Recherche avancée

Sur d’autres sites (8131)

• Recording a webpage stream with multiple requests using PhantomJS & ffmpeg to /dev/stdout leads to ffmpeg error

  2 septembre 2016, par Allisson Ferreira

  First of all, sorry for my english.

  I’m in a quest for days. I’ve researched everywhere and I couldn’t find an answer to my problem.

  I’m using Nodejs, Phantomjs and ffmpeg in this scenary :

  • An user enters the site, login with facebook and he can ask for a video with his name and some random photos (gathered by /me/ & sent via JSON POST) ;
  • Node receive the user data, creates a child process (PhantomJS + ffmpeg) and awaits for a response to send the video URL to the user.

  When I run a single instance of this request, everything is working fine. BUT, when two or more users make the request, only one video is sent and the others process end up in a ffmpeg stream error.

  I think the reason is that all the ffmpeg processes are using the same place (/dev/stdout). Since one process is already using it, the others enters in a "can’t access" error. But it is a assumption, I don’t know how /dev/stdout really works.

  Here are my codes. (I have removed some lines and renamed some variables for better understanding, sorry for any mistake)

  index.js :

  var generateVideo = 'phantomjs phantom.js '+videoID+' '+userID+' | ffmpeg -vcodec png -f image2pipe -r 30 -i - -pix_fmt yuv420p public/videos/'+userID+'/'+videoID+'.mp4 -y';
  
  
  
  childProcess.exec(generateVideo, function(err, stdout, stderr) {
  
      var json    = {};
  
      json.video  = '/videos/'+userID+'/'+videoID+'.mp4';
  
      res.send(json);
  
  });

  phantom.js :

  var page            = require('webpage').create();
  
  page.viewportSize   = { width: 1366, height: 768 };
  
  page.settings.resourceTimeout = 10000;
  
  
  
  var args            = require('system').args;
  
  var videoID         = args[1];
  
  var userID          = args[2];
  
  
  
  page.open('http://localhost:3000/recordvideo/'+videoID, 'post', function(status){
  
      var frame       = 0;
  
      var target_fps  = 30;
  
      var maxframes   = page.evaluate(function () {
  
                          return getTotalDurationInSeconds();
  
                      }) * target_fps;
  
  
  
      setInterval(function(){
  
          page.render('/dev/stdout', { format: "png" });
  
          if( frame >= maxframes ){
  
              phantom.exit();
  
          }
  
          frame++;
  
      }, (1000 / target_fps));
  
  });

  And the error :

  [Error: Command failed: /bin/sh -c phantomjs phantom.js XXXXXXXX XXXXXXXX | ffmpeg -vcodec png -f image2pipe -r 30 -i - -pix_fmt yuv420p public/videos/XXXXXXXX/XXXXXXXX.mp4 -y
  
  www-0 ffmpeg version N-80901-gfebc862 Copyright (c) 2000-2016 the FFmpeg developers
  
  www-0   built with gcc 4.8 (Ubuntu 4.8.4-2ubuntu1~14.04.3)
  
  www-0   configuration: --extra-libs=-ldl --prefix=/opt/ffmpeg --mandir=/usr/share/man --enable-avresample --disable-debug --enable-nonfree --enable-gpl --enable-version3 --enable-libopencore-amrnb --enable-libopencore-amrwb --disable-decoder=amrnb --disable-decoder=amrwb --enable-libpulse --enable-libfreetype --enable-gnutls --enable-libx264 --enable-libx265 --enable-libfdk-aac --enable-libvorbis --enable-libmp3lame --enable-libopus --enable-libvpx --enable-libspeex --enable-libass --enable-avisynth --enable-libsoxr --enable-libxvid --enable-libvidstab
  
  www-0   libavutil      55. 28.100 / 55. 28.100
  
  www-0   libavcodec     57. 48.101 / 57. 48.101
  
  www-0   libavformat    57. 41.100 / 57. 41.100
  
  www-0   libavdevice    57.  0.102 / 57.  0.102
  
  www-0   libavfilter     6. 47.100 /  6. 47.100
  
  www-0   libavresample   3.  0.  0 /  3.  0.  0
  
  www-0   libswscale      4.  1.100 /  4.  1.100
  
  www-0   libswresample   2.  1.100 /  2.  1.100
  
  www-0   libpostproc    54.  0.100 / 54.  0.100
  
  www-0 [png @ 0x3d7c4a0] Invalid PNG signature 0x46726F6D20506861.
  
