calculate date in future?

TheGringo

I'm using this function, copied from php.net, to calculate a date in the future:

function DateAddBusinessDays($i){
$d = mktime(0,0,0,date('n'),date('j')+1,date('Y')) ;
 do {
   $d = mktime(0,0,0,date('n',$d),date('j',$d)+1,date('Y',$d)) ;
   $dow = date('D',$d);
   switch ($dow) {
       case 'Sat':
       case 'Sun':
         break;
       default:
         $i--;
       }
 } while ($i>0);
 return $d;
}

I'm calling it like this: $due = DateAddBusinessDays($turnaround1);
where $turnaround is a number (of days) submitted by the user via a form field.
What is returned looks somethiing like this: 1074488400, instead of a date-formatted string. What am I doing wrong?

paulnaj

The number returned from this function is a UNIX timestamp ... the no. of seconds since some date in the past (doesn't matter).

Use date() function to format the timestamp.

ovisopa

I think this way you should use the date function

echo date("d. m. Y", $d"]);

hope this helps

see ya

TheGringo

thank you... what is the proper syntax? from the manual it looks like:

date("F j, Y, g:i a," [ , $duedate]);

but that isn't right...

paulnaj

Hi Gringo,

In the manual any parameters to functions that are enclosed in square brackets are optional ... you don't have to include.

If you do want to include them, you don't include the brackets ...

date("F j, Y, g:i a," , $duedate);

TheGringo

gotcha... thanks. Here's another question: if I want to include the time (hour, minutes, am or pm), would I add this to the function:

$d = mktime(0,0,0,date('n',$d),date('j',$d)+1,date('Y',$d), date('g:i a',$d));

instead of

$d = mktime(0,0,0,date('n',$d),date('j',$d)+1,date('Y',$d));

(obviously, I know it's wrong, since it doesn't return the correct time )

paulnaj

Your try was:

$d = mktime(0, 0, 0, date('n',$d), date('j',$d)+1, date('Y', $d), date('g:i a',$d));

It should be:

$d = mktime(
  date('g',$d) // This is the hours
, date('i',$d) // This is the minutes
, 0 // You didn't wan't to consider the seconds
, date('n',$d)
, date('j',$d) + 1
, date('Y', $d)
);

==================
By the way, did you made a change to that the DateAddBusinessDays() function?

The first line has the '+1' in the 'days' parameter where the $d variable is set up. I would have thought that $d should start off as today ... is that right? At the moment, with the '+1', $d starts off as tomorrow. So, if you want to 'add one business to 'today' (and get 'tomorrow') then, at the moment, you would have to call ...

$tomorrow = DateAddBusinessDays(0);

... instead of ...

$tomorrow = DateAddBusinessDays(1);

.. which is what I would have expected.