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 Posted Friday, March 1, 2013 4:40 AM
 Valued Member Group: General Forum Members Last Login: Tuesday, May 7, 2013 3:30 AM Points: 62, Visits: 155
 This is my criteria for the SQL below:From Date 07 Jan 2013 To Date 18 Mar 20131 - First Sunday After 07 Jan 2013 is 13 Jan 2013 - Done (FirstSunday) 2 - First Sunday Before 18 Mar 2013 is 17 Mar 2013 - Done (LastSunday) 3 - How many days between 13 Jan 2013 and 17 Mar 2013 = 63, Divide 63 by 7 = 9, +1 = 10So what I need my datediff to do is use the results from the 1st and 2nd rows above (First and Last Sundays date) so I get the correct result for section 3. Can you help?Declare @From DatetimeDeclare @To DatetimeSet @From = '07 Jan 2013'Set @To = '18 Mar 2013'Select dateadd(dd, CASE WHEN datepart(weekday, @From) = 1 THEN 0 ELSE 8 - datepart(weekday, @From) END,@From) as FirstSunday, dateadd(dd, CASE WHEN datepart(weekday, @To) = 1 THEN 0 ELSE 1 - datepart(weekday, @To) END,@To) as LastSunday,datediff(d,@From, @To)
Post #1425437
 Posted Friday, March 1, 2013 6:34 AM
 SSCrazy Eights Group: General Forum Members Last Login: Thursday, December 1, 2016 10:11 AM Points: 8,587, Visits: 18,753
 What happens if your initial FROMdate or TOdate happen to be a sunday? This batch shows solutions either way:`-- Recalculating startdate;WITH SampleData AS ( SELECT TheDate = CAST('06 Jan 2013' AS DATE) UNION ALL SELECT TheDate = '07 Jan 2013' UNION ALL -- StartDate SELECT TheDate = '08 Jan 2013' UNION ALL SELECT TheDate = '09 Jan 2013' UNION ALL SELECT TheDate = '10 Jan 2013' UNION ALL SELECT TheDate = '11 Jan 2013' UNION ALL SELECT TheDate = '12 Jan 2013' UNION ALL SELECT TheDate = '13 Jan 2013' UNION ALL SELECT TheDate = '14 Jan 2013' UNION ALL SELECT TheDate = '15 Jan 2013' UNION ALL SELECT TheDate = '16 Jan 2013' UNION ALL SELECT TheDate = '17 Jan 2013' UNION ALL SELECT TheDate = '18 Jan 2013' UNION ALL SELECT TheDate = '19 Jan 2013' UNION ALL SELECT TheDate = '20 Jan 2013' UNION ALL SELECT TheDate = '21 Jan 2013' )SELECT DATENAME(weekday,TheDate), TheDate, FirstSundayAfterTheDate1 = DATEADD(dd,6-DATEDIFF(dd,7,TheDate)%7,TheDate), -- use this FirstSundayAfterTheDate2 = DATEADD(dd,7-DATEDIFF(dd,6,TheDate)%7,TheDate) -- or this FROM SampleDataCROSS APPLY ( -- workings - helps to explain the solution SELECT calc1 = DATEDIFF(dd,7,TheDate), calc2 = 6-DATEDIFF(dd,7,TheDate)%7) x-- Recalculating enddate;WITH SampleData AS ( SELECT TheDate = CAST('06 Mar 2013' AS DATE) UNION ALL SELECT TheDate = '07 Mar 2013' UNION ALL SELECT TheDate = '08 Mar 2013' UNION ALL SELECT TheDate = '09 Mar 2013' UNION ALL SELECT TheDate = '10 Mar 2013' UNION ALL SELECT TheDate = '11 Mar 2013' UNION ALL SELECT TheDate = '12 Mar 2013' UNION ALL SELECT TheDate = '13 Mar 2013' UNION ALL SELECT TheDate = '14 Mar 2013' UNION ALL SELECT TheDate = '15 Mar 2013' UNION ALL SELECT TheDate = '16 Mar 2013' UNION ALL SELECT TheDate = '17 Mar 2013' UNION ALL SELECT TheDate = '18 Mar 2013' UNION ALL -- enddate SELECT TheDate = '19 Mar 2013' UNION ALL SELECT TheDate = '20 Mar 2013' UNION ALL SELECT TheDate = '21 Mar 2013' )SELECT DATENAME(weekday,TheDate), TheDate, calc1 = DATEDIFF(dd,6,TheDate), calc2 = DATEDIFF(dd,6,TheDate)%7, LastSundayBeforeEnddate1 = DATEADD(dd,0-DATEDIFF(dd,6,TheDate)%7,TheDate), -- use this LastSundayBeforeEnddate2 = DATEADD(dd,-1-DATEDIFF(dd,7,TheDate)%7,TheDate) -- or thisFROM SampleData` “Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail ShawFor fast, accurate and documented assistance in answering your questions, please read this article.Understanding and using APPLY, (I) and (II) Paul White Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff ModenExploring Recursive CTEs by Example Dwain Camps
Post #1425468
 Posted Friday, March 1, 2013 8:09 AM
 Valued Member Group: General Forum Members Last Login: Tuesday, May 7, 2013 3:30 AM Points: 62, Visits: 155
 Thanks Chris, I have figured it out now:Declare @From DatetimeDeclare @To Datetime Set @From = '07 Jan 2013'Set @To = '18 Mar 2013' SELECT Datediff(Week, dateadd(dd, CASE WHEN datepart(weekday, @From) = 1 THEN 0 ELSE 8 - datepart(weekday, @From) END,@From), dateadd(dd, CASE WHEN datepart(weekday, @To) = 1 THEN 0 ELSE 1 - datepart(weekday, @To) END,@To))+1
Post #1425532
 Posted Friday, March 1, 2013 8:34 AM
 SSCrazy Eights Group: General Forum Members Last Login: Thursday, December 1, 2016 10:11 AM Points: 8,587, Visits: 18,753
 wafw1971 (3/1/2013)Thanks Chris, I have figured it out now:Declare @From DatetimeDeclare @To Datetime Set @From = '07 Jan 2013'Set @To = '18 Mar 2013' SELECT Datediff(Week, dateadd(dd, CASE WHEN datepart(weekday, @From) = 1 THEN 0 ELSE 8 - datepart(weekday, @From) END,@From), dateadd(dd, CASE WHEN datepart(weekday, @To) = 1 THEN 0 ELSE 1 - datepart(weekday, @To) END,@To))+1Be careful with datepart. From BOL: "When datepart is week (wk, ww) or weekday (dw), the return value depends on the value that is set by using SET DATEFIRST." The first day of the week is sunday in the US, monday in the UK. “Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail ShawFor fast, accurate and documented assistance in answering your questions, please read this article.Understanding and using APPLY, (I) and (II) Paul White Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff ModenExploring Recursive CTEs by Example Dwain Camps
Post #1425552

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