117

I have a query that counts member's wedding dates in the database.

SELECT 
  SUM(NumberOfBrides) AS [Wedding Count]
  , DATEPART( wk, WeddingDate) AS [Week Number]
  , DATEPART( year, WeddingDate) AS [Year]
FROM  MemberWeddingDates
GROUP BY DATEPART(year, WeddingDate), DATEPART(wk, WeddingDate)
ORDER BY SUM(NumberOfBrides) DESC

How do I work out when the start and end of each week represented in the result set?

SELECT
  SUM(NumberOfBrides) AS [Wedding Count]
  , DATEPART(wk, WeddingDate) AS [Week Number]
  , DATEPART(year, WeddingDate) AS [Year]
  , ??? AS WeekStart
  , ??? AS WeekEnd
FROM  MemberWeddingDates
GROUP BY DATEPART(year, WeddingDate), DATEPART(wk, WeddingDate)
ORDER BY SUM(NumberOfBrides) DESC
Dale K
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digiguru
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19 Answers19

194

You can find the day of week and do a date add on days to get the start and end dates..

DATEADD(dd, -(DATEPART(dw, WeddingDate)-1), WeddingDate) [WeekStart]

DATEADD(dd, 7-(DATEPART(dw, WeddingDate)), WeddingDate) [WeekEnd]

You probably also want to look at stripping off the time from the date as well though.

Robin Day
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    Bear in mind that setting `DATEFIRST` to anything other than 7 breaks this. – Tomalak Aug 12 '09 at 16:52
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    It won't "break" it, it will use datefirst to set WeekStart = to what DateFirst says the first day of a week is. Your version will always do Monday and Sunday as start and end of a week, not what the server is set to use as start and end of the week – Robin Day Aug 12 '09 at 17:11
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    Hm... That's a valid point, +1. :) I'll delete mine, then (Though for being a shot in the foot, it was extremely well aimed. *g*). – Tomalak Aug 12 '09 at 17:31
  • Not at all, I think yours is perfectly valid as well as it said Monday and Sunday in the columns. That might have been what was required regardless of any localization settings. – Robin Day Aug 12 '09 at 17:37
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    Well, then I undelete it again. I'm not sure who in their right mind would assume anything other than Monday as the start of the week anyway. To start the week on Sunday makes no sense *whatsoever*. :-) – Tomalak Aug 12 '09 at 18:01
  • My organization uses Saturday as the start of the week... so this wouldn't work :( Gotta change it up a bit – LaRae White Apr 17 '14 at 15:49
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    'set datefirst 1' for Monday (http://msdn.microsoft.com/en-ie/library/ms181598.aspx) – Selrac Oct 15 '14 at 08:24
  • @RobinDay - Setting DATEFIRST to some other value DOES change the dates. That's neither right nor wrong if no one is going to define what the first day of the week actually is. The OP didn't so your code works. The problem in the future may be if a company changes DATEFIRST because they suddenly "got religious about ISO" or for any other reason. The OP really needed to define what the first day of the week was so that they could use more bullet-proof code. – Jeff Moden Dec 02 '21 at 05:15
46

Here is a DATEFIRST agnostic solution:

SET DATEFIRST 4     /* or use any other weird value to test it */
DECLARE @d DATETIME

SET @d = GETDATE()

SELECT
  @d ThatDate,
  DATEADD(dd, 0 - (@@DATEFIRST + 5 + DATEPART(dw, @d)) % 7, @d) Monday,
  DATEADD(dd, 6 - (@@DATEFIRST + 5 + DATEPART(dw, @d)) % 7, @d) Sunday
mbuechmann
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Tomalak
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    This is great, but the Monday was not working for me. I had to add "0 - " to get Mondays. My Monday code is now: DATEADD(dd, 0 - (@@DATEFIRST + 5 + DATEPART(dw, @d)) % 7, @d) – Warren Oct 21 '14 at 00:01
  • Upvoted both the answer and Warrens comment. Sql Server Version 11.0.5058.0 at the very least gives the wrong date for Monday without Warrens modification. Was giving me Friday instead. – Morvael Apr 06 '16 at 09:17
  • Note that the offset `+ 5` in the answer may need to change depending on the start-of-week. eg for Sunday as the start-of-week, this needs to be `- 1` – Kevin Swann May 21 '22 at 19:44
20

you can also use this:

