Get only Saturday and Sunday dates of month in SQL Server - sql-server

Question

Can I find the date of a day that is on which dates the Saturdays and Sundays of a specific month fall? For e.g consider the month of JANUARY-2017. The following dates are weekend days:
7/1/2017 - Saturday
14/1/2017 - Saturday
21/1/2017 - Saturday
28/1/2017 - Saturday
1/1/2017 - Sunday
8/1/2017 - Sunday
15/1/2017 - Sunday
22/1/2017 - Sunday
29/1/2017 - Sunday
I want a SQL Server query for this such that when I pass in month and year as input, I should get back all the above dates (only dates of Saturday and Sunday) as output
I do not wish to use any user defined function and want to finish it in a single SELECT statement

Answer 1

Note: As already noted by another user in the comments, this query depends upon your server settings, namely DATEFIRST. If you need alterations to the query because of different settings, just tell me and I can change it around for you.
Using a CTE as dummy data...
/* Ignore this part...*/
WITH CTE AS
(
SELECT CAST('01/01/2017' AS DATE) AS [Date]
UNION ALL
SELECT DATEADD(DAY,1,[Date])
FROM CTE
WHERE DATE <= '12/31/2017'
)
/*Your actual SELECT statement would look like this, from your own table of course*/
SELECT
[Date]
,CASE DATEPART(dw,[Date])
WHEN 1 THEN 'Sunday'
WHEN 2 THEN 'Monday'
WHEN 3 THEN 'Tuesday'
WHEN 4 THEN 'Wednesday'
WHEN 5 THEN 'Thursday'
WHEN 6 THEN 'Friday'
WHEN 7 THEN 'Saturday'
END
FROM CTE
WHERE DATEPART(dw,[Date]) IN (1,7)
AND MONTH([Date]) = 12--<month>
AND YEAR([Date]) = 2017--<year>
OPTION (MAXRECURSION 0) -- You won't need this line if you're querying a real table
;
If running that works for you, then your real query would probably look something like this:
SELECT
[Date]
,CASE DATEPART(dw,[Date])
WHEN 1 THEN 'Sunday'
WHEN 2 THEN 'Monday'
WHEN 3 THEN 'Tuesday'
WHEN 4 THEN 'Wednesday'
WHEN 5 THEN 'Thursday'
WHEN 6 THEN 'Friday'
WHEN 7 THEN 'Saturday'
END
FROM < the table you want >
WHERE DATEPART(dw,[Date]) IN (1,7) -- Only Sundays and Saturdays
AND MONTH([Date]) = < the month you want >
AND YEAR([Date]) = < the year you want >
;
If you want to generate the data, then a CTE is the way to go. If you're passing parameters, it would look something like this:
DECLARE
#MONTH INT
,#YEAR INT
;
SET #MONTH = 1;
SET #YEAR = 2017;
WITH CTE AS
(
SELECT CAST(CAST(#MONTH AS VARCHAR(2)) + '/01/' + CAST(#YEAR AS VARCHAR(4)) AS [Date]) AS DATE
UNION ALL
SELECT DATEADD(DAY,1,[Date])
FROM CTE
WHERE DATE <= CAST(#MONTH AS VARCHAR(2)) +
CASE
WHEN #MONTH IN (9,4,6,11)
THEN '/30/'
WHEN #MONTH IN (1,3,5,7,8,10,12)
THEN '/31/'
WHEN #MONTH = 2 AND #YEAR/4.00 = #YEAR/4
THEN '/29/'
ELSE '/28/'
END
+ CAST(#YEAR AS VARCHAR(4))
)
SELECT
[Date]
,CASE DATEPART(dw,[Date])
WHEN 1 THEN 'Sunday'
WHEN 2 THEN 'Monday'
WHEN 3 THEN 'Tuesday'
WHEN 4 THEN 'Wednesday'
WHEN 5 THEN 'Thursday'
WHEN 6 THEN 'Friday'
WHEN 7 THEN 'Saturday'
END
FROM CTE
WHERE DATEPART(dw,[Date]) IN (1,7)
OPTION (MAXRECURSION 0)
;

