I downloaded a simple countdown PHP script which takes the server time to display the actual date. Could somebody help me on how to modify
$actualDateDisplay = date($dateFormat,$actualDate);
to make it show the time one hour later?
Or do I have to modify this line?:
<?php echo $actualDateDisplay; ?>
The problem is that I am totally new to PHP
Thanks a lot!
Date problem
Moderator: General Moderators
Re: Date problem
Maybe it is better to paste the whole code of this free PHP script, so you know better what to change:
If you post code here in the future, please enclose it in BBcode tags, as I have done above for you.
Code: Select all
<?php
//*****************************************************************************
//
// MICRO COUNTDOWN - Version: 1.0
//
// You may use this code or any modified version of it on your website.
//
// NO WARRANTY
// This code is provided "as is" without warranty of any kind, either
// expressed or implied, including, but not limited to, the implied warranties
// of merchantability and fitness for a particular purpose. You expressly
// acknowledge and agree that use of this code is at your own risk.
//
//*****************************************************************************
?>
<?php
// Define your target date here
$targetYear = 2008;
$targetMonth = 07;
$targetDay = 20;
$targetHour = 15;
$targetMinute= 00;
$targetSecond= 00;
// End target date definition
// Define date format
$dateFormat = "d-m-Y H:i:s";
$targetDate = mktime($targetHour,$targetMinute,$targetSecond,$targetMonth,$targetDay,$targetYear);
$actualDate = time();
$secondsDiff = $targetDate - $actualDate;
$remainingDay = floor($secondsDiff/60/60/24);
$remainingHour = floor(($secondsDiff-($remainingDay*60*60*24))/60/60);
$remainingMinutes = floor(($secondsDiff-($remainingDay*60*60*24)-($remainingHour*60*60))/60);
$remainingSeconds = floor(($secondsDiff-($remainingDay*60*60*24)-($remainingHour*60*60))-($remainingMinutes*60));
$targetDateDisplay = date($dateFormat,$targetDate);
$actualDateDisplay = date($dateFormat,$actualDate);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>MICRO COUNTDOWN</title>
<link href="style/style.css" rel="stylesheet" type="text/css" />
<script type="text/javascript">
var days = <?php echo $remainingDay; ?>
var hours = <?php echo $remainingHour; ?>
var minutes = <?php echo $remainingMinutes; ?>
var seconds = <?php echo $remainingSeconds; ?>
function setCountDown ()
{
seconds--;
if (seconds < 0){
minutes--;
seconds = 59
}
if (minutes < 0){
hours--;
minutes = 59
}
if (hours < 0){
days--;
hours = 23
}
document.getElementById("remain").innerHTML = days+" days, "+hours+" hours, "+minutes+" minutes, "+seconds+" seconds";
setTimeout ( "setCountDown()", 1000 );
}
</script>
</head>
<body onload="setCountDown();">
<div id="container">
<div id="header"><div id="header_left"></div>
<div id="header_main">Countdown script</div><div id="header_right"></div></div>
<div id="content">
<table class="countTable">
<tr>
<td>Target date:</td><td><?php echo $targetDateDisplay; ?></td></tr>
<tr><th colspan="2" id="remain"><?php echo "$remainingDay days, $remainingHour hours, $remainingMinutes minutes, $remainingSeconds seconds";?></th></tr>
<tr>
<td>Actual date:</td><td><?php echo $actualDateDisplay; ?></td></tr>
</table>
</div>
<div id="footer"><a href="http://www.phpf1.com" target="_blank">Powered by PHP F1</a></div>
</div>
</body>
</html>
- JAB Creations
- DevNet Resident
- Posts: 2341
- Joined: Thu Jan 13, 2005 6:44 pm
- Location: Sarasota Florida
- Contact:
Re: Date problem
I had a date problem once too though after several attempts I discovered they wouldn't run if I showed up wearing pants.
...oh wait this is a coding thread? Maybe it belongs in a coding forum.
...oh wait this is a coding thread? Maybe it belongs in a coding forum.