  SELECT DATEADD(day, DATEDIFF(day, 0, WeddingDate) /7*7, 0) AS weekstart,
         DATEADD(day, DATEDIFF(day, 6, WeddingDate-1) /7*7 + 7, 6) AS WeekEnd
hkravitz
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  • I like this version the best for getting an agnostic Mon to Sun date with no time and you only have to enter the date once per line. – ukgav Nov 25 '21 at 10:39
  • @hkravitz when I run your 2nd line `DATEADD(day, DATEDIFF(day, 6, WeddingDate-1) /7*7 + 7, 6) AS WeekEnd` I get an error `date is incompatible with int`. Guessing, you want to subtract 1 day from the date parameter - why? – Kevin Swann May 21 '22 at 19:11
  • You need to make sure your date column is a date data type. – hkravitz May 22 '22 at 20:08
  • @hkravitz would you mind explaining the numbers 0, 6, and the multiplication+division by 7? – Gene Burinsky Aug 16 '23 at 18:42
4

Here is another version. If your Scenario requires Saturday to be 1st day of Week and Friday to be last day of Week, the below code will handle that:

  DECLARE @myDate DATE = GETDATE()
  SELECT    @myDate,
    DATENAME(WEEKDAY,@myDate),
    DATEADD(DD,-(CHOOSE(DATEPART(dw, @myDate), 1,2,3,4,5,6,0)),@myDate) AS WeekStartDate,
    DATEADD(DD,7-CHOOSE(DATEPART(dw, @myDate), 2,3,4,5,6,7,1),@myDate) AS WeekEndDate

Screenshot of Query

Ajay Dwivedi
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3

Expanding on @Tomalak's answer. The formula works for days other than Sunday and Monday but you need to use different values for where the 5 is. A way to arrive at the value you need is

Value Needed = 7 - (Value From Date First Documentation for Desired Day Of Week) - 1

here is a link to the document: https://msdn.microsoft.com/en-us/library/ms181598.aspx

And here is a table that lays it out for you.

          | DATEFIRST VALUE |   Formula Value   |   7 - DATEFIRSTVALUE - 1
Monday    | 1               |          5        |   7 - 1- 1 = 5
Tuesday   | 2               |          4        |   7 - 2 - 1 = 4
Wednesday | 3               |          3        |   7 - 3 - 1 = 3
Thursday  | 4               |          2        |   7 - 4 - 1 = 2
Friday    | 5               |          1        |   7 - 5 - 1 = 1
Saturday  | 6               |          0        |   7 - 6 - 1 = 0
Sunday    | 7               |         -1        |   7 - 7 - 1 = -1

But you don't have to remember that table and just the formula, and actually you could use a slightly different one too the main need is to use a value that will make the remainder the correct number of days.

Here is a working example:

DECLARE @MondayDateFirstValue INT = 1
DECLARE @FridayDateFirstValue INT = 5
DECLARE @TestDate DATE = GETDATE()

SET @MondayDateFirstValue = 7 - @MondayDateFirstValue - 1
SET @FridayDateFirstValue = 7 - @FridayDateFirstValue - 1

SET DATEFIRST 6 -- notice this is saturday

SELECT 
    DATEADD(DAY, 0 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as MondayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as MondayEndOfWeek
   ,DATEADD(DAY, 0 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as FridayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as FridayEndOfWeek


SET DATEFIRST 2 --notice this is tuesday

SELECT 
    DATEADD(DAY, 0 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as MondayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as MondayEndOfWeek
   ,DATEADD(DAY, 0 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as FridayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as FridayEndOfWeek

This method would be agnostic of the DATEFIRST Setting which is what I needed as I am building out a date dimension with multiple week methods included.

