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• How to use (django-celery,RQ) worker to execute a video filetype conversion (ffmpeg) in django on heroku (My code works locally)

  15 janvier 2013, par GetItDone

  One part of my website includes a form that allows users to upload video. I use ffmpeg to convert the video to flv. My media and static files are stored on Amazon S3. I can get everything to work perfectly locally, however I can't seem to figure out how to use a worker to run the video conversion subprocess in production. I have dj-celery and rq installed in my app. The code in my view that I was able to get to work locally is :

  #views.py
  
  def upload_broadcast(request):
  
      if request.method == 'POST':
  
          form = VideoUploadForm(request.POST, request.FILES)
  
          if form.is_valid():
  
              new_video=form.save()
  
              def convert_to_flv(video):
  
                  filename = video.video_upload
  
                  sourcefile = "%s%s" % (settings.MEDIA_ROOT, filename)
  
                  flvfilename = "%s.flv" % video.id
  
                  imagefilename = "%s.png" % video.id
  
                  thumbnailfilename = "%svideos/flv/%s" % (settings.MEDIA_ROOT, imagefilename)
  
                  targetfile = "%svideos/flv/%s" % (settings.MEDIA_ROOT, flvfilename)
  
                  ffmpeg = "ffmpeg -i %s -acodec mp3 -ar 22050 -f flv -s 320x240 %s" % (sourcefile, targetfile)
  
                  grabimage = "ffmpeg -y -i %s -vframes 1 -ss 00:00:02 -an -vcodec png -f rawvideo -s 320x240 %s" % (sourcefile, thumbnailfilename)
  
                  print ("SOURCE: %s" % sourcefile)
  
                  print ("TARGET: %s" % targetfile)
  
                  print ("TARGET IMAGE: %s" % thumbnailfilename)
  
                  print ("FFMPEG TASK CODE: %s" % ffmpeg)
  
                  print ("IMAGE TASK CODE: %s" % grabimage)
  
                  try:
  
                      ffmpegresult = subprocess.call(ffmpeg)
  
                      print "---------------FFMPEG---------------"
  
                      print ffmpegresult
  
                  except:
  
                      print "Not working."
  
                  try:
  
                      videothumbnail = subprocess.call(grabimage)
  
                      print "---------------IMAGE---------------"
  
                      print videothumbnail
  
                  except:
  
                      print "Not working."
  
                  video.flvfilename = flvfilename
  
                  video.videothumbnail = imagefilename
  
                  video.save()
  
  
  
              convert_to_flv(new_video)
  
  
  
              return HttpResponseRedirect('/video_list/')
  
      else:
  
  ...

  This is my first time trying to use a worker (or ever pushing a project to production), so even with the documentation it is still unclear to me what I need to do. I have tried several different things but nothing seems to work. Is there just a simple way to tell celery to run the ffmpegresult = subprocess.call(ffmpeg) ? Thanks in advance for any help or insight.

  EDIT- Added heroku logs

  2013-01-10T20:58:57+00:00 app[web.1]: TARGET: /media/videos/flv/8.flv
  
  2013-01-10T20:58:57+00:00 app[web.1]: IMAGE TASK CODE: ffmpeg -y -i /media/videos/practice.wmv -vframes 1 -ss 00:00:02 - an -vcodec png -f rawvideo -s 320x240 /media/videos/flv/8.png
  
  2013-01-10T20:58:57+00:00 app[web.1]: SOURCE: /media/videos/practice.wmv
  
  2013-01-10T20:58:57+00:00 app[web.1]: FFMPEG TASK CODE: ffmpeg -i /media/videos/practice.wmv -acodec mp3 -ar 22050 -f fl v -s 320x240 /media/videos/flv/8.flv
  
  2013-01-10T20:58:57+00:00 app[web.1]: TARGET IMAGE: /media/videos/flv/8.png
  
  2013-01-10T20:58:57+00:00 app[web.1]: Not working.
  
  2013-01-10T20:58:57+00:00 app[web.1]: Not working.

  NEWER EDIT

  I tried adding a tasks.py and added the task :

  celery = Celery('tasks', broker='redis://guest@localhost//')
  
  
  
  @celery.task
  
  def ffmpeg_task(video):
  
      converted_file = subprocess.call(video)
  
      return converted_file

  then I changed the relevant section of my view to :

  ...
  
  try:
  
      ffmpeg_task.delay(ffmpeg)
  
      print "---------------FFMPEG---------------"
  
      print ffmpegresult
  
  except:
  
      print "Not working."
  
