Listing and previewing images in a directory using PHP - php

Question

I'm trying to improve the administrator panel of my website. I need to preview the images in the thumbnails folder so that when i'm using thumbnails for news I dont have to upload the image for the second time. I found a great script, but I get failed to read the directory error. Here is the script:
<?php
// filetypes to display
$imagetypes = array("image/jpeg", "image/gif", "image/png");
// Original PHP code by Chirp Internet: www.chirp.com.au
// Please acknowledge use of this code by including this header.
function getImages($dir)
{
global $imagetypes;
// array to hold return value
$retval = array();
// add trailing slash if missing
if(substr($dir, -1) != "/") $dir .= "/";
// full server path to directory
$fulldir = "{$_SERVER['DOCUMENT_ROOT']}/$dir";
$d = #dir($fulldir) or die("getImages: Failed opening directory $dir for reading");
while(false !== ($entry = $d->read())) {
// skip hidden files
if($entry[0] == ".") continue;
// check for image files
$f = escapeshellarg("$fulldir$entry");
$mimetype = trim(`file -bi $f`);
foreach($imagetypes as $valid_type) {
if(preg_match("#^{$valid_type}#", $mimetype)) {
$retval[] = array(
'file' => "/$dir$entry",
'size' => getimagesize("$fulldir$entry")
);
break;
}
}
}
$d->close();
return $retval;
}
// fetch image details
$images = getImages("../images/thumbnails");
// display on page
foreach($images as $img) {
echo "<div class=\"photo\">";
echo "<img src=\"{$img['file']}\" {$img['size'][3]} alt=\"\"><br>\n";
// display image file name as link
echo "",basename($img['file']),"<br>\n";
// display image dimenstions
echo "({$img['size'][0]} x {$img['size'][1]} pixels)<br>\n";
// display mime_type
echo $img['size']['mime'];
echo "</div>\n";
}
?>
I really appreciate if someone could help..
EDIT:
<div style=" height: 200px; width: 600px; overflow: auto;">
<?PHP
foreach(glob("../thumbnail/".'*') as $filename){
echo "<div style=\"display:inline-table; font-size:10px; font-family:'Tahoma'; margin:5px;\">";
echo "<img width=\"100px\" height=\"100px\" src=\"../thumbnail/$filename\"/>";
echo "<br>".basename($filename) . "<br>";
echo "</div>";
}
?>
</div>
This method works perfect. No need to use complicated scripts.
Anyway, can somebody please tell me how to check images less than 100px x 100px displayed?

Answer 1

<?PHP
foreach(glob("../thumbnail/".'*') as $filename){
list($width, $height, $type, $attr) = getimagesize("../thumbnail/".$filename);
if($width>=100 || $height >=100) continue;
$rest = substr($filename, 3);
?>
That should do it..