I am writing a query that returns the sum of rows for the last 10 weeks FRI-THURS.
It uses a group by to show the sum of each week:
WITH Vars (Friday) -- get current week Fridays Date
AS (
SELECT CAST(DATEADD(DAY,(13 - (##DATEFIRST + DATEPART(WEEKDAY,GETDATE())))%7,GETDATE()) AS DATE) As 'Friday'
)
SELECT datepart(week, DateField) AS WeekNum, COUNT(*) AS Counts
FROM Table
WHERE DateField >= DATEADD(week,-9, (SELECT Friday from Vars))
GROUP BY datepart(week, DateField)
ORDER BY WeekNum DESC
The problem is every week starts on Monday so the Group By doesn't group the dates on how I want it. I want a week to be defined as FRI-THURS.
One workaround to this is to use DATEFIRST. e.g:
SET DATEFIRST = 5; --set beginning of each week to Friday
WITH Vars (Friday) -- get current week Fridays Date
... rest of query
However due to limitations on the interface I am writing this query I cannot have two separate statements run. It needs to be one query with no semicolons.
How can I achieve this?
This should do it. First pre-compute once the StartingFriday of 9 weeks ago, rather than doing that for each row. Then compute the dfYear and dfWeek giving them alias-es, where their DateField is after the starting friday. Lastly, Count/GroupBy/OrderBy.
Declare #StartingFriday as date =
DATEADD(week,-9, (DATEADD(day, - ((Datepart(WEEKDAY,GETDATE()) +1) % 7) , GETDATE())) ) ;
SELECT dfYear, dfWeek, COUNT(*) AS Counts
FROM
(Select -- compute these here, and use alias in Select, GroupBy, OrderBy
(Datepart(Year,(DATEADD(day, - ((Datepart(WEEKDAY,DateField) +1) % 7) , DateField)) ) )as dfYear
,(Datepart(Week,(DATEADD(day, - ((Datepart(WEEKDAY,DateField) +1) % 7) , DateField)) ) )as dfWeek
From Table
WHERE #StartingFriday <= DateField
) as aa
group by dfYear, dfWeek
order by dfYear desc, dfWeek desc
-- we want the weeknum of the (Friday on or before the DateField)
-- the % (percent sign) is the math MODULO operator.
-- used to get back to the nearest Friday,
-- day= Fri Sat Sun Mon Tue Wed Thu
-- weekday= 6 7 1 2 3 4 5
-- plus 1 = 7 8 2 3 4 5 6
-- Modulo7= 0 1 2 3 4 5 6
-- which are the days to subtract from DateField
-- to get to its Friday start of its week.
I did some testing with this
declare #dt as date = '8/17/18';
select ((DATEPART(WEEKDAY,#dt) +1) % 7) as wd
,(DATEADD(day, - ((Datepart(WEEKDAY,#dt) +1) % 7) , #dt)) as Fri
,(Datepart(Week,(DATEADD(day, - ((Datepart(WEEKDAY,#dt) +1) % 7) , #dt)) ) )as wk
,DATEADD(week,-9, (DATEADD(day, - ((Datepart(WEEKDAY,#dt) +1) % 7) , #dt)) ) as StartingFriday
-- SET DATEFIRST to U.S. English default value of 7.
SET DATEFIRST 7;
SELECT
##DATEFIRST;
SELECT
GETDATE()
, DATEPART(dw , GETDATE()) AS DayOfWeek;
-- January 1, 1999 is a Friday. Because the U.S. English default
-- specifies Sunday as the first day of the week, DATEPART of 1999-1-1
-- (Friday) yields a value of 6, because Friday is the sixth day of the
-- week when you start with Sunday as day 1.
SET DATEFIRST 3;
SELECT
##DATEFIRST;
-- Because Wednesday is now considered the first day of the week,
-- DATEPART now shows that 1999-1-1 (a Friday) is the third day of the
-- week. The following DATEPART function should return a value of 3.
SELECT
GETDATE()
, DATEPART(dw , GETDATE()) AS DayOfWeek;
SET DATEFIRST 7;
How do we handle getting the DATEPART (1 = Sunday always) irregardless of DATEFIRST setting?
I really don't want to do a case and subtract...
this always seems ridiculous to me, how about
select datediff(day,0, getdate()) % 7
where 6 represents Sunday
or you could do
select (datediff(day,0, '2016-07-31') - 5) % 7
to get Sun = 1, Mon = 2, Tue = 3 ... etc
or you could do this fiddle
select (datepart(weekday,get_date()) + ##datefirst - 1) % 7 + 1
seems to work for all datepart
set datefirst 5
select (datepart(weekday,'2016-07-31') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-01') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-02') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-03') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-04') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-05') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-06') + ##datefirst - 1) % 7 + 1
select (datepart(weekday,'2016-08-07') + ##datefirst - 1) % 7 + 1
This code selects Monday as the first day of week whatever the setting in your engine:
((datepart(DW, #YourDateVariable) + ##DATEFIRST + 5) % 7) + 1
by adding ##DATEFIRST value before the modulus operator it neglects the datefirst setting in your SQL engine.
The value 5 is to make Monday as the first day of the week
To make Sunday as the first day of the week add 6
to make saturday as the first day of the week then add 0
and so on the rest of the day weeks.
