如何在Windows XP上的PHP中给定文件夹中的每个文件的扩展名？

E.g:

folder name:

myFonlder

files in myFolder

myFolder.01.mkv
myFolder.02.mkv
myFolder.03.avi
myFolder.04.mts 
...

// each file's extension may be different.

So,how can I extract the extension of each file?

Thank you very much!!

[update]

my own solution; want to know is it fast enough!?

foreach (glob("d:\\myFolder\\*.*") as $filename) {
    //echo "$filename size " . filesize($filename) . "
";

    $path_parts = pathinfo($filename);

    echo $path_parts['dirname'], "
";
    echo $path_parts['basename'], "
";
    echo $path_parts['extension'], "
";
    echo $path_parts['filename'], "
"; // since PHP 5.2.0
}

<?php

foreach (new DirectoryIterator('../moodle') as $fileInfo) {
    if($fileInfo->isDot()) continue;
    echo $fileInfo->getFilename() . "<br>
";
}

?>

And use http://www.php.net/manual/en/function.pathinfo.php on filename

This is the quick and easy solution on windows.

exec("dir d:\directory_name /b" ,$output); // in Linux dir will change to ls
foreach($output as $file_name){
$file_parts = explode(".",$file_name);
echo "File Name : ".$file_parts[0]."
";
echo "File Extension : ".$file_parts[1]."

";
} 

Enjoy..!!