Community
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Matt
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2

If sunday is considered as week start day, then here is the code

Declare @currentdate date = '18 Jun 2020'

select DATEADD(D, -(DATEPART(WEEKDAY, @currentdate) - 1), @currentdate)

select DATEADD(D, (7 - DATEPART(WEEKDAY, @currentdate)), @currentdate)
shyambabu
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1

Let us break the problem down to two parts:

1) Determine the day of week

The DATEPART(dw, ...) returns a number, 1...7, relative to DATEFIRST setting (docs). The following table summarizes the possible values:

                                                   @@DATEFIRST
+------------------------------------+-----+-----+-----+-----+-----+-----+-----+-----+
|                                    |  1  |  2  |  3  |  4  |  5  |  6  |  7  | DOW |
+------------------------------------+-----+-----+-----+-----+-----+-----+-----+-----+
|  DATEPART(dw, /*Mon*/ '20010101')  |  1  |  7  |  6  |  5  |  4  |  3  |  2  |  1  |
|  DATEPART(dw, /*Tue*/ '20010102')  |  2  |  1  |  7  |  6  |  5  |  4  |  3  |  2  |
|  DATEPART(dw, /*Wed*/ '20010103')  |  3  |  2  |  1  |  7  |  6  |  5  |  4  |  3  |
|  DATEPART(dw, /*Thu*/ '20010104')  |  4  |  3  |  2  |  1  |  7  |  6  |  5  |  4  |
|  DATEPART(dw, /*Fri*/ '20010105')  |  5  |  4  |  3  |  2  |  1  |  7  |  6  |  5  |
|  DATEPART(dw, /*Sat*/ '20010106')  |  6  |  5  |  4  |  3  |  2  |  1  |  7  |  6  |
|  DATEPART(dw, /*Sun*/ '20010107')  |  7  |  6  |  5  |  4  |  3  |  2  |  1  |  7  |
+------------------------------------+-----+-----+-----+-----+-----+-----+-----+-----+

The last column contains the ideal day-of-week value for Monday to Sunday weeks*. By just looking at the chart we come up with the following equation:

(@@DATEFIRST + DATEPART(dw, SomeDate) - 1 - 1) % 7 + 1

2) Calculate the Monday and Sunday for given date

This is trivial thanks to the day-of-week value. Here is an example:

WITH TestData(SomeDate) AS (
    SELECT CAST('20001225' AS DATETIME) UNION ALL
    SELECT CAST('20001226' AS DATETIME) UNION ALL
    SELECT CAST('20001227' AS DATETIME) UNION ALL
    SELECT CAST('20001228' AS DATETIME) UNION ALL
    SELECT CAST('20001229' AS DATETIME) UNION ALL
    SELECT CAST('20001230' AS DATETIME) UNION ALL
    SELECT CAST('20001231' AS DATETIME) UNION ALL
    SELECT CAST('20010101' AS DATETIME) UNION ALL
    SELECT CAST('20010102' AS DATETIME) UNION ALL
    SELECT CAST('20010103' AS DATETIME) UNION ALL
    SELECT CAST('20010104' AS DATETIME) UNION ALL
    SELECT CAST('20010105' AS DATETIME) UNION ALL
    SELECT CAST('20010106' AS DATETIME) UNION ALL
    SELECT CAST('20010107' AS DATETIME) UNION ALL
    SELECT CAST('20010108' AS DATETIME) UNION ALL
    SELECT CAST('20010109' AS DATETIME) UNION ALL
    SELECT CAST('20010110' AS DATETIME) UNION ALL
    SELECT CAST('20010111' AS DATETIME) UNION ALL
    SELECT CAST('20010112' AS DATETIME) UNION ALL
    SELECT CAST('20010113' AS DATETIME) UNION ALL
    SELECT CAST('20010114' AS DATETIME)
), TestDataPlusDOW AS (
    SELECT SomeDate, (@@DATEFIRST + DATEPART(dw, SomeDate) - 1 - 1) % 7 + 1 AS DOW
    FROM TestData
)
SELECT
    FORMAT(SomeDate,                            'ddd yyyy-MM-dd') AS SomeDate,
    FORMAT(DATEADD(dd, -DOW + 1, SomeDate),     'ddd yyyy-MM-dd') AS [Monday],
    FORMAT(DATEADD(dd, -DOW + 1 + 6, SomeDate), 'ddd yyyy-MM-dd') AS [Sunday]
FROM TestDataPlusDOW

