capture data for the week

Hello experts,

I need to help with the following syntax.  I need to capture sales for the current and sales for the same week last year.

Current calender  week: ?

Current calender   week Last year: ?
tips54Asked:
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ravikantninaveCommented:
find current calendar week
select DATEPART(WEEK,GETDATE())
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ravikantninaveCommented:
Here is full solution:

DECLARE @datecol datetime = GETDATE();
DECLARE @WeekNum INT
      , @YearNum char(4);

SELECT @WeekNum = DATEPART(WK, @datecol)
     , @YearNum = CAST(DATEPART(YY, @datecol) AS CHAR(4));

-- once you have the @WeekNum and @YearNum set, the following calculates the date range.
SELECT DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-1), 6) AS StartOfWeek;
SELECT DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + @YearNum) + (@WeekNum-1), 5) AS EndOfWeek;

--
Declare @NextDate datetime= DATEADD(YEAR,-1, getdate())
SET @YearNum= CAST(DATEPART(YY, @NextDate) AS CHAR(4));
SELECT DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-1), 6) AS StartOfWeek;
SELECT DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + @YearNum) + (@WeekNum-1), 5) AS EndOfWeek;
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tips54Author Commented:
Hi Ravi,
 
Your code returns the current week number for the year. (10)

what I'm after is something that returns : 3/2/2014 - 3/8/2014

And 3/2/2013  -  3/8/2013
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ravikantninaveCommented:
3/2/2014 - 3/8/2014 is the date range of current year of the 10 th week.

and date of 10th week of last year was 3/2/2013  -  3/8/2013
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tips54Author Commented:
If I have a table with multiple invoices with dates,  how would the syntax look for sum of invoices for just week 10?
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ravikantninaveCommented:
DECLARE @FirstDate as datetime = DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-1), 6) --AS StartOfWeek;
DECLARE @SecondDate as datetime = DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + @YearNum) + (@WeekNum-1), 5) --AS EndOfWeek;

Select sum(Amount) from Invoices Where InvDate > = @FirstDate and InvDate <=@SecondDate
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ravikantninaveCommented:
Same for last year
Declare @ThirdDate as datetime =DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-1), 6) --AS StartOfWeek;
Declare @ForthDate as Datetime =DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + @YearNum) + (@WeekNum-1), 5) --AS EndOfWeek;

Select sum(Amount) from Invoices Where InvDate > = @ThirdDate and InvDate <=@ForthDate
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tips54Author Commented:
where does it differentiate Current and Last year?

I get this error when I run the code:  
Cannot assign a default value to a local variable.
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tips54Author Commented:
Thanks ravikantninave,

I got it.  I had to set those values on the next line. This might be a Sql 2005 issue.
Thanks again.

What I Used:

/*
DECLARE @datecol datetime
set @datecol = GETDATE();
DECLARE @WeekNum INT
set @WeekNum = DATEPART(week,GETDATE());
   DECLARE    @YearNum char(4);
set @YearNum =DATEPART(year,GETDATE());

Declare @NextDate datetime
SET @NextDate= DATEADD(YEAR,-1, getdate())
DECLARE    @YearNum2 char(4);
SET @YearNum2= CAST(DATEPART(YY, @NextDate) AS CHAR(4));

-- once you have the @WeekNum and @YearNum set, the following calculates the date range.
SELECT DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-1), 6) AS StartOfWeek;
SELECT DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + @YearNum) + (@WeekNum-1), 5) AS EndOfWeek;

SELECT DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum2) + (@WeekNum-1), 6) AS StartOfWeek;
SELECT DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + @YearNum2) + (@WeekNum-1), 5) AS EndOfWeek;

*/
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tips54Author Commented:
How would I get the sum for the quarter?
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tips54Author Commented:
I would like to capture the same data quarterly compare to last years quarter.
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ravikantninaveCommented:
first three months of the year? or what? Find quarter :

SELECT DATEADD(mm, (quarter - 1) * 3, year_date) StartDate,
       DATEADD(dd, -1, DATEADD(mm, quarter * 3, year_date)) EndDate,
       quarter QuarterNo
  FROM
(
  SELECT '2013-01-01' year_date
) s CROSS JOIN
(
  SELECT 1 quarter UNION ALL
  SELECT 2 UNION ALL
  SELECT 3 UNION ALL
  SELECT 4
) q
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tips54Author Commented:
Thank you.

the second question was answered here:
http://www.experts-exchange.com/Database/MS-SQL-Server/Q_28381957.html
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