Date dropdown function

[Dillenger]

Ok so I have this code that creates a date for every monday of the year, but what if I wanted to have every monday for 2 or 3 years??? how would I do that?

<? 

/////////////////////////// 

		$startMonday = mktime(12, 0, 0, 1, 5, 2004); 
		$sequence = 0; 
		for ($i = -53; $i <= 53; $i++) { 

			$date2[$i] = date('d-m-Y', $startMonday + ($i * 7 * 24 * 60 * 60)); 




	 		$year1 = "2005";

	 		if (substr($date2[$i], -4) == $year1) { 

				$monday[++$sequence] = $date2[$i]; 
			}





		} 

		echo "<select name='issue_date' size='1' class='boxes'>\n";

		foreach ($monday as $key => $value) 
		{

			list($day, $month, $year) = split('[/.-]', $value);

			$date_display = date('F d, Y', mktime(12, 0, 0, $month, $day, $year));

			$value2 = date('Y-m-d', mktime(12, 0, 0, $month, $day, $year));

			if($value2 == $date)
			{
				echo "<option value='$value2' selected>$date_display</option>\n";
			}

			else
			{			
				echo "<option value='$value2'>$date_display</option>\n";
			}

		}

		echo "</select>\n";

?>

[devinemke]

$years_ahead = 3; // change this to however many years in the future you want to go

$today = strtotime('00:00 today');
$ahead = strtotime('00:00 ' . $years_ahead . ' years');

echo 'here are all of the Mondays for the next ' . $years_ahead . ' years<br><br>';
while ($today <= $ahead)
{
	if (date('l', $today) == 'Monday') {echo date('l Y-m-d', $today) . '<br>';}
	$today += 86400;
}

[Dillenger]

Thanks man,

That works good, but how would i incorporate that into my dropdown?

[devinemke]

Originally posted by Dillenger
That works good, but how would i incorporate that into my dropdown?

use the appropriate HTML tags