Re: Date help needed

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

 



You could loop through the weeks and put those 3 specifically in:
$days = array();
for($i = 0; $i < 365; $i +=7) {
  $days[] = strtotime('next Monday', strtotime('+ '.$i.' days'));
  $days[] = strtotime('next Friday', strtotime('+ '.$i.' days'));
  $days[] = strtotime('next Sunday', strtotime('+ '.$i.' days'));
}

sort($days);

foreach($days as $day) {
  echo date('Y-m-d', $day).'<br/>';
}

(This is not tested, but it *should* work,)

On Thu, 24 Jun 2004 17:07:12 -0400, Chris Payne
<chris_payne@xxxxxxxxxxxxxxxxx> wrote:
> 
> Hi there everyone,
> 
> I have a problem, I currently have some code which populates a dropdown box
> - this code gives me every day for the next x amount of days (EG: a years
> worth of days), however what I really need to be able to do, is to find a
> way to display this data in the dropdown box but ONLY show 3 days a week,
> IE: Mondays, Fridays and Sundays, so it would show the dates for each
> Monday, Friday and Sunday for X amount of days (IE: 365 days in the
> dropdown).
> 
> Does anyone have any idea how to do this?  I would really appreciate any
> help, I'd send my sample code only I'm not at my home/work computer ATM.
> 
> Chris
> 
> !DSPAM:40db40cb34094233914063!
> 


-- 
paperCrane --Justin Patrin--

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php


[Index of Archives]     [PHP Home]     [PHP Users]     [Postgresql Discussion]     [Kernel Newbies]     [Postgresql]     [Yosemite News]

  Powered by Linux