[AccessD] Age calculation function

Good point!  I did not think that you were picking on me.

While thinking about this, you technically have to add 0.0001 (may could go
smaller) day.  

Think about it

12/1/99-12/1/99 is technically saying the same as 
12/1/99 12:00am 0 12/1/99 11:59:59pm, which IMHO is not 1 full day until
that last second has past (and then it is the next day).

When you are using my method, I think that maybe you have to add 0.0001 days
to get the desired results.

I modified the function to take a fraction of a day and entered .0001 as the
fraction to add and I get:

Debug.Print (Greg2JD("03/01/1992",0)-Greg2JD("03/01/1987",0.0001))/365.25=
5.00205311425984 

What do you all think?

Bobby

-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Gustav Brock
Sent: Friday, August 01, 2003 3:29 PM
To: Access Developers discussion and problem solving
Subject: Re: [AccessD] Age calculation function


Hi Bobby

> This function can be made even more generic by passing in the month, day,
> year, and day fraction (or the day as a fraction 12/4/2003 6:00AM would be
> 12, 4.25, 2003).  This way, the function can handle years BC.

> Just in case anyone cared.  :-)

Well, could be interesting ... who knows, but it still fails an
example like this:

? (Greg2jd("03/01/1992") - Greg2jd("03/01/1987")) / 365.25
 4,99931553730322

like any other method relying on dividing by 365,25. This method is
what we call a "shoemaker method" as it is a method working in "most
cases" but not all.

This is not to pick on you - because it is still published many places
- but for the records, so no list member should be tempted to use a
quick and dirty suggestion for serious use.

/gustav


> Public Function Greg2JD(ByVal strDate As String) As Double
> Dim A    As Long
> Dim B    As Long
> Dim MM   As Long
> Dim YY   As Long
> Dim DD   As Single

>    MM = Month(strDate)
>    YY = Year(strDate)
>    DD = Day(strDate)
   
>    'note, you could pass the time of day in as a fraction of a day.
>    'if you do so, simply add the fraction of the day to DD
>    'As is, this code assumes 0 hour, i.e. if it is the 4th,
>    'then it assumed to be the 4th at 12:00:00 AM
>    'DD = DD + sDayFrac     'sDayFrac is the variable holding fraction of
day

>    If MM < 3 Then
>       YY = YY - 1
>       MM = MM + 12
>       End If

>    A = YY \ 100         'integer math
>    B = 2 - A + A \ 4    'integer math

>    Greg2JD = Int(365.25 * (YY + 4716)) + Int(30.6001 * (MM + 1)) + DD + B
-
> 1524.5
   
> End Function

> To work out our example, I called it with:
> Debug.Print (Greg2JD("02/18/2002") - Greg2JD("02/29/1988")) / 365.25

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