[how-to] Display Directory Files Using PHP

 

1. Create PHP Script

Basic Script

<?php function outputFiles($path, $arryFilter) { // Check directory exists or not if (file_exists($path) && is_dir($path)) { // Scan the files in this directory $result = scandir($path); // Filter out the current (.) and parent (..) directories $files = array_diff($result, array( '.', '..' )); if (count($files) > 0) { // Loop through retuned array foreach ($files as $file) { if (is_file("$path/$file")) { if (isset($arryFilter)) { //echo 'got arryFilter<br/>'; $ret = array_keys(array_filter($arryFilter, function ($var) use ($file) { return strpos($file, $var) !== false; })); //echo count($ret).'found <br/>'; if (count($ret) > 0) { //echo 'with filter'; // Display filename echo '<a href="' . $path . "/" . $file . '" target="new">' . $file . '</a><br>'; } } else { //echo 'no filter'; // Display filename echo '<a href="' . $path . "/" . $file . '" target="new">' . $file . '</a><br>'; } } else if (is_dir("$path/$file")) { // Recursively call the function if directories found outputFiles("$path/$file"); } } } else { echo "ERROR: No files found in the directory."; } } else { echo "ERROR: The directory does not exist."; } } if (isset($_GET['folder'])) { $folder = $_GET['folder']; //echo $folder; $filter = NULL; $arryFilter = NULL; if (isset($_GET['filter'])) { if ($_GET['filter'] != "") { $filter = $_GET['filter']; $arryFilter = explode(',', $filter); } } //echo $filter; //var_dump($arryFilter); outputFiles($folder, $arryFilter); } ?>

2. Put the above script at a level higher than the target directory.