Re: Return Month/Day from Day Number

Yes it was.

That won't work.  2000 was not a leap year.

Then the best answer is the one you already noted: DateAdd the serial
day number to a base date of 1999-12-31, and discard the year part of
the result. There's no point, and no benefit, in making it any more
complicated.

Because the purpose of function is to cover EVERY SINGLE DAY possible,
whether or not that day exists for the current year. Note that it
returns Month/Day and not year as it is year neutral. There is really
no point in getting too deep into details why. Just trust that it is
important. :-)

I don't understand the 'lookup' table. You mean to fill-up a complete
table (like an array) that contains every single day, perhaps in
strings? January 1, January 2, January 3...

The the DayNum would index this array (lookup table)?

Would you do it this way rather than adding the Day Num to December
31, 1999?

Thanks.

Webbiz

...

...

OK, I'll bite...for what earthly reason would you not consider the
actual number of days/year?

But, if you're adamant of it always being 366, then sure the lookup
table based on accumulated days is as the logical implementation.  (Of
course, it's also a reasonable implementation w/ the addition of a
LeapYear logical input, too...)

--

Hello Mike-

Since the purpose of my function is to return a date based on a Day
Number as if the year is a leap year (1 to 366), one can't just use
any date reference for that to happen, right?

So the routine MUST use the date of the last day just prior to a leap
year so that by adding 1 to 366, you get January 1 to December 31st,
right?

On another note, wouldn't putting DateAdd inside Format just be adding
more overhead? Since I'm using an actual Date type and pre-loading it
with #December 31, 1999#, using simple addition of 1 to 366 can easily
be done without any conversion (such as Datevalue, etc.) and really no
need for DateAdd.

I completely agree with the formatting issue in as much as limiting
the re-usuability of the function. However, this particular routine
resides in a class so it's contained and single-minded.  It does one
thing. It takes a number from 1 to 366 and returns the verbage of the
month and day, such as "March 14", etc. It's for display purposes as
one moves the mouse across a picturebox.

Thanks for your comments.

Webbiz

I know I would definately use DateAdd:

Print DateAdd("d",DayNum,#12/31/1999#)

Always better to use date functions than to rely on mathematically adding
and subtracting. If you want the date formatted a certain way, then nest
that inside a Format function. Of course, you must have a date to base this
on or devise a default date to use.  For example, a logical choice for a
default would be 12/31 of last year...and you could write that as such:

Print DateAdd("d", DayNum, DateSerial(Year(Date) - 1, 12, 31))

(assuming the PC's clock is correct...at least the year)

I don't know if I'd write this into a wrapper function (your
GetMMDDFromDayNum). I suppose it depends on how many times I'd need to call
it. If only once or twice, I doubt that I would. If a dozen times, probably.
If I WERE going to write a wrapper function, I think I'd make the date an
optional parameter (mind you, I'm pretty much just winging this). Something
like this:

Public Function GetDateFromDayNum(ByVal DayNum As Long, Optional RefDate As
Variant) As Date

   If Not IsDate(RefDate) Then
       GetDateFromDayNum = DateAdd("d", DayNum, DateSerial(Year(Date) - 1,
12, 31))
   Else
       GetDateFromDayNum = DateAdd("d", DayNum, RefDate)
   End If

End Function

(a Variant, in this case, is beneficial)

I would NOT have the function return a specifically formatted string as in
your example because that would limit the function's usefulness (it wouldn't
be generic anymore). Better to have it return a Date data type and let the
calling procedure format it as appropriate.

I'm sure others will have ideas that may be better, but I think this
function would give you a bit of flexibility so that you could use it in
multiple applications.

--
Mike

Could someone help me with this?

How might one convert a day number from 1 to 366 into a Month and Day
string?

Most years are 365, I know. But the assumption will be that each year
contains 366 (is a leap) by this function.

Day Num as Return Values:

1 = January 1
60 = February 29
366 = December 31

The above must always be true.

Now, I was thinking about doing something like this. Maybe this is the
best way to go, maybe not. If you have a better idea, please let me
know.

What I thought I'd do in my GetMMDDFromDayNum function is this:

------------------------
Function GetMMDDFromDayNum(ByVal DayNum as long) as String
Dim dLeapYearRef as Date

   'Ref date is leap year minus 1 day   
   dLeapYearRef = #12/31/1999#

   GetMMDDFromDayNum = Format(dLeapYearRef + DayNum, "MMMM dd")

End Function

Any thoughts?

Thanks.

Webbiz