Calculating Month Start/End With A Fixed Last Day Of Month - sql

Question

I need to loop through each day within a month but with a fixed number as the last day of the month.
E.g. If the last day of the month is fixed as 30,
January would become:
Start of Month: Dec 31 2012 12:00AM
End of Month: Jan 30 2013 12:00AM
I have implemented the following which works perfectly except for the month of February. I can't seem to find a solution that will ensure this always works for every month no matter what the month end date is.
Any suggestions are much appreciated.
DECLARE #dt DATETIME
DECLARE #DayInstance DATETIME
DECLARE #LastDayOfMonth DATETIME
DECLARE #monthEndDate INT = 30
SET #dt = '2012-01-01 00:00:00.000'
WHILE #dt < GETDATE()
BEGIN
SET #DayInstance = #dt
SET #LastDayOfMonth = (SELECT CONVERT(VARCHAR(25),DATEADD(dd,-(DAY(DATEADD(mm,1,#dt))),DATEADD(mm,1,#dt)),101))
IF #monthEndDate > 0
BEGIN
SET #LastDayOfMonth = DATEADD(DAY,(#monthEndDate-DATEPART(dd,#LastDayOfMonth)),#LastDayOfMonth)
SET #DayInstance = DATEADD(MONTH, -1, #LastDayOfMonth)
SET #DayInstance = DATEADD(DAY, 1, #DayInstance)
END
PRINT 'Month Start Date: ' + CAST(#DayInstance AS NVARCHAR(20))
PRINT 'Month End Date: ' + CAST(#LastDayOfMonth AS NVARCHAR(20))
PRINT ''
WHILE #DayInstance <= #LastDayOfMonth
BEGIN
-- Going to do more stuff here
SET #DayInstance = DATEADD(DAY, 1, #DayInstance)
END
SET #dt = DATEADD(MONTH, 1, #dt)
END

Answer 1

I think that this query (it can be simplified still further, but I want to show how I was thinking) sets #StartDate and #EndDate to suitable dates - you can then do your iteration between these two values:
DECLARE #dt DATETIME
DECLARE #DayInstance DATETIME
DECLARE #monthEndDate INT = 30
SET #dt = '2012-01-01 00:00:00.000'
DECLARE #MagicDate1 DATETIME
DECLARE #MagicDate2 DATETIME
SELECT #MagicDate1 = DATEADD(day,#monthEndDate-1,'20010101'),
#MagicDate2 = DATEADD(day,#monthEndDate-1,'20001201')
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SELECT #StartDate = DATEADD(day,1,DATEADD(month,DATEDIFF(month,'20010101',#dt),
#MagicDate2)),
#EndDate = DATEADD(month,DATEDIFF(month,'20010101',#dt),#MagicDate1)
select #StartDate,#EndDate
I first construct two "magic" dates. Those are, respectively the Nth day of January 2001 and the Nth day of December 2000, where N is the desired month end date. Notice that the choice of 2000/2001 was arbitrary, and never need to be changed.1
What we then do in the final expressions is to work out how many months have elapsed between January 2001 and your #dt variable. If we then add that same number of months onto our two "magic" dates then we end up with the Nth day of the same month as #dt and the Nth day of the previous month to #dt (Or the last day of either month if the month has fewer than N days).
Finally, we adjust what we've found as the Nth day of last month by adding 1 to it - which should be the "first" day of the current month.
1 The only important thing is to pick two months which both have 31 days and are consecutive in the calendar. The year is arbitrary.

Answer 2

You've slightly over-complicated things I think the following I believe does what you want it to do:
DECLARE #dt DATETIME
DECLARE #LastDayOfMonth DATETIME
SET #dt = '2012-01-01 00:00:00.000'
WHILE #dt < GETDATE()
BEGIN
SET #LastDayOfMonth = DATEADD(DAY,-1,(DATEADD(MONTH,1,#dt)))
PRINT 'Month Start Date: ' + CAST(#dt AS NVARCHAR(20))
PRINT 'Month End Date: ' + CAST(#LastDayOfMonth AS NVARCHAR(20))
PRINT ''
SET #dt = DATEADD(MONTH, 1, #dt)
END
This will work so long as #dt is the first of the month, otherwise you will have to create a new #FirstDayOfMonth variable and set this above the Setting of `#LastDayOfMonth', but the above should be a good starting point.