Answer 2

Please try this one.
DECLARE #Year AS INT=2017,
#Month AS INT=3,
#FirstDateOfYear DATETIME,
#LastDateOfYear DATETIME
SELECT #FirstDateOfYear = DATEADD(yyyy, #Year - 1900, 0)
SELECT #LastDateOfYear = DATEADD(yyyy, #Year - 1900 + 1, 0)
-- Creating Query to Prepare Year Data
;WITH cte AS (
SELECT 1 AS DayID,
#FirstDateOfYear AS FromDate,
DATENAME(dw, #FirstDateOfYear) AS Dayname
UNION ALL
SELECT cte.DayID + 1 AS DayID,
DATEADD(d, 1 ,cte.FromDate),
DATENAME(dw, DATEADD(d, 1 ,cte.FromDate)) AS Dayname
FROM cte
WHERE DATEADD(d,1,cte.FromDate) < #LastDateOfYear
)
SELECT FromDate AS Date, Dayname
FROM CTE
WHERE DayName IN ('Saturday','Sunday') and month(FromDate) = #Month
OPTION (MaxRecursion 370)

Answer 3

This should do the trick:
DECLARE #month date = '2017-01-01'
SET #month = dateadd(month, datediff(month, 0, #month), 0)
;WITH CTE as
(
SELECT 0 x
FROM (values(1),(1),(1),(1),(1),(1)) x(n)
),
CTE2 as
(
SELECT
top(day(eomonth(#month)))
-- use this syntax for sqlserver 2008
-- top(datediff(d, #month,dateadd(month,1,#month)))
cast(dateadd(d, row_number()over(order by(select 1))-1,#month) as date) cDate
FROM CTE CROSS JOIN CTE C2
)
SELECT
cDate,
datename(weekday, cDate) Weekday
FROM CTE2
WHERE
datediff(d,0,cDate)%7 > 4
Fiddle

Answer 4

From https://www.sqlservercentral.com/articles/finding-the-correct-weekday-regardless-of-datefirst, you simply:
(DATEPART(dw, #your_date) + ##DATEFIRST) % 7 NOT BETWEEN 2 AND 6

Answer 5

As requested, single select, language neutral, dateFirst neutral, almost SQL version neutral:
declare #OneDate datetime = '28/01/2017'; -- Any date from the target month/year
select MyDate -- raw date or ...
-- convert(varchar, MyDate, 103) + ' - ' + dateName(dw, MyDate) -- as Sample
as WeekEndDate
from (
select dateAdd(dd, number, dateAdd(mm, dateDiff(mm, 0, #OneDate), 0)) as MyDate
from master..spt_values
where type = 'P' and number < 31
) j
where 1 + (datePart(dw, MyDate) + ##DATEFIRST + 5) % 7 in (6, 7)
and month(MyDate) = month(#OneDate)
-- order by 1 + (datePart(dw, MyDate) + ##DATEFIRST + 5) % 7, MyDate -- as Sample
;

Answer 6

Another way to solve this problem as follow -
DECLARE #MONTH INT,#YEAR INT
SET #MONTH = 1;
SET #YEAR = 2017;
Declare #StartDate date =CAST(CAST(#MONTH AS VARCHAR(2)) + '/01/' + CAST(#YEAR AS VARCHAR(4)) AS [Date]), #EndDate date
Set #EndDate = EOMONTH(#StartDate)
Declare #Temp table (DateOfDay date, DaysName varchar(50))
While(#StartDate <= #EndDate)
Begin
Insert into #Temp
SELECT #StartDate DateOfMonth,
case when DATENAME(DW, #StartDate) = 'Saturday' then DATENAME(DW, #StartDate)
when DATENAME(DW, #StartDate) = 'Sunday' then DATENAME(DW, #StartDate)
end DaysName
set #StartDate = DATEADD(d,1,#StartDate)
End
select * from #Temp where DaysName is not null order by DaysName, DateOfDay

Answer 7

Can't you do something like this?
SELECT DATENAME(dw,'10/11/2016') AS DATE
WHERE DATE CONTAINS('Saturday') OR DATE CONTAINS('SUNDAY')
and instead of '10/11/2016' you only have to figure out how to generate all the dates in a month/year?