Can someone help me with getting the first and last day of week based on yearweek integer like 201648
without conserning about setting the @@firstdate
attribute. I want iso date
starting on monday in datetime
format.
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S3S
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user5767413
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2Possible duplicate of [Get dates from a week number in T-SQL](http://stackoverflow.com/questions/607817/get-dates-from-a-week-number-in-t-sql) – David Rushton Dec 14 '16 at 16:11
2 Answers
1
After a little consideration, I thought that perhaps my dynamic Date/Time Range UDF may help here. I use this UDF to generate dynamic date/time ranges. You can supply the desired date range, date part and increment. A tally table would do the trick as well
In this case, we are getting the Nth Monday regardless of the datepart(WK,..) as per the requirements.
Declare @YYYYWW int = 201648
Select WkNbr = B.RetSeq
,WkBeg = B.RetVal
,WkEnd = DateAdd(DD,6,B.RetVal)
From (
Select MinDate=Min(RetVal)
From [dbo].[udf-Range-Date](DateFromParts(Left(@YYYYWW,4),1,1),DateFromParts(Left(@YYYYWW,4),1,10),'DD',1)
Where DateName(DW,RetVal)='Monday'
) A
Cross Apply (Select * From [dbo].[udf-Range-Date](A.MinDate,DateFromParts(Left(@YYYYWW,4),12,31),'DD',7) ) B
Where B.RetSeq = Right(@YYYYWW,2)
Returns
WkNbr WkBeg WkEnd
48 2016-11-28 2016-12-04
The UDF if interested
CREATE FUNCTION [dbo].[udf-Range-Date] (@R1 datetime,@R2 datetime,@Part varchar(10),@Incr int)
Returns Table
Return (
with cte0(M) As (Select 1+Case @Part When 'YY' then DateDiff(YY,@R1,@R2)/@Incr When 'QQ' then DateDiff(QQ,@R1,@R2)/@Incr When 'MM' then DateDiff(MM,@R1,@R2)/@Incr When 'WK' then DateDiff(WK,@R1,@R2)/@Incr When 'DD' then DateDiff(DD,@R1,@R2)/@Incr When 'HH' then DateDiff(HH,@R1,@R2)/@Incr When 'MI' then DateDiff(MI,@R1,@R2)/@Incr When 'SS' then DateDiff(SS,@R1,@R2)/@Incr End),
cte1(N) As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)),
cte2(N) As (Select Top (Select M from cte0) Row_Number() over (Order By (Select NULL)) From cte1 a, cte1 b, cte1 c, cte1 d, cte1 e, cte1 f, cte1 g, cte1 h ),
cte3(N,D) As (Select 0,@R1 Union All Select N,Case @Part When 'YY' then DateAdd(YY, N*@Incr, @R1) When 'QQ' then DateAdd(QQ, N*@Incr, @R1) When 'MM' then DateAdd(MM, N*@Incr, @R1) When 'WK' then DateAdd(WK, N*@Incr, @R1) When 'DD' then DateAdd(DD, N*@Incr, @R1) When 'HH' then DateAdd(HH, N*@Incr, @R1) When 'MI' then DateAdd(MI, N*@Incr, @R1) When 'SS' then DateAdd(SS, N*@Incr, @R1) End From cte2 )
Select RetSeq = N+1
,RetVal = D
From cte3,cte0
Where D<=@R2
)
/*
Max 100 million observations -- Date Parts YY QQ MM WK DD HH MI SS
Syntax:
Select * from [dbo].[udf-Range-Date]('2016-10-01','2020-10-01','YY',1)
Select * from [dbo].[udf-Range-Date]('2016-01-01','2017-01-01','MM',1)
*/

John Cappelletti
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declare @yrwk int = 201648
declare @yr int = left(@yrwk,4)
declare @wk int = right(@yrwk,2)
select dateadd (week, @wk, dateadd (year, @yr-1900, 0)) - 4 - datepart(dw, dateadd (week, @wk, dateadd (year, @yr-1900, 0)) - 4) + 1
--returns 11/27/2016 which is Sunday of that week (start of week)
--change +1 to +2 at the end for "Monday"

S3S
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1I'll be dipped! Never considered left/right on an INT. Stealing IT +1 for that alone – John Cappelletti Dec 14 '16 at 16:17
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I can't thank you enough for that little trick. I've twisted myself into knots getting positions of an int. – John Cappelletti Dec 14 '16 at 16:23
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@JohnCappelletti don't forget wrapper... it helps to get those middle digits. `SELECT LEFT(RIGHT(123467,4),1)` – S3S Dec 14 '16 at 16:25
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1This is exactly why I love SO. I learn something new every day. – John Cappelletti Dec 14 '16 at 16:29
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1@JohnCappelletti Hi, this is simple: The function `LEFT()` expects a string as input. `INT` can be translated to a string implicitly. So this is the same as `LEFT(CAST(SomeInt AS VARCHAR(100),4)`. One must be aware, that the returned result is of string type (but will be implicitly casteable to `INT` again. – Shnugo Dec 14 '16 at 16:34
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@Shnugo When I saw it in-action, my actual thought was "I'm a dope" – John Cappelletti Dec 14 '16 at 16:37