How to get week of a mont with input parameter date.

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How to get week of a mont with input parameter date.

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Lynn Pettis

Posted Wednesday, September 12, 2012 7:39 AM

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Lynn Pettis (9/11/2012)

Here is a bit of code that returns what you are looking for:


declare @TestDate date = '2012-09-11';
with SevenRows(n) as (select row_number() over (order by (select null)) - 1 from (values (1),(1),(1),(1),(1),(1),(1))dt(n))
select dateadd(dd, n,dateadd(wk,datediff(wk,0,dateadd(dd,-1,@TestDate)),0)) from SevenRows;

Code rewritten for SQL Server 2005:


declare @TestDate date = '2012-09-11';
with SevenRows(n) as (select
                                      row_number() over (order by (select null)) - 1
                                   from (select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1)dt(n))
select dateadd(dd, n,dateadd(wk,datediff(wk,0,dateadd(dd,-1,@TestDate)),0)) from SevenRows;

Lynn Pettis

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Post #1357971

Jeff Moden

Posted Wednesday, September 12, 2012 4:08 PM

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xRafo (9/11/2012)

Hi,
i have a question,
How to get First week of a actual month?

Please define what you mean by "first week of the month". For example, is it the actual first 7 days of the month regardless of which day of the week it starts on? Is it the week starting on Sunday (for example) that contains the first of the month even if it starts on a Saturday? Is it the week of the month starting on Sunday (for example) that contains at least the first 4 calendar days of the month (like ISO).

What would you define as the first week of September 2012 and why (for example???)

Now... all of that appears to be contrary to the rest of your request...

In my procedure i will send a parameter date: Example

input -> today()

output <-
10/09 | 11/09 | 12/09 | 13/09 | 14/09 | 15/09 | 16/09

any help would be useful.

If that's what you really want, then the other posters have already posted some dandy ideas on how to do that.

--Jeff Moden

"RBAR is pronounced "ree-bar" and is a "Modenism" for "Row-By-Agonizing-Row".

First step towards the paradigm shift of writing Set Based code:
Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column."

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Post #1358248

Jeff Moden

Posted Wednesday, September 12, 2012 8:25 PM

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Lynn Pettis (9/12/2012)

Lynn Pettis (9/11/2012)

Here is a bit of code that returns what you are looking for:


declare @TestDate date = '2012-09-11';
with SevenRows(n) as (select row_number() over (order by (select null)) - 1 from (values (1),(1),(1),(1),(1),(1),(1))dt(n))
select dateadd(dd, n,dateadd(wk,datediff(wk,0,dateadd(dd,-1,@TestDate)),0)) from SevenRows;

Code rewritten for SQL Server 2005:


declare @TestDate date = '2012-09-11';
with SevenRows(n) as (select
                                      row_number() over (order by (select null)) - 1
                                   from (select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1 union all
                                            select 1)dt(n))
select dateadd(dd, n,dateadd(wk,datediff(wk,0,dateadd(dd,-1,@TestDate)),0)) from SevenRows;

I have a general distrust for the "wk" datepart even though it doesn't seem to matter matter here. With that thought in mind, here's a bit of code that uses a zero based Tally Table. It can be easily modified to handle a unit based Tally Table if needed. If nothing else, it makes for some really simple code.

  SELECT DATEADD(dd,DATEDIFF(dd,-1,GETDATE())/7*7+(t.N),-1)
    FROM dbo.Tally t
   WHERE t.N BETWEEN 0 AND 6

--Jeff Moden

Post #1358298

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