Get the week start date and week end date from week number - sql

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 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

Answer 5

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.

Answer 6

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)

Answer 7

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())

Answer 8

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.

Answer 9

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

Answer 10

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:
It works for all weeks, 1st or others.

Answer 11

Week Start & End Date From Date For Power BI Dax Formula
WeekStartDate = [DateColumn] - (WEEKDAY([DateColumn])-1)
WeekEndDate = [DateColumn] + (7-WEEKDAY([DateColumn]))

Answer 12

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]

Answer 13

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]

Answer 14

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'

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

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..