12

Is there a way to grab a video thumbnail in FFmpeg?

I'd like to grab the middle-most frame as the video and use that as the thumbnail. Video duration is unknown.

The ability to specify the dimensions of the thumbnail would also be helpful.

0
11

One line command :

ffmpeg -i input.mp4 -vcodec mjpeg -vframes 1 -an -f rawvideo -ss `ffmpeg -i input.mp4 2>&1 | grep Duration | awk '{print $2}' | tr -d , | awk -F ':' '{print ($3+$2*60+$1*3600)/2}'` output.jpg

The subcommand get the total duration of the input video

ffmpeg -i input.mp4 2>&1 | grep Duration | awk '{print $2}' | tr -d ,

And pipe it to awk to compute duration/2 like this

echo '00:01:30.000' | awk -F ':' '{print ($3+$2*60+$1*3600)/2}'

So it give this among of seconds to -ss option and it's done.

If you would like to specify the size, add -s option like that WxH where W and H are integer values (ex: -s 800x600 or -s svga) or read the ffmpeg man page to see other available options or format.

2
  • 1
    -vcodec mjpeg -an -f rawvideo is not needed when your output is just a .jpg file. I'd actually recommend outputting to PNG instead and doing the compression later.
    – slhck
    Oct 12 '17 at 16:13
  • 2
    The ffmpeg output is only for informative purposes only and not intended for parsing, so therefore processing it may be considered fragile. Use ffprobe to get the duration instead. See FFmpeg Wiki: FFprobe Tips.
    – llogan
    Oct 12 '17 at 18:52
5

First of all, always use the latest version of FFmpeg.

If you have access to PHP, your question is perfectly answered on Stack Overflow: Use FFMpeg to get middle frame of a video?

$output = shell_exec("/usr/local/bin/ffmpeg -i {$path}");
preg_match('/Duration: ([0-9]{2}):([0-9]{2}):([^ ,])+/', $output, $matches);
$time = str_replace("Duration: ", "", $matches[0]);
$time_breakdown = explode(":", $time);
$total_seconds = round(($time_breakdown[0]*60*60) + ($time_breakdown[1]*60) + $time_breakdown[2]);
shell_exec("/usr/local/bin/ffmpeg -y  -i {$path} -f mjpeg -vframes 1 -ss " . ($total_seconds / 2) . " -s {$w}x{$h} {$output_filename}"; 

What it'll do is just extract the duration from FFmpeg's output and use that to determine the timecode of the middle frame.

You can easily adapt that to other shells, and simply insert into the following command, where the middle frame is roughly at 534 seconds:

ffmpeg -y  -i input.mp4 -f mjpeg -vframes 1 -ss 534 thumbnail.jpg

You can always change the size with -s 480x320 or similar, depending on how big you want it, inserted somewhere after the -i input.mp4 option.


The above solution is a little inaccurate. It'll immediately give you a result but you can't specify which frame you want. If you already know the exact frame you want to extract, use the select filter, e.g. with the frame number 12345:

ffmpeg -i input.mp4 -filter:v select="eq(n\,12345)" -vframes 1 thumbnail.jpg

Note that this command can take a while since it needs to skip to the specified frame before it can extract it.

3
  • An alternate way to specify thumbnail dimensions - this example crops it to a 4:3 aspect ratio: ffmpeg -y -i input.mp4 -vf "crop=(ih*4/3):ih" -f mjpeg -vframes 1 -ss 534 thumbnail.jpg
    – rymo
    Jan 5 '14 at 3:47
  • Is there a way to skip the "waiting" part? Fast-forward to the specified frame/time ?
    – Samson
    Nov 20 '15 at 14:38
  • @Samson Which solution exactly are you talking about? If you specify -ss before -i, seeking should be much faster.
    – slhck
    Nov 23 '15 at 12:35
5

Here's a method using:

  • ffprobe to get the duration
  • bc to calculate half the duration
  • ffmpeg to make the thumbnail image

Example:

input=input.mp4; ffmpeg -ss "$(bc -l <<< "$(ffprobe -loglevel error -of csv=p=0 -show_entries format=duration "$input")*0.5")" -i "$input" -frames:v 1 half.png
0
$currentVideoDuration =  shell_exec("$ffmpeg -i $videoPath 2>&1 | grep Duration");
  $actualDuration = substr($currentVideoDuration, 11, 12);
  $arrayWithHoursMinsAndSecs = explode(":", $actualDuration);
  $thisVideoDurationInSeconds = $arrayWithHoursMinsAndSecs[2] + $arrayWithHoursMinsAndSecs[1]*60 + $arrayWithHoursMinsAndSecs[0]*3600;
  $halfVideoDuration = $thisVideoDurationInSeconds/2;
  echo 'midpoint of this video: '. $halfVideoDuration;
  echo shell_exec("$ffmpeg -i $videoPath -y -ss 00:00:$halfVideoDuration -vframes 1 thumbnail.png 2>&1");

replace $videopath with the path to your video

-1

This bash command works like a charm:

avconv -i 'in.mpg' -vcodec mjpeg -vframes 1 -an -f rawvideo \
-s 420x300 -ss `avconv -i in.mpg 2>&1 | grep Duration \
| awk '{print $2}' | tr -d , \ 
| awk -F ':' '{print ($3+$2*60+$1*3600)/2}'` out.jpg
1
-1

Simple solution is :

 ffmpeg -i l1.mp4 -ss 00:00:14.435 -vframes 1 out.png
2
  • 2
    This looks a lot like Prometee’s answer and LordNeckbeard’s answer except you assume that the user already knows the length of the video (the question says that they don’t) and you don’t explain anything.
    – Scott
    Sep 28 '18 at 7:29
  • I gives just example, if no need time then try this ffmpeg -i l1.mp4 -vframes 1 out11.jpg Sep 28 '18 at 8:23

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