[RESOLVED] php warning and invalid argument supplied for forech()

Thikho

Hi i have two question to ask here the warning message on page

Warning: arsort() expects parameter 1 to be array, string given in /var/www/admin/image_list.php on line 42, i have google through the net and i have found a way of removing the warning by placing the @arsort before the function , i was also getting filetime warning and i need to find out the reason why i shud put the @ sign and what cause these error messages
Invalid argument supplied for foreach , this code works perfectly on the local host but as soon as a i upload it on my web server it give me problems , pls help

Warning: Invalid argument supplied for foreach() in /var/www/admin/image_list.php on line 45

here is the code its for viewing the images i have uploaded

<?php

$whatField = $_REQUEST['whatField'];

?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Administrator Control Panel</title>
<meta http-equiv="Content-Type"  content="text/html; charset=iso-8859-1">
<link rel="stylesheet" href="styles/admin.css" type="text/css">

</head>
<body>
<?php

	$banners = '';
	$bannerDir = '../photos/';
	$i = 0;
	array($banners);
	$d = dir($bannerDir);
	$fileArray = '';
	array($fileArray);
	$fileTimeArray = '';
	array($fileTimeArray);
	while (false !== ($entry = $d->read()))
		{
		if (strlen($entry) > 2)
			{
			list($imageName,$imageExtension) = split('[.]',$entry);
			if (eregi('jpg|jpeg',$imageExtension))
				{
				$file = $bannerDir.$entry;
				$fileArray[$entry] = date('U', filemtime($file));
				$fileTimeArray[$entry] = date('j F Y H:i', filemtime($file));
				}
			$banners[$i] = $entry;
			$i++;
			}
		}
	$d->close();
	arsort($fileArray);
	echo '<table cellpadding="3" cellspacing="1" border="0" style="width:100%; margin-top: 10px" style="position:relative; ">'."\n";
	echo '<tr class="tableHeading"><td>Image name</td><td>View image</td><td>Date uploaded</td><td>Use image</td></tr>'."\n";
	foreach ($fileArray as $key => $value) {
		$class      = ($class == "adminRow1") ? "adminRow2" : "adminRow1";
echo '<tr class="'.$class.'"><td>'.$key.'</td><td><a target="_blank" href="../photos/'.$key.'">View image</a></td><td>'.$fileTimeArray[$key].'</td><td><a onclick="opener.document.getElementById(\''.$whatField.'\').value=\'../photos/'.$key.'\'; window.close()" href="javascript:void(0)">Use image</a></td></tr>'."\n";
		}
		echo '</table>'."\n";


?>
</body>
</html>

NogDog

In a number of places you do something like this in an attempt to initialize array variables:

$variable = '';
array($variable);

Instead, do this:

$variable = array();

Thikho

tnks i have done dat, do you think the way im initializing my arrays is causing this kind of warnings?

and after i have initialize in this way it does not show the warning errors but it not showing my pictures either , it is only showing my headings!!!!! please help, I cant see the pictures that i have uploaded

                $banners = array();
		$bannerDir = '../photos/';
		$i = 0;
		$d = dir($bannerDir);
		$fileArray = array();
		$fileTimeArray = array();