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SQL Puzzle | Find every Saturday and their corresponding last months Saturday with in a Year
Puzzle Statement
In this puzzle you have to find every Saturday and their corresponding last months Saturday with in a Year. For example in year 2012 consider first two months January and February following table is required. For January obviously we get all null values, For February first Saturday is on 4th Feb so here we have to pull first Saturday from January and so on. If there is no corresponding record then we have show NULL value. See table below for details.
PrevSaturday | Saturday |
NULL | 07-01-2012 |
NULL | 14-01-2012 |
NULL | 21-01-2012 |
NULL | 28-01-2012 |
07-01-2012 | 04-02-2012 |
14-01-2012 | 11-02-2012 |
21-01-2012 | 18-02-2012 |
28-01-2012 | 25-02-2012 |
Well this question was asked at Noida based IT company. You cannot use a loop or cursor. Please check input and expected output for details.
Sample Input
Year = 2012
Expected output
PrevSaturday | Saturday |
NULL | 07-01-2012 |
NULL | 14-01-2012 |
NULL | 21-01-2012 |
NULL | 28-01-2012 |
07-01-2012 | 04-02-2012 |
14-01-2012 | 11-02-2012 |
21-01-2012 | 18-02-2012 |
28-01-2012 | 25-02-2012 |
04-02-2012 | 03-03-2012 |
11-02-2012 | 10-03-2012 |
18-02-2012 | 17-03-2012 |
25-02-2012 | 24-03-2012 |
NULL | 31-03-2012 |
03-03-2012 | 07-04-2012 |
10-03-2012 | 14-04-2012 |
17-03-2012 | 21-04-2012 |
24-03-2012 | 28-04-2012 |
07-04-2012 | 05-05-2012 |
14-04-2012 | 12-05-2012 |
21-04-2012 | 19-05-2012 |
28-04-2012 | 26-05-2012 |
05-05-2012 | 02-06-2012 |
12-05-2012 | 09-06-2012 |
19-05-2012 | 16-06-2012 |
26-05-2012 | 23-06-2012 |
NULL | 30-06-2012 |
02-06-2012 | 07-07-2012 |
09-06-2012 | 14-07-2012 |
16-06-2012 | 21-07-2012 |
23-06-2012 | 28-07-2012 |
07-07-2012 | 04-08-2012 |
14-07-2012 | 11-08-2012 |
21-07-2012 | 18-08-2012 |
28-07-2012 | 25-08-2012 |
04-08-2012 | 01-09-2012 |
11-08-2012 | 08-09-2012 |
18-08-2012 | 15-09-2012 |
25-08-2012 | 22-09-2012 |
NULL | 29-09-2012 |
01-09-2012 | 06-10-2012 |
08-09-2012 | 13-10-2012 |
15-09-2012 | 20-10-2012 |
22-09-2012 | 27-10-2012 |
06-10-2012 | 03-11-2012 |
13-10-2012 | 10-11-2012 |
20-10-2012 | 17-11-2012 |
27-10-2012 | 24-11-2012 |
03-11-2012 | 01-12-2012 |
10-11-2012 | 08-12-2012 |
17-11-2012 | 15-12-2012 |
24-11-2012 | 22-12-2012 |
NULL | 29-12-2012 |
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 #
-- ;WITH CTE AS ( SELECT DATEFROMPARTS(2012,1,1) Start , DATENAME(DW, DATEFROMPARTS(2012,1,1)) DayNamea UNION ALL SELECT DATEADD(d,1,Start) Start , DATENAME(DW, DATEADD(d,1,Start)) DayNamea FROM CTE WHERE Start < DATEFROMPARTS(2012,12,31) ) ,CTE1 AS ( SELECT Start Saturday , Months , ranker FROM ( SELECT * , ROW_NUMBER() OVER (PARTITION BY Months ORDER BY Months) rnk , ROW_NUMBER() OVER (PARTITION BY Months ORDER BY (SELECT NULL)) ranker FROM ( SELECT * , MONTH(Start) Months FROM CTE WHERE DayNamea = 'Saturday' ) a )a ) SELECT PrevSaturday , Saturday FROM CTE1 c1 OUTER APPLY ( SELECT Saturday PrevSaturday FROM CTE1 c WHERE c.Months = c1.Months - 1 AND c.ranker = c1.ranker )z OPTION ( MAXRECURSION 0 ) -- |
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