  ...

  My new logs are :

  2013-01-15T13:19:52+00:00 app[web.1]: TARGET IMAGE: /media/videos/flv/12.png
  
  2013-01-15T13:19:52+00:00 app[web.1]: SOURCE: /media/videos/practice.wmv
  
  2013-01-15T13:19:52+00:00 app[web.1]: FFMPEG TASK CODE: ffmpeg -i /media/videos/practice.wmv -acodec mp3 -ar 22050 -f fl v -s 320x240 /media/videos/flv/12.flv
  
  2013-01-15T13:19:52+00:00 app[web.1]: IMAGE TASK CODE: ffmpeg -y -i /media/videos/practice.wmv -vframes 1 -ss 00:00:02 -an -vcodec png -f rawvideo -s 320x240 /media/videos/flv/12.png
  
  2013-01-15T13:19:52+00:00 app[web.1]: TARGET: /media/videos/flv/12.flv
  
  2013-01-15T13:20:17+00:00 app[web.1]: 2013-01-15 13:20:17 [2] [CRITICAL] WORKER TIMEOUT (pid:12)
  
  2013-01-15T13:20:17+00:00 app[web.1]: 2013-01-15 13:20:17 [2] [CRITICAL] WORKER TIMEOUT (pid:12)
  
  2013-01-15T13:20:17+00:00 app[web.1]: 2013-01-15 13:20:17 [19] [INFO] Booting worker with pid: 19

  Am I completely missing something ? I'll keep trying, but will be very appreciative of any direction or assistance.

• Creating a convertAll() function that converts all .filetype in working directory

  5 janvier 2019, par Risviltsov

  I seem to not know proper bash syntax ; despite this, I’ve tried to create a tool that changes the dimensions of all files of a ffmpeg-accepted filetype in the working directory and converts it to another ffmpeg-accepted filetype. In this instance, this tool converts all .webm files over 1080x720 into 1080x-1 or -1x720 .mp4 files. If the .webm file is under 1080x720, the new .mp4 file will have the same dimensions.

  However, there’s a wrench in the tool.

  convertAll () {
  
  local wantedWidth = 1080
  
  local wantedHeight = 720
  
  for i in *.webm; do
  
  local newWidth = $i.width
  
  local newHeight = $i.height
  
  until [$newWidth <= $wantedWidth && $newHeight <= $wantedHeight]; do
  
  if [$videoWidth > $wantedWidth]; then
  
  newHeight = $newWidth*($wantedWidth/$newWidth)
  
  newWidth = $newWidth*($wantedWidth/$newWidth)
  
  fi
  
  if [$videoHeight > $wantedHeight]; then
  
  newWidth = $newWidth*($wantedHeight/$newHeight)
  
  newHeight = $newHeight*($wantedHeight/$newHeight)
  
  fi
  
  done
  
  ffmpeg -i "$i" -vf scale=$newWidth:$newHeight "${i%.*}.mp4";
  
  done
  
  echo "All files have been converted."
  
  }

  What this returns is a bunch of lines that look like this :

  bash: [: missing ']'
  
  bash: [: missing ']'
  
  bash: =: No such file or directory

  My best guess is that BASH can’t do mathematics, and that I’m declaring and editing my variables incorrectly.

  I’d like some input on this --- my lack of experience is really getting me here.

• look for a specific filetype in a directory in php and send it to a different directory after conversion

  12 juin 2019, par flash

  I have a directory in which there is a mp4 file (including other files as well) which I want to convert into mp3 and then send it to different directory. I have used the following command line command to covert into mp3 and its working perfectly fine.

  ffmpeg -i 36031P.mp4 -map 0:2 -ac 1 floor_english.mp3 

  mp4 file is inside in_folder. Using ffmpeg, I want to convert mp4 file into mp3 and send it to out_folder.

  <?php
  
  $dir    = 'in_folder';
  
  $files1 = scandir($dir);
  
  print_r($files1);    /* It lists all the files in a directory including mp4 file*/
  
  ?>

  print_r($files1) lists all the file in a directory including mp4file.

  Problem Statement :

  I am wondering what php code I need to write so that it looks for only mp4 file inside the directory and send it to different directory (let say out_folder) after converting into mp3.

  The pictorial representation of what I want :