countdown

[clowngirl]

can anyone tell me how to get this to show a message if countdown time has been reached. Seems no matter what I try it isnt going my way..
I thought by making a if statment so that if day,hour and minute are at zero it would work, but NOT........
So please any help would be great..
Or if you have a better way to count down to a date using the server time...

// parameters: (year, month, day, hour, minute)
countdown(2005,11,14,23,59);

function countdown($year, $month, $day, $hour, $minute)
{
  // make a unix timestamp for the given date
  $the_countdown_date = mktime($hour, $minute, 0, $month, $day, $year, -1);

  // get current unix timestamp
  $today = time();

  $difference = $the_countdown_date - $today;
  if ($difference < 0) $difference = 0;

  $days_left = floor($difference/60/60/24);
  $hours_left = floor(($difference - $days_left*60*60*24)/60/60);
  $minutes_left = floor(($difference - $days_left*60*60*24 - $hours_left*60*60)/60);

  // OUTPUT
  echo "Today's date ".date("F j, Y, g:i a")."<br/>";
  echo "Countdown date ".date("F j, Y, g:i a",$the_countdown_date)."<br/>";
  echo "Countdown ".$days_left." days ".$hours_left." hours ".$minutes_left." minutes left";
}

[rikmoncur]

Here's one I have:

<?php 
/* function */
    $countdown = countdown("May 22 2006"); /* the date is assigned to $dateto in the function below */
    if ($countdown != "") 
    {echo "There are just ".$countdown." days remaining to get your ticket!";} 
/* This PHP function is designed to return the number of days until a particular date. 
The only parameter passed to this function is the date of the event. It will return a string containing the number of days until then. If the date is today then this figure will be zero. If it's passed then the result will be blank.  */
    function countdown ($dateto){ 
    $tstampfrom = time(); /* gets current timestamp */
    $tstampto = strtotime($dateto); /* gets future timestamp */
    $datediff = $tstampto - $tstampfrom; /* difference in timestamps */
    $daysdiff = $datediff / 86400; /* divided by no. of seconds in a day */
    if (round($daysdiff,0) > $daysdiff) 
    {$numdays = round($daysdiff,0);} /* rounds down */
    else 
    {$numdays = round($daysdiff,0)+1;} /* rounds up */
    if ($daysdiff < 0) 
    {$numdays = "";} 
    return $numdays; /* ends function and returns argument as value of the function call */
	} 
    ?>