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
| 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. | |
|---|---|
| 4.0 | |
| PHP 5.0: path parameter support | ftp:// URL encapsulation protocol. |
The above is the detailed content of How to use opendir in PHP. For more information, please follow other related articles on the PHP Chinese website!
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
