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• What the correct args for ffmpeg conversion

  8 novembre 2013, par Andrew Simpson

  I have a C# app. I am using ffmpge to convert jpgs to an OGG file. I use the process class to write to a stream buffer within my app using piping without having to write the ogg file to a disk. I then upload these bytes to my server where it is written to the hard drive.

  On the server side I want to convert it to say an mp4 format. Again, I want to pipe it out to a stream buffer and allow my User to download the file.

  This is my code :

  Process _serverBuild = null ;

  public byte[] EncodeAndUploadImages(string _args1)
  
  {
  
      byte[] _data = null;
  
      try
  
      {
  
          _args1 = "ffmpeg.exe -i \"c:\\Cloud\\Zipped\\000EC902F17F\\f3e45e44-b61c-472b-8dc1-a8e9b9511497_00007.ogg\" avi -";
  
          _serverBuild = new Process();
  
  
  
          _serverBuild.StartInfo.WorkingDirectory = Environment.CurrentDirectory;
  
          _serverBuild.StartInfo.Arguments = _args1;
  
          _serverBuild.StartInfo.FileName =  @"C:\inetpub\wwwroot\bin\ffmpeg.exe";
  
          _serverBuild.StartInfo.UseShellExecute = false;
  
          _serverBuild.StartInfo.RedirectStandardOutput = true;
  
          _serverBuild.StartInfo.RedirectStandardError = true;
  
          _serverBuild.StartInfo.RedirectStandardInput = true;
  
          _serverBuild.StartInfo.CreateNoWindow = true;
  
          _serverBuild.StartInfo.LoadUserProfile = false;
  
          _serverBuild.ErrorDataReceived += _server_ErrorDataReceived;
  
          _serverBuild.OutputDataReceived += _server_OutputDataReceived;
  
          _serverBuild.Exited += new EventHandler(_server_Exited);
  
          _serverBuild.EnableRaisingEvents = true;
  
          _serverProcessHasFinished = false;
  
          _serverBuild.Start();
  
  
  
          mStandardOutput = _serverBuild.StandardOutput.BaseStream;
  
          mStandardOutput.BeginRead(mReadBuffer, 0, mReadBuffer.Length, StandardOutputReadCallback, null);            
  
          _serverBuild.WaitForExit();
  
  
  
  
  
  
  
  
  
      }
  
      catch (Exception _ex)
  
      {
  
  
  
      }
  
      finally
  
      {
  
          _serverBuild.ErrorDataReceived -= _server_ErrorDataReceived;
  
          _serverBuild.OutputDataReceived -= _server_OutputDataReceived;
  
          _serverBuild.Dispose();
  
      }
  
      return _data;
  
  }
  
  
  
  byte[] mReadBuffer = new byte[4096];
  
  MemoryStream mStandardOutputMs = new MemoryStream();
  
  Stream mStandardOutput;
  
  
  
  private void StandardOutputReadCallback(IAsyncResult ar)
  
  {
  
      try
  
      {
  
          int numberOfBytesRead = mStandardOutput.EndRead(ar);
  
          {
  
              mStandardOutputMs.Write(mReadBuffer, 0, numberOfBytesRead);
  
              mStandardOutput.BeginRead(mReadBuffer, 0, mReadBuffer.Length, StandardOutputReadCallback, null);
  
          }
  
      }
  
      catch (Exception _ex)
  
      {
  
  
  
      }
  
  }

  The error I get (when I test with a bat file and run within a DOS prompt) is :

  

  Obviously my parameters are at fault. I want to keep it simple as I may want to use other formats apart from MP4.

  Thanks

• How much the duration of an mp3 file given by Ffprobe is precise ?

  24 septembre 2024, par Julien Larget-Piet

  I have an mp3 file at a sample rate value of 44100, let's name it a.mp3.

  


  So i get the duration of a.mp3 with the following ffprobe command :

  


  ffprobe -v error -show_entries format=duration -of default=noprint_wrappers=1:nokey=1 a.mp3

  


  I get 7.752 seconds.

  


  Because the sample rate is per second, in theory the total amount of samples should be,

  


  44100 * 7.752 = 341863.2

  


  Let's round it to 341863

  


  But using python library ffmpegio with the following code, i get a total amount of sample equal to 340560.

  


  with ffmpegio.open(file, 'ra', blocksize = 16, sample_fmt = 'dbl') as file_opened:

        for i, indata in enumerate(file_opened):

            do some stuff
    print(i * 16)


  


  So there is a non-negligible difference between 341863 and 340560.

  


  I think it comes from the duration value given by Ffmpeg.
What do you thin ?

  


  I search in ffmpegio documentation, but found nothing that trigered my attention :

  


  https://python-ffmpegio.github.io/python-ffmpegio/

  


  Thanks.

  


  Search for answers about the precision of the ffprobe command.

  


• Container does not reflect MediaReader setting

  8 mars 2014, par Mondain

  I would like my reader to be able to handle audio, when it does not start with the video ; hence the setting of "AddDynamicStreams". How do I get the container to recognize the setting of the reader from which it originated ? Is there a flag or some other property that I need to set ?

  String inputUrl = "rtmp://127.0.0.1:1935/myapp/mystream live=1 buffer=1";
  
  IMediaReader reader = ToolFactory.makeReader(inputUrl);
  
  reader.setCloseOnEofOnly(false);
  
  reader.setQueryMetaData(false);
  
  // we want dynamic addition of streams
  
  reader.setAddDynamicStreams(true);
  
  // this shows "true"
  
  log.debug("Reader can dynamically add streams: {}", reader.canAddDynamicStreams());
  
  // get the container
  
  IContainer container = reader.getContainer();
  
  // this shows "false"
  
  log.debug("Container can dynamically add streams: {}", container.canStreamsBeAddedDynamically());