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SQL Select end of week date from last year

Posted on 2016-09-12
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Last Modified: 2016-09-12
Hello EE,

if today we are 09/16/2016, I would like to be able from the same day last year to get the Friday of the week

for that example, i woud like to return 09/18/2015
if I chose 09/09/2016   i would need to return 09/11/2015

the user will always chose a date that is a friday.  09/16/2016 is a friday and 09/09/2016 as well.

i need to return the friday's date of last year

can you help me ?
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Question by:PhilippeRenaud
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6 Comments
 
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Expert Comment

by:zephyr_hex
Comment Utility
To get Friday of the current week:  

SELECT DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 4) AS Friday

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Just replace GETDATE() with a different date to get the Friday of that week.
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Expert Comment

by:zephyr_hex
Comment Utility
Oh wait, you want last year's Friday... hold on...
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Assisted Solution

by:zephyr_hex
zephyr_hex earned 250 total points
Comment Utility
So, to get last year's Friday, you would use DATEADD to get the date from last year, and then figure out the Friday from that date:
SELECT DATEADD(wk, DATEDIFF(wk,0,DATEADD(YEAR,-1,GETDATE())), 4) AS Friday

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or, using a date other than today:

SELECT DATEADD(wk, DATEDIFF(wk,0,DATEADD(YEAR,-1,'2016-09-16')), 4) AS Friday

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Accepted Solution

by:
Éric Moreau earned 250 total points
Comment Utility
or something like this which can be embedded in a UDF;
DECLARE @date DATETIME = '2016-09-09'

SET @date = DATEADD(YEAR, -1, @date)

WHILE DATEPART(WEEKDAY, @date) <> 6
	SET @date = DATEADD(DAY, 1, @date)

SELECT @date

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Expert Comment

by:ScottPletcher
Comment Utility
Code below will work with any DATEFIRST setting and does not require a loop.  I use a CROSS APPLY just to assign a name to the date calc --  you could instead just repeat the expression in the main code, as the second code block shows.

--more "self-documented" code
SELECT
    user_date_entered,
    /* back up next years max possible date to the last Friday */
    DATEADD(DAY, -(DATEDIFF(DAY, Friday, prev_year_max_date) % 7), prev_year_max_date) AS last_years_Fri_date
FROM (
    VALUES('20160916'),('20160909'),
          ('20160910'),/*test a Saturday input date to verify the code always returns Friday*/
          ('20160911'),/*test a Sunday input date to verify the code always returns Friday*/
          ('20160912')/*test a Monday input date to verify the code always returns Friday*/
) AS test_data(user_date_entered)
CROSS APPLY (
    /*push the date forward 6 days so we can "back up" to Friday*/
    SELECT 4 AS Friday, DATEADD(DAY, +6, DATEADD(YEAR, -1, user_date_entered)) AS prev_year_max_date
) AS assign_alias_names


--minimum amount of code
SELECT
    user_date_entered,
    DATEADD(DAY, -(DATEDIFF(DAY, 4,DATEADD(DAY, +6, DATEADD(YEAR, -1, user_date_entered))) % 7),
        DATEADD(DAY, +6, DATEADD(YEAR, -1, user_date_entered))) AS last_years_Fri_date
FROM (
    VALUES('20160916'),('20160909'),
          ('20160910'),/*test a Saturday input date to verify the code always returns Friday*/
          ('20160911'),/*test a Sunday input date to verify the code always returns Friday*/
          ('20160912')/*test a Monday input date to verify the code always returns Friday*/
) AS test_data(user_date_entered)
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Expert Comment

by:ScottPletcher
Comment Utility
Just don't use that code in Europe :-) :

SET LANGUAGE ENGLISH

DECLARE @date DATETIME = '2016-09-09'

SET @date = DATEADD(YEAR, -1, @date)

WHILE DATEPART(WEEKDAY, @date) <> 6
      SET @date = DATEADD(DAY, 1, @date)

SELECT @date


SET LANGUAGE GERMAN

SET @date = DATEADD(YEAR, -1, @date)

WHILE DATEPART(WEEKDAY, @date) <> 6
      SET @date = DATEADD(DAY, 1, @date)

SELECT @date
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