I have fallen a problem to set date first into my SQL query. Here is the code below where I have set by using IF, Else IF. But I need one sql statement not IF, Else IF.
DECLARE #CurrentDate DATE, #FirstDayOfWeek INT
SELECT #CurrentDate = DATEADD(yy, DATEDIFF(yy, 0, GETDATE()), 0)
SELECT #FirstDayOfWeek = DATEPART(DW,#CurrentDate)
IF #FirstDayOfWeek = 6
SET DATEFIRST 5; --friday
ELSE IF #FirstDayOfWeek = 7
SET DATEFIRST 6; --sunday
--AND SO ON
The result of DATEPART(DW, . . .) depends on the current value of ##DATEFIRST, so you need to take that into account when setting the new value of ##DATEFIRST. The simplest thing would be to set ##DATEFIRST back to 1, get the current day of the week, and then set ##DATEFIRST to that.
By the way, if you just want the date of a date-time, you can just cast it to DATE. Ultimately, though, there's no need to convert the date-time to a date, because the result of DATEPART(DW, . . .) will be the same either way.
DECLARE #CurrentDate DATE, #CurrentDayOfWeek INT
SET #CurrentDate = CAST(GETDATE() AS DATE)
SET DATEFIRST 1 -- normalize first day of week
SET #CurrentDayOfWeek = DATEPART(DW, #CurrentDate)
SET DATEFIRST #CurrentDayOfWeek
Or
DECLARE #CurrentDayOfWeek INT
SET DATEFIRST 1 -- normalize first day of week
SET #CurrentDayOfWeek = DATEPART(DW, GETDATE())
SET DATEFIRST #CurrentDayOfWeek
Example:
DECLARE #CurrentDayOfWeek INT
SET DATEFIRST 1 -- normalize first day of week to Monday
SET #CurrentDayOfWeek = DATEPART(DW, '2016-01-01') -- a Friday, which will be 5 because first day of week is currently 1
SET DATEFIRST #CurrentDayOfWeek -- first day of week is now 5 (Friday)
I know your question mentions DATEFIRST, but is there any reason you don't just calculate it based on how many days it has been since the first of the year?
SELECT ((DATEPART(DY, GETDATE()) - 1) % 7) + 1
old answer:
Just got to mention that if DATEFIRST has been changed from the default of 7, this will no longer work since DATEPART(dw,...) will return a different value. The value used with SET DATEFIRST is always 1 = Monday, 7 = Sunday, etc. You can always to a SET DATEFIRST 7 ahead of time...
Basically what you are doing is subtracting 1 and rolling back over to 7 if you hit 0. You can just use the mod operator (%) for that:
See if this works for you
DECLARE #CurrentDate DATETIME, #FirstDayOfWeek INT
SELECT #CurrentDate = DATEADD(yy, DATEDIFF(yy, 0, GETDATE()), 0)
SELECT #FirstDayOfWeek = DATEPART(DW,#CurrentDate)
SELECT #FirstDayOfWeek = (SELECT CASE #FirstDayOfWeek
WHEN 6 THEN 5
WHEN 7 THEN 6
END )
SET DATEFIRST #FirstDayOfWeek
I am trying to get the specific day of a Year.
Here's what I have tried till now:-
-- Declare few variables
DECLARE #Currentdate AS DATETIME
DECLARE #DueDate AS DATETIME
DECLARE #NewDate AS DATETIME
-- Set the variables properly, just for testing
SET #Currentdate = GETDATE()
SET #DueDate = DATEADD(MONTH, 2, DATEADD(YEAR, 1, #Currentdate))
-- Check the output
SELECT #Currentdate -- 2013-09-30 00:00:00.000
SELECT #DueDate -- 2014-11-30 00:00:00.000
So, I want to get the #NewDate based on the #Currentdate year.
For this I tried:-
SELECT #NewDate = DATEADD(DAY, DAY(DATEDIFF(day, 1, #DueDate)), DATEADD(MONTH, DATEDIFF(MONTH, 0, #Currentdate), 0))
SELECT #NewDate -- 2013-09-30 00:00:00.000
But it didn't worked. :(
My expected result is like:
-- 2013-11-30 00:00:00.000
-- Having the due date month and date same, but the year as current date one.
Any help is appreciated!
UPDATE
Sorry for all the confusion I have created. My question in simple words is:-
I want to get the a new date variable having the date and the month same as #DueDate variable but the year as given in the #Currentdate variable.
I hope that would clear things up a bit.
If the question is "given I have a particular datetime value in one variable, can I set another variable to be for the same day and month but in the current year" then the answer would be:
declare #DueDate datetime
declare #NewDate datetime
set #DueDate = '20141130'
--Need to set #NewDate to the same month and day in the current year
set #NewDate = DATEADD(year,
--Here's how you work out the offset
DATEPART(year,CURRENT_TIMESTAMP) - DATEPART(year,#DueDate),
#DueDate)
select #DueDate,#NewDate
I want to get the a new date variable having the date and the month same as #DueDate variable but the year as given in the #Currentdate variable.
Well, that's simply the above query with a single tweak:
set #NewDate = DATEADD(year,
--Here's how you work out the offset
DATEPART(year,#Currentdate) - DATEPART(year,#DueDate),
#DueDate)
Try this instead ?
DECLARE #Currentdate AS DATETIME
DECLARE #DueDate AS DATETIME
-- Set the variables properly, just for testing
SET #Currentdate = GETDATE()
SET #DueDate = DATEADD(MONTH, 2, DATEADD(YEAR, 1,
DateAdd(day, datediff(day, 0, #currentDate), 0)))
-- Check the output
SELECT #Currentdate -- 2013-09-30 18:32:35.310
SELECT #DueDate
Using DateAdd(day, datediff(day, 0, #DateTime), 0) strips off the time portion. You should also check out this SO Question/answer.
Try this one:
CAST(CAST( -- cast INT to VARCHAR and then to DATE
YEAR(GETDATE()) * 10000 + MONTH(#DueDate) * 100 + DAY(#DueDate) -- convert current year + #DueDate's month and year parts to YYYYMMDD integer representation
+ CASE -- change 29th of February to 28th if current year is a non-leap year
WHEN MONTH(#DueDate) = 2 AND DAY(#DueDate) = 29 AND ((YEAR(GETDATE()) % 4 = 0 AND YEAR(GETDATE()) % 100 <> 0) OR YEAR(GETDATE()) % 400 = 0) THEN 0
ELSE -1
END
AS VARCHAR(8)) AS DATE)
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
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