I have a week number and year and i need to display "total for mm/dd/yy to mm/dd/yy in a row of my ssrs report. My week starts with Monday. For example if my week number is '2' and year is '2010' then I have to display "total for 01/04/2010 to 01/10/2010 in my ssrs column. how to do this?
Try this
declare #year char(4) = '2010'
declare #week int = 2
declare #fromdate datetime
declare #todate datetime
set #fromdate = DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + #year) + (#week-1), 7);
set #todate = DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + #year) + (#week-1), 6) ;
;WITH dates AS
(
SELECT CONVERT(datetime,#fromDate) as Date
UNION ALL
SELECT DATEADD(d,1,[Date])
FROM dates
WHERE DATE < #toDate
)
select * from dates
SQL Server has a DATEPART function which calculates the ordinal week number of a year. However, you have to call DATEFIRST before this to define which day of the week represents the start of the week. In your case, you have stated that the start of your week is Monday (i.e. 1).
SET DATEFIRST 1;
SELECT SUM([your data column])
FROM [your table]
WHERE DATEPART(WEEKNUM, [your date column])=[your week parameter]
AND DATEPART(YEAR, [your date column])=[your year parameter]
Your description is not american standard nor isoweek. Seems like a mix of those. I never heard of that as a standard. It is nearly isoweek. So that is what this answer is based on.
Calculating iso year is a bit tricky, you can read about it here:
This is the syntax you need:
DECLARE #year int = 2010
DECLARE #week int = 2
;WITH CTE AS
(
SELECT
dateadd(wk, datediff(wk, - #week * 7,
cast(cast(#year as char(4)) as datetime) - 5), 0) startofweek
)
SELECT
replace('total for ' + convert(char(10), startofweek, 110)
+ ' to ' + convert(char(10), dateadd(day, 6, startofweek) , 110), '-', '/')
FROM CTE
Result:
total for 01/11/2010 to 01/17/2010
Isoweek 2 in 2010 is 2010-01-11
Try setting DATEFIRST (https://msdn.microsoft.com/en-ie/library/ms181598.aspx)
SET DATEFIRST 7
declare #wk int
declare #yr int
declare #EndOfWeek as datetime
set #wk = 2
set #yr = 2010
SET #EndOfWeek = dateadd (week, #wk, dateadd (year, #yr-1900, 0)) + 1 - datepart(dw, dateadd (week, #wk, dateadd (year, #yr-1900, 0)) )
SELECT
replace('total for ' + convert(char(10), dateadd(day, -6, #EndOfWeek) , 110)
+ ' to ' + convert(char(10), #EndOfWeek, 110), '-', '/')
The result:
total for 01/04/2010 to 01/10/2010
I hope it helps:
declare #year char(4) = '2014'
declare #week int = 2
select dateadd(week,#week,convert(date,#year+'-01-01',121))
Change the date format appropriate for you from this list
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())
I need to write a function or SP that will return the first occurance of the 15th. For example, if I pass the date as May 8th, then it should return May 15th. If I pass May 30th, then it should return June 15th.
One way
DECLARE #d DATETIME
SELECT #d = '20110508'
--SELECT #d = '20110530'
SELECT CASE WHEN DAY(#d) > 15
THEN dateadd(mm, datediff(mm, 0, #d)+1, 0) + 14
ELSE dateadd(mm, datediff(mm, 0, #d)+0, 0)+ 14 end
How about;
create function udf_getNextDate(#base datetime, #day int) returns datetime as begin
set #base = case when day(#base) > #day
then dateadd(month, 1, #base)
else #base
end
return dateadd(day, -day(#base) + #day, #base)
end
select
dbo.udf_getNextDate('08 may 2011', 15),
dbo.udf_getNextDate('30 may 2011', 15),
dbo.udf_getNextDate('16 dec 2011', 15),
dbo.udf_getNextDate('01 may 2011', 15)
2011-05-15 00:00:00.000
2011-06-15 00:00:00.000
2012-01-15 00:00:00.000
2011-05-15 00:00:00.000
Just another way of doing it:
Declare #d datetime
Set #d = getdate()
Select Case
When DateDiff(Day, Day(#d), 15) < 0 then
DateAdd(month, 1, DateAdd(Day, DateDiff(Day, Day(#d), 15), #d))
Else DateAdd(Day, DateDiff(Day, Day(#d), 15), #d)
End as [Next15th]
this function may helps you
create function Get15th(#date datetime)
returns datetime
as
begin
declare #resultdate datetime
declare #y int
declare #m int
declare #d int
set #y = datepart(year,#date)
set #m = datepart(month,#date)
set #d = datepart(day,#date)
if( #d<=15)
set #resultdate =cast((str(#y)+'-'+str(#m)+'-15') as datetime)
else
set #resultdate =cast((str(#y)+'-'+str(#m+1)+'-15') as datetime)
return #resultdate
end
try
DATEADD(Day, DATEDIFF(Day, 15, Created), 0) AS CreatedDay
explained here http://improve.dk/archive/2006/12/13/sql-server-datetime-rounding-made-easy.aspx
using that link you can achive what you want
UPDATE: got it wrong ... see Dems answer
Rob's answer is close...
if you take the datediff in months, rather than days, and make your base the 15'th of a month, then add one extra month...
DATEADD(MONTH, DATEDIFF(MONTH, 14, Created) + 1, 14)
EDIT
Modified to use DATEDIFF MONTH how it works, not how I thought it should work ;)
DATEADD(MONTH, DATEDIFF(MONTH, 0, Created - 15) + 1, 14)
I usually do something like this:
declare
#date datetime ,
#target_day int
set #date = 'June 16, 2009'
set #target_day = 15
select date = #date ,
next_date = case when day(#date) <= #target_day
then dateadd(day,15-day(#date),#date)
else dateadd(day,15,dateadd(month,1,dateadd(day,-day(#date),#date)))
end
dealing with the last day of the month is a wee bit trickier, since month's have variable numbers of days, and SQL Server's date math is sometimes a wee bit baroque.
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