Finding Closest future date - sql

Question

Is it possible to find the closest future date (datetime) by a date varchar value?
Given,
DECLARE #DayValue VARCHAR(3)
, #DateValue DATETIME
SET #DayValue = 'Tue' -- Values could be 'Mon', 'Tue', 'Wed' and etc.
SET #DateValue = '10/15/2014' -- Format is MM/dd/yyyy
I want to get:
Oct 21 2014 12:00AM

Answer 1

Using Loop,
DECLARE #DayValue VARCHAR(3)
,#DateValue DATETIME
SET #DayValue = 'tue'
SET #DateValue = '10/15/2014'
declare #i int= 1 ,#day varchar(3) = null
while (#i<=7 )
begin
Select #day = left(datename (dw,#DateValue),3)
if #day = #DayValue
begin
Select #DateValue
break
end
Select #DateValue = #DateValue+ 1
Select #i = #i+1
end

Answer 2

You could use this function if you had a date-table:
CREATE FUNCTION [dbo].[GetNextDayOfWeek]
( #DayOfWeek VARCHAR(3),
#DateValue datetime
)
RETURNS SmallDateTime
AS
BEGIN
DECLARE #NextDayOfWeek smalldatetime
SET #NextDayOfWeek = (
SELECT
MIN(d.Date)
FROM
tDefDate d
WHERE
d.Date > #DateValue
AND LEFT(DATENAME(Weekday, d.Date), 3) = #DayOfWeek);
RETURN #NextDayOfWeek
END
Then it's simple as:
select [dbo].[GetNextDayOfWeek]('Tue', Getdate()) -- next tuesday=> 2014-10-21
Note that it takes the language of the database into account. So if it's in german:
select [dbo].[GetNextDayOfWeek]('Die', Getdate()) -- next tuesday(Dienstag)
Here's a version that works also without a date-table (but is less efficient).
CREATE FUNCTION [dbo].[GetNextDayOfWeek]
( #DayOfWeek VARCHAR(3),
#DateValue datetime
)
RETURNS SmallDateTime
AS
BEGIN
DECLARE #NextDayOfWeek smalldatetime
;WITH CTE as
(
SELECT GetDate() DateValue, DayNum=0
UNION ALL
SELECT DateValue + 1, DayNum=DayNum+1
FROM CTE
WHERE DayNum <=7
)
SELECT #NextDayOfWeek = (
SELECT
MIN(d.DateValue)
FROM
CTE d
WHERE d.DateValue > #DateValue
AND LEFT(DATENAME(Weekday, d.DateValue), 3) = #DayOfWeek
)OPTION (MAXRECURSION 8);
RETURN #NextDayOfWeek
END

Answer 3

If you could define DayValue as an integer, you solve this problem with more elegant way:
DECLARE #DayValue int, #DateValue DATETIME 
SET #DayValue = 3 -- Values could be 1-Sun, 2-Mon, 3-Tue, 4-Wed and etc. 
SET #DateValue = '10/15/2014' -- Format is MM/dd/yyyy
select dateadd(day,(7 + #DayValue - datepart(w,#DateValue)), #DateValue)
TRY SQL FIDDLE DEMO

Answer 4

No loops and will work in selects with multiple rows. :)
DECLARE #DayValue CHAR(3)
DECLARE #DateValue DATETIME
DECLARE #FutureDate DATE
SET #DayValue='MON'
SET #DateValue='10/12/2014'
DECLARE #Days TABLE
(
[DayOfWeek] TINYINT,
[DayValue] CHAR(3)
)
INSERT INTO #Days([DayOfWeek],[DayValue])
SELECT 0,'SUN' UNION
SELECT 1,'MON' UNION
SELECT 2,'TUE' UNION
SELECT 3,'WED' UNION
SELECT 4,'THU' UNION
SELECT 5,'FRI' UNION
SELECT 6,'SAT'
SET #FutureDate=
DATEADD(DAY,
--Skip to next week if we are already on the desired day or past it
+ CASE WHEN ((SELECT [DayOfWeek] FROM #Days WHERE [DayValue]=#DayValue)<DATEPART(WEEKDAY,#DateValue)) THEN 7 ELSE 0 END
--reset to start of week (add one as DATEPART is base 1, not base 0)
- DATEPART(WEEKDAY,#DateValue) + 1
--Add the desired day of the week
+ (SELECT [DayOfWeek] FROM #Days WHERE [DayValue]=#DayValue)
,#DateValue)
SELECT #FutureDate

Answer 5

This is a bit chunky solution, but it works. :)
SET DATEFIRST 1
DECLARE #DateValue DateTime
, #DayValue VARCHAR(3)
, #tmp INT
SET #DateValue = '09/30/2014'
SET #DayValue = 'wed'
SET #tmp = CASE #DayValue
WHEN 'Mon' THEN (1 - DATEPART(dw, #DateValue) + 7) % 7
WHEN 'Tue' THEN (2 - DATEPART(dw, #DateValue) + 7) % 7
WHEN 'Wed' THEN (3 - DATEPART(dw, #DateValue) + 7) % 7
WHEN 'Thu' THEN (4 - DATEPART(dw, #DateValue) + 7) % 7
WHEN 'Fri' THEN (5 - DATEPART(dw, #DateValue) + 7) % 7
WHEN 'Sat' THEN (6 - DATEPART(dw, #DateValue) + 7) % 7
WHEN 'Sun' THEN (7 - DATEPART(dw, #DateValue) + 7) % 7
END
SELECT
CASE
WHEN #tmp = 0 THEN DATEADD (DAY, 7, #DateValue)
ELSE DATEADD (DAY, #tmp, #DateValue)
END