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• How to set a timeout for FFmpegFrameGrabber grab() method in JavaCV

  18 janvier 2019, par Null Pointer

  I’m trying to create a player in Java using JavaCV. But .grab() method takes different amounts of time for each grab. How can I discard a grab() operation when it takes too much time ?

  In detail : I’m working on a project to display rtsp streams. To do this, I use JavaCV and its FFmpegFrameGrabber class. Inside a loop I call its .grab() method to grab frames and then I print them on a panel.
  (Not knowing the reason) grab() method takes much more time in some calls. For example :

  1. grab : 15ms,
  2. grab : 13ms,
  3. grab : 14ms,
  4. grab : 95ms,
  5. grab : 12ms, ...

  I wish not to wait for a grab, in case it is taking too long like in this fourth call.

  (Sorry for not providing a working code, it’s all over different huge classes)

  I use the following grabber options (selected by some other colleague) :

  grabber.setImageHeight(480);
  
  grabber.setImageWidth(640);
  
  grabber.setOption("reconnect", "1");
  
  grabber.setOption("reconnect_at_eof", "1");
  
  grabber.setOption("reconnect_streamed", "1");
  
  grabber.setOption("reconnect_delay_max", "2");
  
  grabber.setOption("preset", "veryfast");
  
  grabber.setOption("probesize", "192");
  
  grabber.setOption("tune", "zerolatency");
  
  grabber.setFrameRate(30.0);
  
  grabber.setOption("buffer_size", "" + this.bufferSize);
  
  grabber.setOption("max_delay", "500000");
  
  grabber.setOption("stimeout", String.valueOf(6000000));
  
  grabber.setOption("loglevel", "quiet");
  
  grabber.start();

  Also I tried to do the grab() in another thread and not wait for that to end in case it takes too long. But this time I get an EXCEPTION_ACCESS_VIOLATION when I call .grab() method before the previous one did not end yet. "Calling .release() ; and .restart() ; methods before trying to grab again" also did not solve the access violation.

  Is there a setting option to cancel unordinarily long grabs ? (Or wouldn’t it work without these frames (because of a reason that I don’t know))

  Thanks

• Working directory in Dockerfile is not what is expected to be

  31 mai 2022, par user1765862

  I have following Dockerfile

  


  FROM public.ecr.aws/lambda/dotnet:6 AS base

FROM mcr.microsoft.com/dotnet/sdk:6.0-bullseye-slim as build
WORKDIR /src
COPY ["AWSServerless.csproj", "AWSServerless/"]
RUN dotnet restore "AWSServerless/AWSServerless.csproj"

WORKDIR "/src/AWSServerless"
COPY . .
RUN dotnet build "AWSServerless.csproj" --configuration Release --output /app/build

FROM build AS publish

RUN apt-get update && apt-get install -y apt-utils libgdiplus libc6-dev

RUN apt-get install -y ffmpeg

RUN dotnet publish "AWSServerless.csproj" \
            --configuration Release \ 
            --runtime linux-x64 \
            --self-contained false \ 
            --output /app/publish \
            -p:PublishReadyToRun=true  

FROM base AS final
WORKDIR /var/task

CMD ["AWSServerless::AWSServerless.LambdaEntryPoint::FunctionHandlerAsync"]
COPY --from=publish /app/publish .


  


  As per ffmpeg docs In order to use ffmpeg I need to set its binary folder
for example with

  


  {
  "BinaryFolder": "/var/task",
  "TemporaryFilesFolder": "/tmp"
}


  


  Error

  


  


  An error occurred trying to start process './ffmpeg' with working
directory '/var/task'. No such file or directory

  


  


  What is the working directory here ?

  


  Update :
Actual error is being thrown when I try to execute following command (FFMpegCore package) from my api controller

  


  public class MyController : ControllerBase&#xA;{&#xA;    public async Task<string> Get()&#xA;    {    &#xA;         await FFMpegArguments&#xA;               .FromPipeInput(new StreamPipeSource(myfile))&#xA;               .OutputToPipe(new StreamPipeSink(outputStream), options => options&#xA;                  .WithVideoCodec("vp9")&#xA;                  .ForceFormat("webm"))&#xA;                  .ProcessAsynchronously();&#xA;         ...&#xA;    }    &#xA;}&#xA;</string>

  


• Cannot find the class in php oop [duplicate]

  30 décembre 2018, par mr stark

  This question already has an answer here :

  • PHP class not found when using namespace
    
    2 answers
    

  I’m trying to make a class for video streaming with FFMpeg Php package.

  I’m getting this error on browser :

  Fatal error : Uncaught Error : Class ’Classes\Video’ not found in C :\xampp\htdocs\php\index.php:5 Stack trace : #0 main thrown in C :\xampp\htdocs\php\index.php on line 5

  this is my index.php :

    use Classes\Video;
  
  
  
    $video = new Video();
  
  
  
    $video->videoStreaming("video.mp4",320,240,"myVideo","WMV");

  and Video in Classes directory :

      namespace Classes;
  
      use FFMpeg\Coordinate\Dimension;
  
      use FFMpeg\FFMpeg;
  
      use FFMpeg\Format\Video\WMV;
  
      require "../vendor/autoload.php";
  
  
  
      class Video
  
      {
  
          private $ffmpeg;
  
          private $video;
  
  
  
  public function __construct()
  
  {
  
      $this->ffmpeg = FFMpeg::create([
  
          'ffmpeg.binaries' => 'C:\ffmpeg\ffmpeg.exe',
  
          'ffprobe.binaries' => 'C:\ffmpeg\ffprobe.exe',
  
          'timeout' => 3600,
  
          'ffmpeg.threads' => 12,
  
      ]);
  
  }
  
  
  
  public function videoStreaming($videoPath,$width,$height,$newName,$newExtension)
  
  {
  
      $this->video = $this->ffmpeg->open($videoPath);
  
      if ($width && $height) {
  
          $this->video->filters()->resize(new Dimension($width,$height))->synchronize();
  
      }
  
      $this->video->save(new WMV(), $newName . $newExtension);
  
  }
  
  
  
      }