Use FFMpeg to get middle frame of a video?

Question

I'm wondering how to use FFMpeg to grap the middle frame of a video. I've already written the C# to grab a frame at a certain time (ie pull one frame at second 3). But I've yet to figure out how to find the middle of video using the FFMpeg commands.

Answer 1

This could be simplified, but here's some old PHP code I had lying around that should do the trick. (Add the location to ffmpeg if it's not in your path)

$output = shell_exec("ffmpeg -i {$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) + $time_breakdown[2]);
shell_exec("ffmpeg -y  -i {$path} -f mjpeg -vframes 1 -ss " . ($total_seconds / 2) . " -s {$w}x{$h} {$output_filename}");   

Answer 2

FFmpeg helps you get the framerate and the length of the video, so you can multiply one by the other and divide by 2 to get the number of the middle frame.

ie for a 30 seconds video running at 15 frames per second : 30 * 15 = 450 / 2 = 225, meaning you need to grab the 225th frame.

Answer 3

With simple shell scripting you can use ffprobe to get a machine readable duration output, bc to calculate the half point, and ffmpeg to make the frame:

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

This eliminates any need for PHP, echo, awk, tr, grep, sed, etc.

Answer 4

This bash command works like a charm (tested):

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