  www-0 [image2pipe @ 0x3d72780] decoding for stream 0 failed
  
  www-0 [image2pipe @ 0x3d72780] Could not find codec parameters for stream 0 (Video: png, none(pc)): unspecified size
  
  www-0 Consider increasing the value for the 'analyzeduration' and 'probesize' options
  
  www-0 Input #0, image2pipe, from 'pipe:':
  
  www-0   Duration: N/A, bitrate: N/A
  
  www-0     Stream #0:0: Video: png, none(pc), 30 tbr, 30 tbn, 30 tbc
  
  www-0 [buffer @ 0x3d81540] Unable to parse option value "0x0" as image size
  
  www-0 [buffer @ 0x3d81540] Unable to parse option value "-1" as pixel format
  
  www-0 [buffer @ 0x3d81540] Unable to parse option value "0x0" as image size
  
  www-0 [buffer @ 0x3d81540] Error setting option video_size to value 0x0.
  
  www-0 [graph 0 input from stream 0:0 @ 0x3d72600] Error applying options to the filter.
  
  www-0 Error opening filters!
  
  www-0 ]

  I really hope that I can find an answer here !
  And sorry if there already is an answer for this. But I researched for days.

  Thank you in advance !

• Java - RTSP save snapshot from Stream Packets

  9 août 2016, par Guerino Rodella

  I’m developing an application which requests snapshots to DVR and IP Cameras. The device I’m working on just offer RTSP requests to do so. Then I implemented the necessary RTSP methods to start receiving the stream packets and I started receiving then via UDP connection established. My doubt is, how can I save the received data to a jpeg file ? Where’s the begging and end of the image bytes received ?

  I searched a lot libraries which implement this type of service in Java, like Xuggler ( which it’s maintained no more ), javacpp-presets - has ffmpeg and opencv libraries included - I had some environment problems with it. If someone know an easy and good one which saves snapshots from the streams, let me know.

  My code :

  final long timeout = System.currentTimeMillis() + 3000;
  
  
  
  byte[] fullImage = new byte[ 1024 * 1024 ];
  
  DatagramSocket udpSocket = new DatagramSocket( 8000 );
  
  int lastByte = 0;
  
  
  
  // Skip first 2 packets because I think they are HEADERS
  
  // Since I don't know what they mean, I just print then in hexa
  
  for( int i = 0; i < 2; i++ ){
  
  
  
      byte[] buffer = new byte[ 1024 ];
  
      DatagramPacket dataPacket = new DatagramPacket( buffer, buffer.length );
  
      udpSocket.receive( dataPacket );
  
  
  
      int dataLenght = dataPacket.getLength();
  
      buffer = Arrays.copyOf( buffer, dataLenght );
  
  
  
      System.out.println( "RECEIVED[" + DatatypeConverter.printHexBinary( buffer ) + " L: " + dataLenght );
  
  
  
  }
  
  
  
  do{
  
  
  
      byte[] buffer = new byte[ 1024 ];
  
      DatagramPacket dataPacket = new DatagramPacket( fullImage, fullImage.length );
  
      udpSocket.receive( dataPacket );
  
  
  
      System.out.println( "RECEIVED: " + new String( fullImage ) );
  
  
  
      for( int i = 0; i < buffer.length; i++ ){
  
          fullImage[ i + lastByte ] = buffer[ i ];
  
          lastByte ++;
  
  
  
      }
  
  
  
  } while( System.currentTimeMillis() < timeout );
  
  // I know this timeout is wrong, I should stop after getting full image bytes

  The output :

  RECEIVED : 80E0000100004650000000006742E01FDA014016C4 L : 21
  RECEIVED : 80E00002000046500000000068CE30A480 L : 17
  RECEIVED : Tons of data from the streaming...
  RECEIVED : Tons of data from the streaming...
  RECEIVED : Tons of data from the streaming...
  [...]

  As you might suppose, the image I’m saving into a file is not readable because I’m doing it wrong. I think the header provide me some info about the next packets the server will sent me telling the start and the end of the image from the streaming. But I don’t understood them. Someone know how to solve it ? Any tips are welcome !

• streaming webcam via rtp protocol

  18 août 2016, par vasu gupta

  i am trying to stream and receive my webcam feed on two terminal on same laptop.For this purpose I am using the following commands :-

  foo.sdp :

  SDP:
  
  v=0
  
  o=- 0 0 IN IP4 127.0.0.1
  
  s=No Name
  
  c=IN IP4 127.0.0.1
  
  t=0 0
  
  a=tool:libavformat 55.2.100
  
  m=video 1235 RTP/AVP 96
  
  a=rtpmap:96 H264/90000
  
  a=fmtp:96 packetization-mode=1

  Transmitting :

  ffmpeg -re -i /dev/video0 -r 24 -b 50k -s 858x500 -f mulaw -f rtp rtp://127.0.0.1:3000> foo.sdp

  Receiving :

  ffplay -i foo.sdp

  While transmission seems to be working fine , but when i am using receiving command I am getting en error :

  Protocol not on whitelist 'file,crypto'!/0   
  
  foo.sdp: Invalid data found when processing input