastSundayInMonth(getdate()-31,7)
*/
BEGIN
DECLARE #MonthsSince1900 INTEGER
DECLARE #NextMonth INTEGER
DECLARE #DaysToSubtract INTEGER
DECLARE #FirstDayOfNextMonth DATETIME
DECLARE #LastDayOfMonthDayOfWeek INTEGER
DECLARE #LastDayOfMonth DATETIME
DECLARE #ReturnValue DATETIME
SET #MonthsSince1900=DateDiff(month, 0, #StartDate)
SET #NextMonth=#MonthsSince1900+1
SET #FirstDayOfNextMonth = DateAdd(month,#NextMonth, 0)
SET #LastDayOfMonth = DateAdd(day, -1, #FirstDayOfNextMonth)
SET #ReturnValue = #LastDayOfMonth
WHILE DATEPART(dw, #ReturnValue) <> #RequiredDayOfWeek
BEGIN
SET #ReturnValue = DATEADD(DAY,-1, #ReturnValue)
END
RETURN #ReturnValue
END
DECLARE #LastDateOfMonth smalldatetime
SELECT #LastDateOfMonth = DATEADD(month, DATEDIFF(month, -1, GETDATE()), 0) -1
Select DATEADD(dd,-( CASE WHEN DATEPART(weekday,#LastDateOfMonth) = 1 THEN 0 ELSE DATEPART(weekday,#LastDateOfMonth) - 1 END ),#LastDateOfMonth)
Holy cow, this is ugly, but here goes:
DECLARE #dtDate DATETIME
SET #dtDate = '2009-11-05'
SELECT DATEADD(dd, -1*(DATEPART(dw, DateAdd(day, -1, DateAdd(month, DateDiff(month, 0, #dtDate)+1, 0)))-1),
DateAdd(day, -1, DateAdd(month, DateDiff(month, 0, #dtDate)+1, 0)))
First built a tally table.
http://www.sqlservercentral.com/articles/T-SQL/62867/
then get what you want..
http://www.sqlservercentral.com/Forums/Topic515226-1291-1.aspx
DECLARE #DateStart DATETIME,
#DateEnd DATETIME
SELECT #DateStart = '20080131',
#DateEnd = '20101201'
SELECT DATEADD(wk,DATEDIFF(wk,6,DATEADD(mm,DATEDIFF(mm,-1,DATEADD(mm,t.N-1,#DateStart)),-1)),6)
FROM dbo.Tally t
WHERE t.N <= DATEDIFF(mm,#DateStart,#DateEnd)
Here's the correct way, accounting for ##DATEFIRST
IF NOT EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[fu_dtLastSundayInMonth]') AND type in (N'FN', N'IF', N'TF', N'FS', N'FT'))
BEGIN
EXECUTE(N'CREATE FUNCTION [dbo].[fu_dtLastSundayInMonth]() RETURNS int BEGIN RETURN 0 END ')
END
GO
/*
SET DATEFIRST 3; -- Monday
WITH CTE AS (
SELECT 1 AS i, CAST('20190101' AS datetime) AS mydate
UNION ALL
SELECT i+1 AS i, DATEADD(month, 1, CTE.mydate) AS mydate
FROM CTE WHERE i < 100
)
SELECT -666 AS i, dbo.fu_dtLastSundayInMonth('17530101') AS lastSundayInMonth, dbo.fu_dtLastSundayInMonth('17530101') AS Control
UNION ALL
SELECT -666 AS i, dbo.fu_dtLastSundayInMonth('99991231') AS lastSundayInMonth, dbo.fu_dtLastSundayInMonth('99991231') AS Control
UNION ALL
SELECT
mydate
,dbo.fu_dtLastSundayInMonth(mydate) AS lastSundayInMonth
,dbo.fu_dtLastSundayInMonth(mydate) AS lastSundayInMonth
,DATEADD(day,DATEDIFF(day,'19000107', DATEADD(MONTH, DATEDIFF(MONTH, 0, mydate, 30))/7*7,'19000107') AS Control
FROM CTE
*/
-- =====================================================================
-- Description: Return date of last sunday in month
-- of the same year and month as #in_DateTime
-- =====================================================================
ALTER FUNCTION [dbo].[fu_dtLastSundayInMonth](#in_DateTime datetime )
RETURNS DateTime
AS
BEGIN
-- Abrunden des Eingabedatums auf 00:00:00 Uhr
DECLARE #dtReturnValue AS DateTime
-- 26.12.9999 SO
IF #in_DateTime >= CAST('99991201' AS datetime)
RETURN CAST('99991226' AS datetime);
-- #dtReturnValue is now last day of month
SET #dtReturnValue = DATEADD
(
DAY
,-1
,DATEADD
(
MONTH
,1
,CAST(CAST(YEAR(#in_DateTime) AS varchar(4)) + RIGHT('00' + CAST(MONTH(#in_DateTime) AS varchar(2)), 2) + '01' AS datetime)
)
)
;
-- SET DATEFIRST 1 -- Monday - Super easy !
-- SET DATEFIRST != 1 - PHUK THIS !
SET #dtReturnValue = DATEADD
(
day
,
-
(
(
-- DATEPART(WEEKDAY, #lastDayofMonth) -- with SET DATEFIRST 1
DATEPART(WEEKDAY, #dtReturnValue) + ##DATEFIRST - 2 % 7 + 1
)
%7
)
, #dtReturnValue
);
RETURN #dtReturnValue;
END
GO
select next_day(last_day(sysdate)-7, 'Sunday') from dual
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