Spent the better half of the day trying to figure this one out.
I want to get the datediff of two dates, based on their ISO week.
Here is my code:
SET DATEFIRST 1 ;
DECLARE #A date, #B date;
SET #A = '20180829'; -- August 29th
SET #B = '20180902'; -- September 2nd
SELECT DATEDIFF(WW, #A, #B )
If you check: http://whatweekisit.org/ (Week 35, Year 2018) you can see that it runs August 27 to September 2.
The code above will return a DateDiff = 1, which should be 0. Trying to run DateDiff on ISO week just returns the following error:
The datepart iso_week is not supported by date function datediff
I've tried taking out the week dateparts from the dates, but then I get the problem when comparing dates from different years.
Is there a way around this?
So after some reading, there's not a native way to do it. But here's a workaround:
DECLARE #A DATE = '20180829' -- August 29th
DECLARE #B DATE = '20180902' -- September 2nd
--We need to back each date up to the first day of its week.
DECLARE #A_FIRSTWEEKDAY DATE = DATEADD(DAY, -(DATEPART(WEEKDAY, #A) - 1), #A)
DECLARE #B_FIRSTWEEKDAY DATE = DATEADD(DAY, -(DATEPART(WEEKDAY, #B) - 1), #B)
/*The WEEKDAY part counts Sunday as day 1. If the original date was
a Sunday, it backed up zero days and it still needs to back up six
days. If it was any other day, it backed all the way up to Sunday
and now it needs to move forward one day.*/
IF DATEPART(WEEKDAY, #A) = 1
BEGIN SET #A_FIRSTWEEKDAY = DATEADD(DAY, -6, #A_FIRSTWEEKDAY) END
ELSE BEGIN SET #A_FIRSTWEEKDAY = DATEADD(DAY, 1, #A_FIRSTWEEKDAY) END
IF DATEPART(WEEKDAY, #B) = 1
BEGIN SET #B_FIRSTWEEKDAY = DATEADD(DAY, -6, #B_FIRSTWEEKDAY) END
ELSE BEGIN SET #B_FIRSTWEEKDAY = DATEADD(DAY, 1, #B_FIRSTWEEKDAY) END
--Now we can just difference the weeks.
SELECT DATEDIFF(DAY, #A_FIRSTWEEKDAY, #B_FIRSTWEEKDAY) / 7
This is because DATEDIFF always uses Sunday as the first day of the week to ensure the function operates in a deterministic way: https://learn.microsoft.com/datediff-transact-sql
So 20180902 (Sunday) is first day of next week.
SELECT DATEDIFF(ww, DATEADD(dd,-1, #A ), DATEADD(dd,-1,#B))
--Seems to do the trick?
Taken from: Number of weeks and partial weeks between two days calculated wrong
Though I cannot see why the other post in the link adds a 1 at the end.
I have a complicated calendar report coming from a SQL query where I have the year, month and WeekNumOfMonth (meaning the week number from the beginning of the month).
I need a function to return the DayOfMonth (ie: values of between 1-28,29,30 or 31 depending on the month) given Year, month and WeekNumOfMonth.
I have seen functions that do this given WeekNum from beginning of year but I only have WeekNum of the given month. I just can't seem to give the datePart and dateAdd functions right. Any help would be appreciated.
Note : Combination of Year ,Month ,Week Number will not give you the
day name because a week contain more than one day so how can you find
which day is it ? it is not possible
example : year 2015 month 06 week number 1
then it has a combination of
1 monday
2 tuesday
3 wednesday
4 thursday
5 friday
6 saturday
then with these results how can you predit the exact day ??
Else need some conditions or restrictions
If you have Year, Month , and date then you can find the day name using the below query it will give the start and end date of the week then you can take the dates from start date and end with end date
SELECT (
DATENAME(dw,
CAST(DATEPART(m, GETDATE()) AS VARCHAR)
+ '/'
+ CAST(DATEPART(d, '2015-06-23') AS VARCHAR)
+ '/'
+ CAST(DATEPART(yy, getdate()) AS VARCHAR))
)
Update
If you need all days between the current week then use the following query
DECLARE
#Year INT,
#Week INT,
#FirstDayOfYear DATETIME,
#FirstMondayOfYear DATETIME,
#StartDate DATETIME,
#EndDate DATETIME
SET #Year = 2015
SET #Week = 24// week number is based on year and it is not based on month
-- Get the first day of the provided year.
SET #FirstDayOfYear = CAST('1/1/' + CAST(#YEAR AS VARCHAR) AS DATETIME)
-- Get the first monday of the year, then add the number of weeks.
SET #FirstMondayOfYear = DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, #FirstDayOfYear), #FirstDayOfYear)), 0)
SET #StartDate = DATEADD(WEEK, #Week - 1, #FirstMondayOfYear)
-- Set the end date to one week past the start date.
