PHP opendir() Function
Example
Open a directory, read its contents, then close:
<?php
$dir = "/images/";
// Open a directory, and read its contents
if
(is_dir($dir)){
if ($dh = opendir($dir)){
while (($file = readdir($dh)) !== false){
echo "filename:" . $file . "<br>";
}
closedir($dh);
}
}
?>
Result:
filename: cat.gif
filename: dog.gif
filename: horse.gif
Definition and Usage
The opendir() function opens a directory handle.
Syntax
opendir(path,context);
| Parameter | Description |
|---|---|
| path | Required. Specifies the directory path to be opened |
| context | Optional. Specifies the context of the directory handle. Context is a set of options that can modify the behavior of a stream |
Technical Details
| Return Value: | Returns the directory handle resource on success. FALSE on failure. Throws an error of level E_WARNING if path is not a valid directory, or if the directory cannot be opened due to permission restrictions or filesysytem errors. You can hide the error output of opendir() by adding '@' to the front of the function name |
|---|---|
| PHP Version: | 4.0+ |
| PHP Changelog: | PHP 5.0: The path parameter now supports the ftp:// URL wrapper |
PHP Directory Reference