Output:

+------------------+------------------+------------------+
|  SomeDate        |  Monday          |    Sunday        |
+------------------+------------------+------------------+
|  Mon 2000-12-25  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Tue 2000-12-26  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Wed 2000-12-27  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Thu 2000-12-28  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Fri 2000-12-29  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Sat 2000-12-30  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Sun 2000-12-31  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Mon 2001-01-01  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Tue 2001-01-02  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Wed 2001-01-03  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Thu 2001-01-04  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Fri 2001-01-05  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Sat 2001-01-06  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Sun 2001-01-07  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Mon 2001-01-08  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Tue 2001-01-09  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Wed 2001-01-10  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Thu 2001-01-11  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Fri 2001-01-12  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Sat 2001-01-13  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Sun 2001-01-14  |  Mon 2001-01-08  |  Sun 2001-01-14  |
+------------------+------------------+------------------+

* For Sunday to Saturday weeks you need to adjust the equation just a little, like add 1 somewhere.

Salman A
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1

Below query will give data between start and end of current week starting from sunday to saturday

SELECT DOB FROM PROFILE_INFO WHERE DAY(DOB) BETWEEN
DAY( CURRENT_DATE() - (SELECT DAYOFWEEK(CURRENT_DATE())-1))
AND
DAY((CURRENT_DATE()+(7 - (SELECT DAYOFWEEK(CURRENT_DATE())) ) ))
AND
MONTH(DOB)=MONTH(CURRENT_DATE())
Anders Wendt
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riya
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0

I just encounter a similar case with this one, but the solution here seems not helping me. So I try to figure it out by myself. I work out the week start date only, week end date should be of similar logic.

Select 
      Sum(NumberOfBrides) As [Wedding Count], 
      DATEPART( wk, WeddingDate) as [Week Number],
      DATEPART( year, WeddingDate) as [Year],
      DATEADD(DAY, 1 - DATEPART(WEEKDAY, dateadd(wk, DATEPART( wk, WeddingDate)-1,  DATEADD(yy,DATEPART( year, WeddingDate)-1900,0))), dateadd(wk, DATEPART( wk, WeddingDate)-1, DATEADD(yy,DATEPART( year, WeddingDate)-1900,0))) as [Week Start]

FROM  MemberWeddingDates
Group By DATEPART( year, WeddingDate), DATEPART( wk, WeddingDate)
Order By Sum(NumberOfBrides) Desc
Amber
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0

The most voted answer works fine except for the 1st week and last week of a year. For example, if the value of WeddingDate is '2016-01-01', the result will be 2015-12-27 and 2016-01-02, but the right answer is 2016-01-01 and 2016-01-02.

Try this:

Select 
  Sum(NumberOfBrides) As [Wedding Count], 
  DATEPART( wk, WeddingDate) as [Week Number],
  DATEPART( year, WeddingDate) as [Year],
  MAX(CASE WHEN DATEPART(WEEK, WeddingDate) = 1 THEN CAST(DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0) AS date) ELSE DATEADD(DAY, 7 * DATEPART(WEEK, WeddingDate), DATEADD(DAY, -(DATEPART(WEEKDAY, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0)) + 6), DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0))) END) as WeekStart,
  MAX(CASE WHEN DATEPART(WEEK, WeddingDate) = DATEPART(WEEK, DATEADD(DAY, -1, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate) + 1, 0))) THEN DATEADD(DAY, -1, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate) + 1, 0)) ELSE DATEADD(DAY, 7 * DATEPART(WEEK, WeddingDate) + 6, DATEADD(DAY, -(DATEPART(WEEKDAY, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0)) + 6), DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0))) END) as WeekEnd
FROM  MemberWeddingDates
Group By DATEPART( year, WeddingDate), DATEPART( wk, WeddingDate)
Order By Sum(NumberOfBrides) Desc;

The result looks like: enter image description here

It works for all weeks, 1st or others.

