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SQL Puzzle | Get Weeks (Start Date – End Date) for a particular month & year
Puzzle Statement
Recently I was asked by a developer how we can find weeks of a particular month and year. Say you want to find out weeks for December month for 2015 year.
Rules
1. Week starts from a Sunday and ends on a Saturday
2. First week start date should be first day of the month
3. Last week end date should be last day of the month
Sample Input
Month = 12 i.e. December
Year = 2015
Expected output
WeekStart | WeekEnd |
01-12-2015 | 06-12-2015 |
07-12-2015 | 13-12-2015 |
14-12-2015 | 20-12-2015 |
21-12-2015 | 27-12-2015 |
28-12-2015 | 31-12-2015 |
Rules/Restrictions
The solution should be should use “SELECT” statement or “CTE”.
Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com
SOLUTION #
-- DECLARE @Mon AS INT = 12 , @Year AS INT = 2015 ;WITH CTE AS ( SELECT DISTINCT number, DATEFROMPARTS(@Year,@Mon,Number) Dt ,DATEADD(d, (8 - datepart(WEEKDAY, DATEFROMPARTS(@Year,@Mon,Number))), DATEFROMPARTS(@Year,@Mon,Number)) Wk FROM MASTER..SPT_Values WHERE Number > 0 AND number < DAY(EOMONTH(CONCAT(@Year,'/',@Mon,'/','01'))) ), CTE1 AS ( SELECT CASE WHEN number = 1 THEN DATEADD(d,-1,Dt) ELSE Dt End Dt ,CASE WHEN Wk > EOMONTH(dt) THEN EOMONTH(dt) ELSE WK END Wk FROM CTE ) SELECT MIN(DATEADD(d,1,Dt)) WeekStart , Wk WeekEnd FROM CTE1 GROUP BY Wk ORDER BY WeekStart -- |
Add a comment if you have any other solution in mind. I would love to learn it.
We all need to learn.
Enjoy !!! Keep Learning
Pawan Kumar Khowal
Http://MSBISkills.com
I have a question. Given that the week ends on a Saturday and starts on a Sunday, shoudn’t the expected results look like the results below? In otherwords, the WeekStart would always fall on a Sunday.
WeekStart WeekEnd
2015-12-01 2015-12-05
2015-12-06 2015-12-12
2015-12-13 2015-12-19
2015-12-20 2015-12-26
2015-12-27 2015-12-31
Based on my interpretation of the rules, here is my solution:
declare @month tinyint = 12, @year int = 2015;
with cte
as
(
select weekStart, weekEnd
from
(
select datefromparts(@year, @month, 1) weekStart,
dateadd(day,7 – datepart(weekday,datefromparts(@year, @month, 1)),
datefromparts(@year, @month, 1)) weekEnd
) d
union all
select dateadd(day,1,weekEnd) weekStart,
case when dateadd(day,7,weekEnd) > eomonth(datefromparts(@year, @month, 1))
then eomonth(datefromparts(@year, @month, 1))
else dateadd(day,7,weekEnd) end weekendEnd
from cte
where dateadd(day,1,weekEnd) <= eomonth(datefromparts(@year, @month, 1))
)
select *
from cte;
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Well we can assume weekstart depending on our business or scenario. Thats perfectly fine. Also great solution sir !
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