Perhaps not the most elegant, but instead of doing the subtraction you can just set it to 7 and then back to what ever it was after your DATEPART
SET DATEFIRST 7;
SELECT DATEPART(dw , GETDATE()) --6
SET DATEFIRST 3;
DECLARE #currentDatefirst int = ##DATEFIRST
SELECT ##DATEFIRST --3
SET DATEFIRST 7;
SELECT DATEPART(dw , GETDATE()) --6
SET DATEFIRST #currentDatefirst
SELECT ##DATEFIRST --3
I have to get/create date from the user input of week of month (week number in that month - 1st,2nd,3rd,4th and last) and day of week (sunday,monday..) in SQL server.
Examples:
4th Sunday of every month, Last Friday of every month, First Monday etc.
I was able to do it easily in .net but SQL server does seem limited in the date functions.
I am having to use lot of logic to get the date. To calculate the date using the above two parameters I had to use lot of datepart function.
Any suggestions on how to come up with the optimal SQL query for such a function?
I created a function other day for another OP GET Month, Quarter based on Work Week number
This function takes the current year as default it can be further modified to take Year as a parameter too.
an extension to that function can produce the results you are looking for ....
WITH X AS
(
SELECT TOP (CASE WHEN YEAR(GETDATE()) % 4 = 0 THEN 366 ELSE 365 END)-- to handle leap year
DATEADD(DAY
,ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1
, CAST(YEAR(GETDATE()) AS VARCHAR(4)) + '0101' )
DayNumber
From master..spt_values
),DatesData AS(
SELECT DayNumber [Date]
,DATEPART(WEEKDAY,DayNumber) DayOfTheWeek
,DATEDIFF(WEEK,
DATEADD(WEEK,
DATEDIFF(WEEK, 0, DATEADD(MONTH,
DATEDIFF(MONTH, 0, DayNumber), 0)), 0)
, DayNumber- 1) + 1 WeekOfTheMonth
FROM X )
SELECT * FROM DatesData
WHERE DayOfTheWeek = 6 -- A function would expect these two parameters
AND WeekOfTheMonth = 4 -- #DayOfTheWeek and #WeekOfTheMonth
Here is a general formula:
declare #month as datetime --set to the first day of the month you wish to use
declare #week as int --1st, 2nd, 3rd...
declare #day as int --Day of the week (1=sunday, 2=monday...)
--Second monday in August 2015
set #month = '8/1/2015'
set #week = 2
set #day = 2
select dateadd(
day,
((7+#day) - datepart(weekday, #month)) % 7 + 7 * (#week-1),
#month
)
You can also find the last, 2nd to last... etc with this reverse formula:
--Second to last monday in August 2015
set #month = '8/1/2015'
set #week = 2
set #day = 2
select
dateadd(
day,
-((7+datepart(weekday, dateadd(month,1,#month)-1)-#day)) % 7 - 7 * (#week-1),
dateadd(month,1,#month)-1
)
I have odd requirement of calculating multiplier for non trading day. So say for Example: If Holiday is on Wednesday then next day (Thursday) will have multiply factor of 2 (for Wed + Thursday) If Holiday is on Monday then Tuesday should have multiply factor of 4 (i.e Saturday + Sunday + Monday + Tuesday) For normal weekend, Monday should have multiple factor of 3 (i.e Sat +Sun + Mon) and for normal weekdays, multiply factor is 1
And to resolve this issue, I have created standard Date dimension, with Holiday flag (which will be 1 for weekend and public holiday) and created column named MULTIPLIER which should have value of For normal weekdays - should have multiplier value of 1 and if Holiday is on Wednesday (28th May) then on Thursday (29th May), multiplier factor will be 2 (i.e sum of IsHoliday column count for Wed + 1 for current day (i.e 1 + 1) and if holiday is on Monday (2nd June) then on 3rd June it will be Sum of IsHoliday flag (i.e 1 for Sat + 1 for Sun + 1 for Mon and summing this 3 holiday flag it will be 3 + 1 for current day so it will be 4)
It would be really great, if someone can help me to resolve this mystery
Note: I am using SQL Server 2012
CREATE Table DateTable (
[date] datetime NOT NULL
,IsHoliday int NOT NULL
);
INSERT INTO DateTable
VALUES ('20140101', 1) -- Just example, not real holidays
,('20140102', 0)
,('20140103', 0)
,('20140104', 0)
,('20140105', 1)
,('20140106', 1)
,('20140107', 1)
,('20140108', 0);
WITH HolidayGaps AS (
SELECT [Date]
,DaysFromStart
,DaysFromStart - ROW_NUMBER() OVER (ORDER BY Date) AS Gap
FROM DateTable
CROSS APPLY (
SELECT DATEDIFF(day, '20140101', [Date]) AS DaysFromStart
) AS CA1
WHERE IsHoliday = 1
)
,DayAfterHoliday AS (
SELECT DATEADD(day, 1, MAX([date])) AS [date]
,COUNT(1) AS Multiplier
FROM HolidayGaps
GROUP BY Gap
)
SELECT *
FROM DayAfterHoliday
Your update statement would look something like
UPDATE DateTable
SET Multiplier = CASE WHEN DateTable.IsHoliday = 1 THEN 0
ELSE COALESCE(DayAfterHoliday.Multiplier, 1)
END
FROM DateTable
LEFT JOIN DayAfterHoliday
ON DayAfterHoliday.Date = DateTable.Date
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'