escape single qoute in ffmpeg filename varible php - php

Question

code works fine but if the filename has a single qoute just as "Britney's video.mp4" it does not work.
$ffmpeg = "/usb/bin/local/ffmpeg";
$videos = "/videos/*.mp4";
$ouput_path = "/videos/thumbnails/";
foreach(glob($videos) as $video_file){
$lfilename = basename($video_file);
$filename = basename($video_file, ".mp4");
$thumbnail = $ouput_path.$filename.'.jpg';
if (!file_exists($filename)) {
#$thumbnail = str_replace("'", "%27", $thumbnail);
exec("/usr/local/bin/ffmpeg -i '$video_file' -an -y -f mjpeg -ss 00:00:30 -vframes 1 '$thumbnail'");
}
echo "<a href='$lfilename'>$filename<img src='thumbnails/$filename.jpg' width='350'>";

Answer 1

i got it working but not using something overly complicated.
thanks all
$comd = "/usr/local/bin/ffmpeg -i \"$video_file\" -y -f mjpeg -ss 00:00:30 -vframes 1 \"$thumbnail\" 2>&1"; shell_exec($comd);
shell_exec($comd);

Answer 2

As suggested in the comments, you can just wrap all your shell-command String within the escapeshellarg() as shown below.
<?php
foreach(glob($videos) as $video_file){
$lfilename = basename($video_file);
$filename = basename($video_file, ".mp4");
$thumbnail = $ouput_path.$filename.'.jpg';
if (!file_exists($filename)) {
$cmd = "/usr/local/bin/ffmpeg -i ";
$cmd .= $video_file . " -an -y -f mjpeg -ss 00:00:30 ";
$cmd .= "-vframes 1 " . $thumbnail;
exec( escapeshellarg($cmd) );
}