Generate thumbnail for a bunch of mp4 video in a folder

Question

I have read this but didnt help too much.

I have a folder called videos and another folder called thumbnails. I have many mp4 videos in video folder and want to catch thumbnails at 4th second to the thumbnails folder using ffmpeg and php.

I am using Wamp server 2.2 on windows whit php 5.3.8 and Apache 2.2.21
I downloaded ffmpeg from FFmpeg Windows Builds section of ffmpeg download page and the static 32 build from this link.

I extracted the 7z file to my website root
here is my php code:

$ffmpeg = "includes/ffmpeg/bin/ffmpeg";
foreach(glob('files/videos/*.mp4') as $pathname){
  $filename = substr($pathname,13,strripos($pathname,'.mp4')-13);
  $thumbnail = 'files/thumbnails/'.$filename.'.jpg';
  exec("ffmpeg -i $pathname -an -y -f mjpeg -ss 00:00:04 -vframes 1 $thumbnail");
}

but nothing happens and the thumbnails folder is always empty!
- How can I find out is ffmpeg installed on my server or not?
- How can I get my script to work?
Please help

@DoSparKot what permissions? I am using wamp server and I have windows folders. I have everyone full access to videos and thumbnails. but nothing changed — , Feb 02 '13 at 13:57

score 4 · Accepted Answer · answered Jun 04 '14 at 11:05

try this:

$ffmpeg = "c:/wamp/www/includes/ffmpeg/bin/ffmpeg";
$videos = "c:/wamp/www/files/videos/*.mp4";
$ouput_path = "c:/wamp/www/files/thumbnails/";
foreach(glob($videos) as $v_file){
    $fname = basename($v_file, ".mp4");
    $thmb = $ouput_path.$fname.'_tn.jpg';
    $cmd = "$ffmpeg -i $v_file -an -y -f mjpeg -ss 00:00:04 -vframes 1 $thmb";
    $stat = system ($cmd);
}

जलजनक · Answer 2 · 2013-02-02T14:52:34.607

1

Try with absolute paths in commands instead of depending on PATH ENV variable: Both exec() and system() works. Resolve the path definitions.

/* Using Absolute paths */
$ffmpeg = "c:/wamp/www/includes/ffmpeg/bin/ffmpeg";
$videos = "c:/wamp/www/files/videos/*.mp4";
$ouput_path = "c:/wamp/www/files/thumbnails/";

foreach(glob($videos) as $video_file){

    $filename = basename($video_file, ".mp4");
    $thumbnail = $ouput_path.$filename.'_tn.jpg';
    $command = "$ffmpeg -i $video_file -an -y -f mjpeg -ss 00:00:04 -vframes 1 $thumbnail";

    $status = system ($command);
/*or
    $status = exec($command);

    if ($status === false) {
        var_dump("ERROR: Conversion Failed!!!!");
    } else
        var_dump($status);
*/
}

edited Feb 02 '13 at 14:52

answered Feb 02 '13 at 14:05

जलजनक

3,072
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Interesting, can you put single file in `C:\temp\` and try with modified paths? – जलजनक Feb 02 '13 at 14:49
It depends on the internal constant DIRECTORY_SEPARATOR it's / in code and outputs `\\` for windows platforms. – जलजनक Feb 02 '13 at 14:53
@PedarJepeto anyways, thanks to you, now I know how to generate thumbnails of videos on the fly. :-) – जलजनक Feb 02 '13 at 14:54
does it really work for you? I exactly copy and paste the code but can not get it to work!! :( – Feb 03 '13 at 03:47

user2217108 · Answer 3 · 2018-11-27T22:12:06.493

0

I´m not allowed to add a comment, so here (as a Post) is what I found out while dealing with this: The white space in the file name prevents ffmpeg from finding the right file. So one way is to change the file name.

edited Nov 27 '18 at 22:12

answered Nov 27 '18 at 21:58

user2217108

1
3

The `ffmpeg` cli tool handles spaces just fine with proper quoting or escaping: `ffmpeg -i input "output 1.mp4" 'output 2.mp4' output\ 3.mp4` – llogan Nov 27 '18 at 22:10
Thanks, I´ll try that – user2217108 Nov 27 '18 at 22:13
I'm not a PHP user (the question is tagged [tag:php]), so my three examples may or may not work for PHP. – llogan Nov 27 '18 at 22:16
Oh, it worked just fine, I used the "output 1.mp4" escaping the quotes, in the string, like this: - \"output 1.mp4\" - and voila!!! :) thanks for the tip, I believe this is the reason that it wasn´t working for Pedar Jepeto. – user2217108 Nov 30 '18 at 06:15

Generate thumbnail for a bunch of mp4 video in a folder

3 Answers3