How to get date from day name? - sql-server

Question

How can I get the date of specific day ? Like if I have Thursday or month number ?
If I give 12 for instance I want to get the date of 12th day of this month. Or if I give 'Sun' or 'Sat' is it possible to get the dates of these days ?

Answer 1

DATEFROMPARTS function can construct a date from day, month and year.
DATEPARTS does the opposite - gives you the day, month, year, hour, etc. of a date. Or you can use functions like YEAR, MONTH and DAY.
You can deconstruct the value returned by GETDATE function and construct whatever date you want. Here is for example how to get the date for 12th day of the current month:
select DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), 12)
Converting 'Sun' or 'Sat' to date is a bit more difficult. First, they aren't quite deterministic. If today is Friday, "Sunday this week" means "next Sunday" in some parts of the world and "last Sunday" in others. You should implement your own logic based on the value returned by DATEPART(dw, GETDATE()) (which will give you the day of the week).

Answer 2

To find the weekday of the current month
DECLARE #daynumber INT = 12
SELECT datename(weekday, dateadd(d, #daynumber - 1, getdate()))
To find the dates of the current month of a given weekday
DECLARE #dayname char(3) = 'sat'
;WITH CTE as
(
SELECt TOP
(datediff(D, eomonth(getdate(), -1),eomonth(getdate())))
dateadd(d,row_number()over(ORDER BY 1/0),
eomonth(getdate(),-1))date
FROM
(values(1),(2),(3),(4),(5),(6))x(x),
(values(1),(2),(3),(4),(5),(6))y(x)
)
SELECT day(date) monthday, date
FROM CTE
WHERE left(datename(weekday, date),3) = #dayname

Answer 3

select sysdatetime(); --2018-12-13 16:29:56.0560574
---If I give 12 for instance I want to get the date of 12th day of this month.
declare #numDate int = 12;
select dateadd(m, datediff(m,0,getdate()),#numDate - 1 ); --2018-12-12 00:00:00.000
--Or if I give 'Sun' or 'Sat' is it possible to get the dates of these days ?
declare #text nvarchar(20) = 'Sunday';
declare #dateStart date = dateadd(month, datediff(month, 0, sysdatetime()), 0),
#days int =( select (DAY(dateadd(dd,-1,DATEADD(m,1,cast(2018 as varchar(4)) + '-' + cast(12 as varchar(2)) +'-01')))));
declare #dateEnd date = DATEADD(day,#days-1,#dateStart);
;WITH CTE (Dates,EndDate) AS
(
SELECT #dateStart AS Dates,#dateEnd AS EndDate
UNION ALL
SELECT DATEADD(day,1,Dates),EndDate
FROM CTE
WHERE DATEADD(day,1,Dates) <= EndDate
)
SELECT CTE.Dates, DATENAME(DW, CTE.Dates)
FROM CTE
where DATENAME(DW, CTE.Dates) = #text;
Result:
Dates,Day
2018/12/2,Sunday
2018/12/9,Sunday
2018/12/16,Sunday
2018/12/23,Sunday
2018/12/30,Sunday

Answer 4

-- Here is how to get week day name to week day number
DECLARE #T TABLE (Dow INT, NameOfDay VARCHAR(15), ShortName CHAR(3));
WITH Days AS
(
SELECT TOP 7
ROW_NUMBER() OVER(PARTITION BY object_id ORDER BY object_id) AS RowNo
FROM
sys.all_columns
)
INSERT INTO #T
SELECT
RowNo,
DATENAME(WEEKDAY, RowNo - 1),
LEFT(DATENAME(WEEKDAY, RowNo - 1), 3)
FROM
Days
SELECT
*
FROM
#T;
-- Here is how to get start of period
SELECT
DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0) AS StartOfDay,
DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0) AS StartOfWeek,
DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) AS StartOfMonth,
DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0) AS StartOfYear;
-- An example
WITH
StartPeriods AS
(
SELECT DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0) AS StartOfWeek
),
SelectedDay AS
(
SELECT
Dow - 1 AS Dow,
(SELECT StartOfWeek FROM StartPeriods) AS StartOfWeek
FROM
#T
WHERE
ShortName = 'Wed'
)
SELECT
DATEADD(DAY, Dow, StartOfWeek)
FROM
SelectedDay;