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 Values

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