SET #EndDate = DATEADD(WEEK, 1, #StartDate)
SELECT #StartDate AS StartDate, DATEADD(SECOND, -1, #EndDate) AS EndDate
Update
if do not have the year week number then use the below query
DATEPART(WEEK,'2015-06-01')
The above query is one example
In the query you have two values that is Year and Month
you need only the third value that means date
as a technique you can give 01 as the date then it will give the
starting week number your selected month
here it will result 23 so you got the idea in 2015 - month 06 starts with week number 23 so then you can easily do your stuff
Idea : if your week number is 1 then take 23 as week if it is 2 then 24 like wise..
How do i get the date for last friday of the month in T-SQL?
I will be passing the year and month as parameter,e.g, 201211.
If I pass '201211' as parameter it should return me '20121130' as answer as it's the date of last friday of month of november'12.
The 5 January 1900 was a Friday. This uses that a base date and calculates the last Friday in any given month though you must give it a date during the month rather than just the month itself. Replace the 2012-12-01 in this with a date in your month
SELECT DATEADD(DY,DATEDIFF(DY,'1900-01-05',DATEADD(MM,DATEDIFF(MM,0,'2012-12-01'),30))/7*7,'1900-01-05')
You can also use this to get the last Saturday by replacing the 1900-01-05 WITH 1900-01-06 etc.
This would be much simpler using a calendar table; after creating the appropriate columns for your own needs you can just write this:
select
max([Date])
from
dbo.Calendar
where
YearAndMonth = 201211 and
DayOfWeek = 'Friday'
A calendar table is generally a much better solution for determining dates than using functions because the code is much more readable and you can use your own definition of things like WeekNumber, FinancialQuarter etc. that vary widely between countries and even companies.
I created a scalar function for this:
create function [dbo].[lastDWMonth]
(
#y int
,#m int
,#dw int
)
returns date
as
begin
declare #d date
;with x as
(
select datefromparts(#y,#m,1) d
union all
select dateadd(day,1,d) from x where d < eomonth(datefromparts(#y,#m,1))
)
select
#d = max(d)
from
x
where
datepart(dw,d) = #dw
return #d
end
Declare #d1 datetime = '2019-12-23'
Declare #searchDay int = 2 -- monday
select DATEADD(DAY, #searchDay-DATEPART(WEEKday, DateADD(day,-1, DATEADD(month, DATEDIFF(MONTH, 0, #d1)+1, 0))),DateADD(day,-1, DATEADD(month, DATEDIFF(MONTH, 0, #d1)+1, 0)))
This will give you Date on last Monday of the month, you can change your search by changing value in #searchDay
I am trying to group records by week, storing the aggregated date as the first day of the week. However, the standard technique I use for rounding off dates does not appear to work correctly with weeks (though it does for days, months, years, quarters and any other timeframe I've applied it to).
Here is the SQL:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), 0);
This returns 2011-08-22 00:00:00.000, which is a Monday, not a Sunday. Selecting ##datefirst returns 7, which is the code for Sunday, so the server is setup correctly in as far as I know.
I can bypass this easily enough by changing the above code to:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), -1);
But the fact that I have to make such an exception makes me a little uneasy. Also, apologies if this is a duplicate question. I found some related questions but none that addressed this aspect specifically.
To answer why you're getting a Monday and not a Sunday:
You're adding a number of weeks to the date 0. What is date 0? 1900-01-01. What was the day on 1900-01-01? Monday. So in your code you're saying, how many weeks have passed since Monday, January 1, 1900? Let's call that [n]. Ok, now add [n] weeks to Monday, January 1, 1900. You should not be surprised that this ends up being a Monday. DATEADD has no idea that you want to add weeks but only until you get to a Sunday, it's just adding 7 days, then adding 7 more days, ... just like DATEDIFF only recognizes boundaries that have been crossed. For example, these both return 1, even though some folks complain that there should be some sensible logic built in to round up or down:
SELECT DATEDIFF(YEAR, '2010-01-01', '2011-12-31');
SELECT DATEDIFF(YEAR, '2010-12-31', '2011-01-01');
To answer how to get a Sunday:
If you want a Sunday, then pick a base date that's not a Monday but rather a Sunday. For example:
DECLARE #dt DATE = '1905-01-01';
SELECT [start_of_week] = DATEADD(WEEK, DATEDIFF(WEEK, #dt, CURRENT_TIMESTAMP), #dt);
This will not break if you change your DATEFIRST setting (or your code is running for a user with a different setting) - provided that you still want a Sunday regardless of the current setting. If you want those two answers to jive, then you should use a function that does depend on the DATEFIRST setting, e.g.
SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, CURRENT_TIMESTAMP), CURRENT_TIMESTAMP);
So if you change your DATEFIRST setting to Monday, Tuesday, what have you, the behavior will change. Depending on which behavior you want, you could use one of these functions:
CREATE FUNCTION dbo.StartOfWeek1 -- always a Sunday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(WEEK, DATEDIFF(WEEK, '19050101', #d), '19050101'));
END
GO
...or...
CREATE FUNCTION dbo.StartOfWeek2 -- always the DATEFIRST weekday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, #d), #d));
END
GO
Now, you have plenty of alternatives, but which one performs best? I'd be surprised if there would be any major differences but I collected all the answers provided so far and ran them through two sets of tests - one cheap and one expensive. I measured client statistics because I don't see I/O or memory playing a part in the performance here (though those may come into play depending on how the function is used). In my tests the results are:
"Cheap" assignment query:
Function - client processing time / wait time on server replies / total exec time
Gandarez - 330/2029/2359 - 0:23.6
me datefirst - 329/2123/2452 - 0:24.5
me Sunday - 357/2158/2515 - 0:25.2
trailmax - 364/2160/2524 - 0:25.2
Curt - 424/2202/2626 - 0:26.3
"Expensive" assignment query:
Function - client processing time / wait time on server replies / total exec time
Curt - 1003/134158/135054 - 2:15
Gandarez - 957/142919/143876 - 2:24
me Sunday - 932/166817/165885 - 2:47
me datefirst - 939/171698/172637 - 2:53
trailmax - 958/173174/174132 - 2:54
I can relay the details of my tests if desired - stopping here as this is already getting quite long-winded. I was a bit surprised to see Curt's come out as the fastest at the high end, given the number of calculations and inline code. Maybe I'll run some more thorough tests and blog about it... if you guys don't have any objections to me publishing your functions elsewhere.
For these that need to get:
Monday = 1 and Sunday = 7:
SELECT 1 + ((5 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Sunday = 1 and Saturday = 7:
SELECT 1 + ((6 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Above there was a similar example, but thanks to double "%7" it would be much slower.
For those who need the answer at work and creating function is forbidden by your DBA, the following solution will work:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-1), YourDate) as DATE) as WeekStart
From.....
This gives the start of that week. Here I assume that Sundays are the start of weeks. If you think that Monday is the start, you should use:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-2), YourDate) as DATE) as WeekStart
From.....
This works wonderfully for me:
CREATE FUNCTION [dbo].[StartOfWeek]
(
#INPUTDATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
-- THIS does not work in function.
-- SET DATEFIRST 1 -- set monday to be the first day of week.
DECLARE #DOW INT -- to store day of week
SET #INPUTDATE = CONVERT(VARCHAR(10), #INPUTDATE, 111)
SET #DOW = DATEPART(DW, #INPUTDATE)
-- Magic convertion of monday to 1, tuesday to 2, etc.
-- irrespect what SQL server thinks about start of the week.
-- But here we have sunday marked as 0, but we fix this later.
SET #DOW = (#DOW + ##DATEFIRST - 1) %7
IF #DOW = 0 SET #DOW = 7 -- fix for sunday
RETURN DATEADD(DD, 1 - #DOW,#INPUTDATE)
END
Maybe you need this:
SELECT DATEADD(DD, 1 - DATEPART(DW, GETDATE()), GETDATE())
Or
DECLARE #MYDATE DATETIME
SET #MYDATE = '2011-08-23'
SELECT DATEADD(DD, 1 - DATEPART(DW, #MYDATE), #MYDATE)
Function
CREATE FUNCTION [dbo].[GetFirstDayOfWeek]
( #pInputDate DATETIME )
RETURNS DATETIME
BEGIN
SET #pInputDate = CONVERT(VARCHAR(10), #pInputDate, 111)
RETURN DATEADD(DD, 1 - DATEPART(DW, #pInputDate),
#pInputDate)
END
GO
Googled this script:
create function dbo.F_START_OF_WEEK
(
#DATE datetime,
-- Sun = 1, Mon = 2, Tue = 3, Wed = 4
-- Thu = 5, Fri = 6, Sat = 7
-- Default to Sunday
#WEEK_START_DAY int = 1
)
/*
Find the fisrt date on or before #DATE that matches
day of week of #WEEK_START_DAY.
