如何使用php列出目录中的文件

Question

如何列出目录中的所有文件。我只想要根目录中的文件。如果根目录中有任何目录，我想跳过这些目录和文件。
现在我正在使用此代码

 $folderPath  = file_directory_path().'/lexalytics/';
    if ($handle = opendir($folderPath)) {
        $result .=  '<div><ul>';
        while (false !== ($entry = readdir($handle))) {
            if ($entry != "." && $entry != "..") {
                $result .=  "<li><a href=../".$folderPath.$entry.">".$entry."</a>\n</li>";
            }
        }
        $result .= '</ul></div>';
        closedir($handle);
    }

但是它列出了子目录和文件。 怎么能避免那些？请帮助我

Answer 1

请使用PHP5s新DirectoryIterator课程：

这只会列出文件并排除文件夹：

$directory  = file_directory_path().'/lexalytics/';
$filenames = array();
$iterator = new DirectoryIterator($directory);
foreach ($iterator as $fileinfo) {
    if ($fileinfo->isFile()) {
        $filenames[$fileinfo->getMTime()] = $fileinfo->getFilename();
    }
}

Answer 2

$path = "your-path";

    // Open the folder
    $dir_handle = @opendir($path) or die("Unable to open $path");

    // Loop through the files
    while ($file = readdir($dir_handle)) {

    if($file == "." || $file == ".." || $file == "index.php" )

        continue;
        echo "<a href=\"$file\">$file</a><br />";

    }
    // Close
    closedir($dir_handle); 

Answer 3

试试此代码

$folderPath  = file_directory_path().'/lexalytics/';
    $handle = @opendir($folderPath) or die("Unable to open $path");
    $result .=  '<div><ul>';
    // Loop through the files
    while ($entry = @readdir($handle)) {
        if(is_file($folderPath.$entry)) {
           $result .= "<li><a href=../".$folderPath.$entry.">".$entry."</a>\n</li>";
        }
    }
    $result .= '</ul></div>';
    closedir($handle);