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• FFMPEG Save stream frames to single image file (not sequence) [duplicate]

  5 mars 2021, par Nikola Knežević

  Is there a way to open a stream (usb webcam for example) and save each frame to the same image file ?

  


  That image would be later accessed as a simple resource from whichever protocol/service.

  


  I tried this :

  


  ffmpeg.exe -f dshow -i video="ASUS USB2.0 Webcam" image.png


  


  but it fails to write second frame. Error :

  


  Could not get frame filename number 2 from pattern 'image.png'


  


  Sure, I can make a script to periodically starts ffmpeg and save only one frame, but that's a huge performance overhead.

  


• Revision f3d8e315ac : Fix a bug in neon that has not save and restore q4-q7 registers. Change-Id : Ie2

  27 février 2014, par hkuang

  Changed Paths :
   Modify /vp9/common/arm/neon/vp9_iht8x8_add_neon.asm

  
  
  Fix a bug in neon that has not save and restore q4-q7 registers.

  Change-Id : Ie21b5ae89100389b80f919710839084f935a8545

• 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 !