On Tue, Jan 13, 2009 at 2:14 PM, <ceo@xxxxxxxxx> wrote:
>
> Hard to say without knowing the data structures...
>
> You can't do a simple count of the holidays and add that, because you might
> end up with yet another holiday in the result.
>
> Start at 12/23 and want to add 6 business days.
>
> You find 1 holiday in between, so you add 7 business days and end up on New
> Year's Day, a non-business day.
>
> 12/23
> 12/24
> 12/25 X
> 12/26
> 12/27
> 12/28
> 12/29
> 12/30
> 12/31
> 01/01 X
>
> I think MySQL has built-in holiday / work day calculations, come to think
> of it...
>
Exactly, ceo!
That's why I'm thinking of creating the array and then running a foreach()
loop on the array.
$day = date("w");
$today = date("m-d-Y");
if ($day == 4 || $day == 5) {
$two_day = mktime(0, 0, 0, date("m"), date("d")+4, date("Y"));
$two_day = date("m-d-Y", $two_day);
} else {
$two_day = mktime(0, 0, 0, date("m"), date("d")+2, date("Y"));
$two_day = date("m-d-Y", $two_day);
}
foreach ($holiday as $h) {
if ($h >= $today && $h <= $two_day) {
$two_day = mktime(0, 0, 0, date("m", $two_day), date("d", $two_day)+1,
date("Y", $two_day));
$two_day = date("m-d-Y", $two_day);
}
}
That should add a day for each instance of a holiday that falls between the
dates, right?
