List a Directory's Contents Using PHP
To list the content of a directory you just need to use the combination of opendir() and readdir() functions.
// directory path can be either absolute or relative
$dirPath = '.';
// open the specified directory and check if it's opened successfully
if ($handle = opendir($dirPath)) {
// keep reading the directory entries 'til the end
while (false !== ($file = readdir($handle))) {
// just skip the reference to current and parent directory
if ($file != "." && $file != "..") {
if (is_dir("$dirPath/$file")) {
// found a directory, do something with it?
echo "[$file]<br>";
} else {
// found an ordinary file
echo "$file<br>";
}
}
}
// ALWAYS remember to close what you opened
closedir($handle);
}
When you list the files surely you need to know which of these files are actually a sub-directory. Using the is_dir()function you can test if the files is really a directory or not. What you want to do later with that directory is up to you. On the code above directories are simply printed in a square brackets to differentiate them with ordinary files
MySql Hosting India
IndoUs Hosting is the leading MySql Hosting companies in India. The company provides dedicated as well as shared hosting services of MySql hosting. If you are looking for guaranteed MySql Hosting India, we can be the best option for you.