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• avfilter/vf_bwdif : Add neon for filter_line3

  4 juillet 2023, par John Cox

  avfilter/vf_bwdif : Add neon for filter_line3
  Signed-off-by : John Cox <jc@kynesim.co.uk>
  
  Signed-off-by : Martin Storsjö <martin@martin.st>
  

  • [DH] libavfilter/aarch64/vf_bwdif_init_aarch64.c
  • [DH] libavfilter/aarch64/vf_bwdif_neon.S

• ffmpeg throwing "errorMessage" : "spawn ffmpeg ENOENT"

  4 juin 2024, par ian

  My ffmpeg function thats running as a shell command isnt working. I think its because 'ffmpeg' isnt really referring to anything. I have the ffmpeg node module in my bundle, but i dont know the execFile command is referring to it here.

  



  Im following aws-lambda-ffmpeg as an example of how to call this particular function. They are referring to 'ffmpeg' as a 64-bit linux build they created from John Vansickle's static FFMPEG builds in their gulp function.

  



  I want to know how to replace 'ffmpeg' with something that will just recognize it like a node_module without having to do the whole gulp static build process. To my understanding the only reason they are doing that is to get the latest build which i really dont need.

  



  If I am wrong and a static build using gulp is needed for another reason please let me know.

  



  function ffmpegProcess(description, cb) {
console.log('Starting FFmpeg');

child_process.execFile(
    'ffmpeg',
    [
        '-y',
        '-loglevel', 'warning',
        '-i', 'download',
        '-c:a', 'copy',
        '-vf', scaleFilter,
        '-movflags', '+faststart',
        '-metadata', 'description=' + description,
        'out.' + config.format.video.extension,
        '-vf', 'thumbnail',
        '-vf', scaleFilter,
        '-vframes', '1',
        'out.' + config.format.image.extension
    ],
    {
        cwd: tempDir
    },
    function(err, stdout, stderr) {
        console.log('FFmpeg done.');
        return cb(err, 'FFmpeg finished:' + JSON.stringify({ stdout: stdout, stderr: stderr}));
    }
);
}


  


• ffmpeg throwing "errorMessage" : "spawn ffmpeg ENOENT"

  25 avril 2016, par user2792129

  My ffmpeg function thats running as a shell command isnt working. I think its because ’ffmpeg’ isnt really referring to anything. I have the ffmpeg node module in my bundle, but i dont know the execFile command is referring to it here.

  Im following aws-lambda-ffmpeg as an example of how to call this particular function. They are referring to ’ffmpeg’ as a 64-bit linux build they created from John Vansickle’s static FFMPEG builds in their gulp function.

  I want to know how to replace ’ffmpeg’ with something that will just recognize it like a node_module without having to do the whole gulp static build process. To my understanding the only reason they are doing that is to get the latest build which i really dont need.

  If I am wrong and a static build using gulp is needed for another reason please let me know.

  function ffmpegProcess(description, cb) {
  
  console.log('Starting FFmpeg');
  
  
  
  child_process.execFile(
  
      'ffmpeg',
  
      [
  
          '-y',
  
          '-loglevel', 'warning',
  
          '-i', 'download',
  
          '-c:a', 'copy',
  
          '-vf', scaleFilter,
  
          '-movflags', '+faststart',
  
          '-metadata', 'description=' + description,
  
          'out.' + config.format.video.extension,
  
          '-vf', 'thumbnail',
  
          '-vf', scaleFilter,
  
          '-vframes', '1',
  
          'out.' + config.format.image.extension
  
      ],
  
      {
  
          cwd: tempDir
  
      },
  
      function(err, stdout, stderr) {
  
          console.log('FFmpeg done.');
  
          return cb(err, 'FFmpeg finished:' + JSON.stringify({ stdout: stdout, stderr: stderr}));
  
      }
  
  );
  
  }