*/
returns datetime
as
begin
declare #START_OF_WEEK_DATE datetime
declare #FIRST_BOW datetime
-- Check for valid day of week
if #WEEK_START_DAY between 1 and 7
begin
-- Find first day on or after 1753/1/1 (-53690)
-- matching day of week of #WEEK_START_DAY
-- 1753/1/1 is earliest possible SQL Server date.
select #FIRST_BOW = convert(datetime,-53690+((#WEEK_START_DAY+5)%7))
-- Verify beginning of week not before 1753/1/1
if #DATE >= #FIRST_BOW
begin
select #START_OF_WEEK_DATE =
dateadd(dd,(datediff(dd,#FIRST_BOW,#DATE)/7)*7,#FIRST_BOW)
end
end
return #START_OF_WEEK_DATE
end
go
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47307
CREATE FUNCTION dbo.fnFirstWorkingDayOfTheWeek
(
#currentDate date
)
RETURNS INT
AS
BEGIN
-- get DATEFIRST setting
DECLARE #ds int = ##DATEFIRST
-- get week day number under current DATEFIRST setting
DECLARE #dow int = DATEPART(dw,#currentDate)
DECLARE #wd int = 1+(((#dow+#ds) % 7)+5) % 7 -- this is always return Mon as 1,Tue as 2 ... Sun as 7
RETURN DATEADD(dd,1-#wd,#currentDate)
END
For the basic (the current week's Sunday)
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) as date)
If previous week:
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) -7 as date)
Internally, we built a function that does it but if you need quick and dirty, this will do it.
Since Julian date 0 is a Monday just add the number of weeks to Sunday
which is the day before -1 Eg. select dateadd(wk,datediff(wk,0,getdate()),-1)
I found some of the other answers long-winded or didn't actually work if you wanted Monday as the start of the week.
Sunday
SELECT DATEADD(week, DATEDIFF(week, -1, GETDATE()), -1) AS Sunday;
Monday
SELECT DATEADD(week, DATEDIFF(week, 0, GETDATE() - 1), 0) AS Monday;
Set DateFirst 1;
Select
Datepart(wk, TimeByDay) [Week]
,Dateadd(d,
CASE
WHEN Datepart(dw, TimeByDay) = 1 then 0
WHEN Datepart(dw, TimeByDay) = 2 then -1
WHEN Datepart(dw, TimeByDay) = 3 then -2
WHEN Datepart(dw, TimeByDay) = 4 then -3
WHEN Datepart(dw, TimeByDay) = 5 then -4
WHEN Datepart(dw, TimeByDay) = 6 then -5
WHEN Datepart(dw, TimeByDay) = 7 then -6
END
, TimeByDay) as StartOfWeek
from TimeByDay_Tbl
This is my logic. Set the first of the week to be Monday then calculate what is the day of the week a give day is, then using DateAdd and Case I calculate what the date would have been on the previous Monday of that week.
This is a useful function for me
/* MeRrais 211126
select [dbo].[SinceWeeks](0,NULL)
select [dbo].[SinceWeeks](5,'2021-08-31')
*/
alter Function [dbo].[SinceWeeks](#Weeks int, #From datetime=NULL)
Returns date
AS
Begin
if #From is null
set #From=getdate()
return cast(dateadd(day, -(#Weeks*7+datepart(dw,#From)-1), #From) as date)
END
I don't have any issues with any of the answers given here, however I do think mine is a lot simpler to implement, and understand. I have not run any performance tests on it, but it should be neglegable.
So I derived my answer from the fact that dates are stored in SQL server as integers, (I am talking about the date component only). If you don't believe me, try this SELECT CONVERT(INT, GETDATE()), and vice versa.
Now knowing this, you can do some cool math equations. You might be able to come up with a better one, but here is mine.
/*
TAKEN FROM http://msdn.microsoft.com/en-us/library/ms181598.aspx
First day of the week is
1 -- Monday
2 -- Tuesday
3 -- Wednesday
4 -- Thursday
5 -- Friday
6 -- Saturday
7 (default, U.S. English) -- Sunday
*/
--Offset is required to compensate for the fact that my ##DATEFIRST setting is 7, the default.
DECLARE #offSet int, #testDate datetime
SELECT #offSet = 1, #testDate = GETDATE()
SELECT CONVERT(DATETIME, CONVERT(INT, #testDate) - (DATEPART(WEEKDAY, #testDate) - #offSet))
I had a similar problem. Given a date, I wanted to get the date of the Monday of that week.
I used the following logic: Find the day number in the week in the range of 0-6, then subtract that from the originay date.
I used: DATEADD(day,-(DATEPART(weekday,)+5)%7,)
Since DATEPRRT(weekday,) returns 1 = Sundaye ... 7=Saturday,
DATEPART(weekday,)+5)%7 returns 0=Monday ... 6=Sunday.
Subtracting this number of days from the original date gives the previous Monday. The same technique could be used for any starting day of the week.
I found this simple and usefull. Works even if first day of week is Sunday or Monday.
DECLARE #BaseDate AS Date
SET #BaseDate = GETDATE()
DECLARE #FisrtDOW AS Date
SELECT #FirstDOW = DATEADD(d,DATEPART(WEEKDAY,#BaseDate) *-1 + 1, #BaseDate)
Maybe I'm over simplifying here, and that may be the case, but this seems to work for me. Haven't ran into any problems with it yet...
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7 - 7) as 'FirstDayOfWeek'
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7) as 'LastDayOfWeek'