If path is not a valid directory or the
directory can not be opened due to permission restrictions or
filesystem errors, opendir() returns FALSE and
generates a PHP error of level
E_WARNING. You can suppress the error output of
opendir() by prepending
'@' to the
front of the function name.
Example 1. opendir() example
<?php $dir = "/etc/php5/";
// Open a known directory, and proceed to read its contents if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo "filename: $file : filetype: " . filetype($dir . $file) . "\n"; } closedir($dh); } } ?>
The above example will output
something similar to:
filename: . : filetype: dir
filename: .. : filetype: dir
filename: apache : filetype: dir
filename: cgi : filetype: dir
filename: cli : filetype: dir
As of PHP 4.3.0 path can also
be any URL which supports directory listing, however only the file://
URL wrapper supports this in PHP 4.3. As of
PHP 5.0.0, support for the ftp://
URL wrapper is included as well.
opendir php code on this is provided for your study purpose, it will guide you to know how create and design a website using php. use it to practice and train your self online
Php opendir syntax tutorial
php tutorial guide and code design are for easy learning and programming. The code practice section provided at the top is for practising of this syntax. Use the code section up to practice your php programming online. Learning php is very easy, all you need is to use the examples on this site and practice them to perfect your skills.