Leo
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0

This doesn't come from me, but it got the job done regardless:

SELECT DATEADD(wk, -1, DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --first day previous week
SELECT DATEADD(wk, 0, DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --first day current week
SELECT DATEADD(wk, 1, DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --first day next week

SELECT DATEADD(wk, 0, DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --last day previous week
SELECT DATEADD(wk, 1, DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --last day current week
SELECT DATEADD(wk, 2, DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --last day next week

I found it here.

RoastBeast
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Francesco Mantovani
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0

Week Start & End Date From Date For Power BI Dax Formula

WeekStartDate = [DateColumn] - (WEEKDAY([DateColumn])-1)
WeekEndDate = [DateColumn] + (7-WEEKDAY([DateColumn]))
V-rund Puro-hit
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0

This is my solution


    SET DATEFIRST 1;    /* change to use a different datefirst  */
    DECLARE @date DATETIME
    SET @date = CAST('2/6/2019' as date)

    SELECT  DATEADD(dd,0 - (DATEPART(dw, @date) - 1) ,@date) [dateFrom], 
            DATEADD(dd,6 - (DATEPART(dw, @date) - 1) ,@date) [dateTo]

0

Get Start Date & End Date by Custom Date


   DECLARE @Date NVARCHAR(50)='05/19/2019' 
   SELECT
      DATEADD(DAY,CASE WHEN DATEPART(WEEKDAY, @Date)=1 THEN -6 ELSE 2 - DATEPART(WEEKDAY, @Date) END, CAST(@Date AS DATE)) [Week_Start_Date]
     ,DATEADD(DAY,CASE WHEN DATEPART(WEEKDAY, @Date)=1 THEN 0 ELSE  8 - DATEPART(WEEKDAY, @Date) END, CAST(@Date AS DATE)) [Week_End_Date]
Ram Maurya
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  • While this code may solve the question, [including an explanation](//s.tk/meta/questions/114762/explaining-entirely-code-based-answers) really helps to improve the quality of your post. – 4b0 May 06 '19 at 06:58
0

I have a way other, It is select day Start and day End of Week Current:

DATEADD(d, -(DATEPART(dw, GETDATE()-2)), GETDATE()) is date time Start

and

DATEADD(day,7-(DATEPART(dw,GETDATE()-1)),GETDATE()) is date time End
RF1991
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0

Another way to do it:

declare @week_number int = 6280 -- 2020-05-07
declare @start_weekday int = 0 -- Monday
declare @end_weekday int = 6 -- next Sunday

select 
    dateadd(week, @week_number, @start_weekday), 
    dateadd(week, @week_number, @end_weekday)

Explanation:

  • @week_number is the week number since the initial calendar date '1900-01-01'. It can be computed this way: select datediff(week, 0, @wedding_date) as week_number
  • @start_weekday for the week first day: 0 for Monday, -1 if Sunday
  • @end_weekday for the week last day: 6 for next Sunday, 5 if Saturday
  • dateadd(week, @week_number, @end_weekday): adds the given number of weeks and the given number of days into the initial calendar date '1900-01-01'
hd84335
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0

You can get all starting day or ending day of week or any nth day of week with this sql code:

SELECT DateList.MyDay, DateAdd("d",1-Weekday([MyDay],2),[MyDay]) AS DateAdd FROM DateList;
-3

Not sure how useful this is, but I ended up here from looking for a solution on Netezza SQL and couldn't find one on stack overflow.

For IBM netezza you would use something (for week start mon, week end sun) like:

select
next_day (WeddingDate, 'SUN') -6 as WeekStart,

next_day (WeddingDate, 'SUN') as WeekEnd
RF1991
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DekuranC
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-4

for Access Queries, you can use in the below format as a field

"FirstDayofWeek:IIf(IsDate([ForwardedForActionDate]),CDate(Format([ForwardedForActionDate],"dd/mm/yyyy"))-(Weekday([ForwardedForActionDate])-1))"

direct Calculation allowed..

xlembouras
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Labeeb
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