Get every 2 weeks on particular days SQL Server - sql

Question

I have a query getting the days in particular day that I selected and show the days in a month. My problem is that I need every two weeks of the month. And I also need to get long date format: dddd, dd MMMM yyyy and eliminate the time. Please help me.
DECLARE #startDate DATETIME = '9/1/2019'
DECLARE #endDate DATETIME = '12/31/2019'
DECLARE #dayOfWeek INT = 1;
WITH cte_Recursion AS
(
SELECT #startDate AS [Date]
UNION ALL
SELECT DATEADD(DAY, 1, [Date])
FROM cte_Recursion
WHERE [Date] < #endDate
)
SELECT [Date]
FROM cte_Recursion
WHERE DATEPART(WEEKDAY, [Date]) = #dayOfWeek
OPTION (MAXRECURSION 0)
The data returned looks like this:
2019-09-01 00:00:00.000
2019-09-08 00:00:00.000
2019-09-15 00:00:00.000
2019-09-22 00:00:00.000
2019-09-29 00:00:00.000
2019-10-06 00:00:00.000
2019-10-13 00:00:00.000
2019-10-20 00:00:00.000
2019-10-27 00:00:00.000
2019-11-03 00:00:00.000
2019-11-10 00:00:00.000
2019-11-17 00:00:00.000
2019-11-24 00:00:00.000
And I need to get like this (every other week):
2019-09-01 00:00:00.000
2019-09-15 00:00:00.000
2019-09-29 00:00:00.000
2019-10-13 00:00:00.000
2019-10-27 00:00:00.000
2019-11-10 00:00:00.000
2019-11-24 00:00:00.000

Answer 1

By changing your CTE to start on the correct day of the week, you can then change the recursion to add 14 days instead of 1, giving you the desired result without further manipulation:
DECLARE #startDate DATETIME = '9/1/2019'
DECLARE #endDate DATETIME = '12/31/2019'
DECLARE #dayOfWeek INT = 1;
WITH cte_Recursion AS
(SELECT DATEADD(DAY, ((#dayOfWeek - DATEPART(WEEKDAY, #startDate)) % 7 + 7) %7, #startDate) AS [Date]
UNION ALL SELECT DATEADD(DAY, 14, [Date])
FROM cte_Recursion
WHERE DATEADD(DAY, 14, [Date]) < #endDate)
SELECT [Date] FROM cte_Recursion
Output:
Date
01/09/2019 00:00:00
15/09/2019 00:00:00
29/09/2019 00:00:00
13/10/2019 00:00:00
27/10/2019 00:00:00
10/11/2019 00:00:00
24/11/2019 00:00:00
08/12/2019 00:00:00
22/12/2019 00:00:00
Demo on dbfiddle
To change the date format to dddd, dd MMMM yyyy, simply replace the final SELECT with:
SELECT FORMAT([Date], 'dddd, dd MMMM yyyy') AS [Date] FROM cte_Recursion
Output:
Date
Sunday, 01 September 2019
Sunday, 15 September 2019
Sunday, 29 September 2019
Sunday, 13 October 2019
Sunday, 27 October 2019
Sunday, 10 November 2019
Sunday, 24 November 2019
Sunday, 08 December 2019
Sunday, 22 December 2019
Demo on dbfiddle

Answer 2

you can use it :
DECLARE #startDate DATETIME = '9/1/2019'
DECLARE #endDate DATETIME = '12/31/2019'
DECLARE #dayOfWeek INT = 1;
WITH cte_Recursion AS
(
SELECT #startDate AS [Date]
UNION ALL SELECT DATEADD(DAY, 1, [Date])
FROM cte_Recursion WHERE [Date] < #endDate
)
SELECT [Date] FROM cte_Recursion
WHERE DATEPART(WEEKDAY, [Date]) = #dayOfWeek
and DATEPART(WEEK, [Date]) % 2 = 0
OPTION (MAXRECURSION 0)

Answer 3

You could use a recursive CTE as
WITH CTE AS
(
SELECT '2019-09-01 00:00:00.000' ADate
UNION SELECT '2019-09-08 00:00:00.000'
UNION SELECT '2019-09-15 00:00:00.000'
UNION SELECT '2019-09-22 00:00:00.000'
UNION SELECT '2019-09-29 00:00:00.000'
UNION SELECT '2019-10-06 00:00:00.000'
UNION SELECT '2019-10-13 00:00:00.000'
UNION SELECT '2019-10-20 00:00:00.000'
UNION SELECT '2019-10-27 00:00:00.000'
UNION SELECT '2019-11-03 00:00:00.000'
UNION SELECT '2019-11-10 00:00:00.000'
UNION SELECT '2019-11-17 00:00:00.000'
UNION SELECT '2019-11-24 00:00:00.000'
),
F AS
(
SELECT CAST('2019-09-01 00:00:00.000' AS DATETIME) ADate
UNION ALL
SELECT DATEADD(Day, 14, F.ADate)
FROM F
WHERE ADate < '2019-11-24 00:00:00.000'
)
SELECT *
FROM F;