ssrs return first monday of month

Hi I am wanting to set a parameter in SSRS, Report Builder 3.0  that returns the first Monday of April in the current year.  Could anyone help me with this expression please, I cannot seem to get my head around it.
deborahhowson00Asked:
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deborahhowson00Author Commented:
I want to return the date of that first Monday of April in the current year, sorry, just to be clear.
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
I have an article out there called How to Build Your Own SQL Calendar Table that handles situations like this.

Otherwise, you'll probably need to build some kind of loop to start at CAST(CAST(YEAR(GETDATE) as char(4)) + '-04-01' as date), then incriment one day at a time until SELECT DATEPART(dw, {the date} ) = 1 for Monday.

So ... Create a SP that contains the below code, attach it as a data set, and in your parameter set all values to this:
Declare @dt datetime
SET @dt = CAST(CAST(YEAR(GETDATE()) as char(4)) + '-04-01' as date) 

WHILE (DATEPART(dw, @dt) <> 2)
	begin
	SET @dt = @dt + 1
	end

-- Return set
SELECT @dt

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Harish VargheseProject LeaderCommented:
Try this:
=DateAdd("d", 1 - WeekDay (Cstr(Year(Today)) + "/04/07", 2), Cstr(Year(Today)) + "/04/07")

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Vikas GargBusiness Intelligence DeveloperCommented:
Hi,

You can try this one

=DATEADD("ww",DATEDIFF("ww", DATEADD("d",0,"1900-01-01"), DATEADD("d", 6 - DATEPART("d", Parameters!DTS.Value), Parameters!DTS.Value)),"1900-01-01")

Here you can use the Parameter DTS value as per the date for which month you want first monday
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Scott PletcherSenior DBACommented:
You can use a simple date calculation to get the first Monday (or given day) of a month -- no looping and definitely no I/O needed.

I think you can use a SELECT to provide a value for SSRS.

Change the "MONTH, 3" to get a different month.  For example, to get first Monday of Jan, use "MONTH, 0", for Jun, "MONTH, 5", etc..


SELECT DATEADD(DAY, -DATEDIFF(DAY, 0, month_day_7) % 7, month_day_7)
FROM (
    SELECT DATEADD(DAY, 6, DATEADD(MONTH, 3, DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0))) AS month_day_7
) AS month_day_7
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deborahhowson00Author Commented:
Cheers, worked a treat!
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
deborahhowson00

Did you try the first solutions provided you?  If yes, please comment on whether they worked.  

Not to take anything away from Scott's answer, which I'm kind of liking btw, but it is not good EE etiquette to ignor multiple earlier comments.

Just curious.
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Scott PletcherSenior DBACommented:
Perhaps.  But I think the overall quality of the answer should matter more than going strictly by the order in which answers were posted.

FWIW, I went back and checked both the "=DATEADD" codes -- neither was fully accurate.

As to the looping code, I would never consider looping if a simple SELECT could do the task flexibly and efficiently.  And it's not accurate unless the DATEFIRST setting is a specific value.

The SELECT code works regardless of any date setting.  And it can easily be completely generalized to work for any year, month and weekday.  For example, for the first Friday in June of the prior year:


DECLARE @day_of_week int --0=Mon,1=Tue,...,6=Sun.
DECLARE @month int --1=Jan,2=Feb,...,12=Dec.
DECLARE @year_adjustment int --0=current yr; -1=prior yr; 1=next yr; etc.
SET @day_of_week = 4
SET @month = 6
SET @year_adjustment = -1

SELECT DATEADD(DAY, -DATEDIFF(DAY, @day_of_week, month_day_7) % 7, month_day_7)
 FROM (
     SELECT DATEADD(DAY, 6, DATEADD(MONTH, (@month - 1), DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()) + @year_adjustment, 0))) AS month_day_7
 ) AS month_day_7
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