How to use opendir in PHP

PHP opendir() function

Example

<?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);

Parameters	Description
path	Required. Specifies the directory path to be opened.
context	Optional. Specifies the environment for directory handles. context is a set of options that modify the behavior of the directory stream.

Technical details

##Return value: If successful, the directory handle resource is returned. Returns FALSE on failure. If the path is not a legal directory, or the directory cannot be opened due to licensing restrictions or file system errors, an error of level E_WARNING is thrown. You can hide the error output of opendir() by adding '@' before the function name. PHP Version: 4.0 PHP Change Log: PHP 5.0: path parameter support

	ftp:// URL encapsulation protocol.

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