Im get error "using outdated pixel format, make sure you set the range correctly with ffmpeg" .. can anyone check my code below?

Question

This is my code using ffmpeg. I want to have a thumbnail of the video, but I don’t know that ffmpeg can someone know the error I received.

[swscaler @ 0x7ff8da028c00] deprecated pixel format used, make sure you did set range correctly
Output #0, mjpeg, to 'image.jpg':
Metadata:
major_brand     : mp42
minor_version   : 0
compatible_brands: isommp42
encoder         : Lavf56.36.100
Stream #0:0(und): Video: mjpeg, yuvj420p(pc), 320x240 [SAR 4:3 DAR 16:9], q=2-31, 200 kb/s, 1 fps, 1 tbn, 1 tbc (default)
Metadata:
  creation_time   : 2016-11-06 09:40:22
  handler_name    : ISO Media file produced by Google Inc.
  encoder         : Lavc56.41.100 mjpeg
Stream mapping:
Stream #0:0 -> #0:0 (h264 (native) -> mjpeg (native))
Press [q] to stop, [?] for help
frame=    1 fps=0.0 q=4.8 Lsize=      16kB time=00:00:01.00 bitrate= 129.5kbits/s    
video:16kB audio:0kB subtitle:0kB other streams:0kB global headers:0kB muxing overhead: 0.000000%

Also this is my code:

$video_url ='https://URL/upload/4b8acab123563649f19e07450d810df6.mp4';


 $ffmpeg = '/opt/local/bin/ffmpeg';
 $image = 'image.jpg';
 $interval = 5;
 $size = '320x240';
 shell_exec($tmp = "$ffmpeg -i $video_url -deinterlace -an -ss $interval -f mjpeg -t 1 -r 1 -y -s $size $image 2>&1");  

Answer 1

According to the ffmpeg forum, this can be ignored when invoked from the command line, and I always ignored it. One thing you can try (test with a dummy PLEASE file) is to add -pix_fmt yuvj422pto your last line so that it looks like this:

shell_exec($tmp = "$ffmpeg -i $video_url -pix_fmt yuvj422p -deinterlace -an -ss $interval -f mjpeg -t 1 -r 1 -y -s $size $image 2>&1");

, , , , , ffmpeg , .

https://lists.ffmpeg.org/pipermail/ffmpeg-user/2014-February/020151.html

Afaict, ,    ffmpeg .

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Answer 2

:

# path of image on the server
$image = '/users/xx/xxxx/sample/upload/image.jpg'; 

:

# $ffmpeg - path of ffmpeg on the server
# $video_url - path of the video file on the server

shell_exec($tmp = "$ffmpeg -i $video_url -pix_fmt yuvj422p -deinterlace -an -ss $interval -f mjpeg -t 1 -r 1 -y -s $size $image 2>&1");