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• Start Android Java VM from native C code NDK (ffmpeg mediacodec use)

  19 avril 2021, par eusoubrasileiro

  I managed to cross-compile ffmpeg using the NDK for armv8a api 27 with MediaCodec hardware acceleration support.

  


  Using root after setting permissions, folders and setting properly LD_LIBRARY_PATH etc. I can run it without problems on a terminal session (ssh). Only if I don't try to use the -hwaccel option.

  


  If try to run something using -hwaccel, like :

  


  ffmpeg -rtsp_transport tcp -an -hwaccel mediacodec -c:v hevc_mediacodec -i rtsp://user:pass@192.168.0.100:554/onvif1 -f null - -benchmark


  


  I get the error bellow about No Java virtual machine.

  


  ...
Input #0, rtsp, from 'rtsp://user:pass@192.168.0.100:554/onvif1':
  Metadata:
    title           : H.265 Video, RtspServer_0.0.0.2
  Duration: N/A, start: 0.000000, bitrate: N/A
    Stream #0:0: Video: hevc (Main), yuv420p(tv, bt470bg), 1920x1080 [SAR 1:1 DAR 16:9], 10 fps, 10 tbr, 90k tbn, 10 tbc
    Stream #0:1: Audio: pcm_alaw, 8000 Hz, mono, s16, 64 kb/s
[amediaformat @ 0x7e2ea27300] No Java virtual machine has been registered
[hevc_mediacodec @ 0x7e2eb44f00] Failed to create media format
Stream mapping:
  Stream #0:0 -> #0:0 (hevc (hevc_mediacodec) -> wrapped_avframe (native))
Error while opening decoder for input stream #0:0 : Generic error in an external library


  


  Would it be possible to start (create or launch ?) the Dalvik Java VM directly from the C code ? I don't even know if those are correct terms. Make it visible for ffmpeg ?

  


  Any information that will help a Android Newbie get on his feet will be greatly appreciated. If that is possible I would write a little patch on the ffmpeg code.

  


  I really would not like to package this in an app only to be able to test around with this.

  


• FFmpeg command works in terminal but gives an error in php code

  21 mai 2021, par Rider

  Use this command :

  


  ffmpeg -i lecture.mp4 -codec: copy -start_number 0 -hls_time 10 -hls_list_size 0 -f hls filename.m3u8


  


  It works.

  


  $path = '1/';
  $inputFile = 'lecture.mp4';
  $filePath = 'filename.m3u8';
  $commandOutput = exec ('/opt/ffmpeg/ffmpeg -I '.$path .$inputFile. ' -Codec: copy -start_number 0 -hls_time 10 -hls_list_size 0 -f hls'.$path .$filePath);


  


  Using this PHP code I get the error : At least one output file must be specified

  


  What is causing this error ?

  


• Can I use a config file within an embedded youtube-dl code ?

  10 mai 2017, par Stack Johan

  Right now, I’m having issues with ffmpeg not being recognized. I have the ffmpeg files stored in a folder, and I normally call ffmpeg using —ffmpeg-location.

  I don’t know how to do that within Python code, and I would prefer to use config files to call all the options.

  This is what I have right now :

  from __future__ import unicode_literals
  
  import youtube_dl
  
  
  
  
  
   ydl_opts = {
  
      'format': 'bestaudio/best',
  
      'outtmpl': u'%(id)s.%(ext)s',
  
      'postprocessors': [{
  
      'key': 'FFmpegExtractAudio',
  
      'preferredcodec': 'mp3',
  
      'preferredquality': '192',
  
  }]
  
  }
  
  with youtube_dl.YoutubeDL(ydl_opts) as ydl:
  
  ydl.download(['https://youtu.be/Y_n0UwQMz5o'])