python - 无法连接以在openzmq中的现有tcp连接中打开另一个tcp连接

Question

我正在从客户端1和客户端2到服务器连接到客户端2。我正在将帧从客户端1发送到客户端2，然后在客户端2上执行预测并将结果发送到服务器。我有以下代码。
客户端1代码:

  context = zmq.Context()
  footage_socket = context.socket(zmq.PUB)
  footage_socket.connect('tcp://172.168.1.2:5555')
  videoFile = 'data.mp4'
  camera = cv2.VideoCapture(videoFile) 
  length=int(camera.get(cv2.CAP_PROP_FRAME_COUNT))
  print(length)
  count=0
  #time.sleep(2)
  while True:        
    grabbed, frame = camera.read()
    count+=1
    print(count)
    try:
       frame = cv2.resize(frame, (224, 224))
    except cv2.error:
        break
    encoded, buffer = cv2.imencode('.jpg', frame)
    jpg_as_text = base64.b64encode(buffer)
    footage_socket.send(jpg_as_text)

客户端2代码:

context = zmq.Context()
footage_socket = context.socket(zmq.SUB)
footage_socket.bind('tcp://0.0.0.0:5555')
footage_socket.setsockopt_string(zmq.SUBSCRIBE, np.unicode(''))
while True:    
    frame = footage_socket.recv_string()
    img = base64.b64decode(frame)
    npimg = np.fromstring(img, dtype=np.uint8)
    source = cv2.imdecode( npimg, 1 )
    frame=cv2.resize(source,(224,224)).astype("float32")
    image = img_to_array( source)
    image = image.reshape( (1, image.shape[0], image.shape[1], image.shape[2]) )
    image = preprocess_input( image )
    preds = model.predict(image)
    ##connecting to server##        
    context1=zmq.Context()
    footage_socket=context1.socket(zmq.PUB)
    footage_socket.connect('tcp://192.168.56.103:9999')
    footage_socket.send(preds)
    print('sending to server')

服务器代码:

   context = zmq.Context()
   footage_socket = context.socket(zmq.SUB)
   footage_socket.bind('tcp://0.0.0.0:9999')
   footage_socket.setsockopt_string(zmq.SUBSCRIBE, np.unicode(''))
   while True:
      frame = footage_socket.recv_string()
      img = base64.b64decode(frame)
      #print(img)

在client-2上，我收到以下错误

    frame = footage_socket.recv_string()
  File "/usr/local/lib/python3.5/dist-packages/zmq/sugar/socket.py", line 583, in recv_string
    msg = self.recv(flags=flags)
  File "zmq/backend/cython/socket.pyx", line 790, in zmq.backend.cython.socket.Socket.recv
  File "zmq/backend/cython/socket.pyx", line 826, in zmq.backend.cython.socket.Socket.recv
  File "zmq/backend/cython/socket.pyx", line 193, in zmq.backend.cython.socket._recv_copy
  File "zmq/backend/cython/socket.pyx", line 188, in zmq.backend.cython.socket._recv_copy
  File "zmq/backend/cython/checkrc.pxd", line 25, in zmq.backend.cython.checkrc._check_rc
zmq.error.ZMQError: Operation not supported

Answer 1

几种罪过，让我们一遍又一遍地揭开面纱:
Client-1可以进行改进，但是Client-2存在大多数问题:

################################################################### FOOTAGE ~ <SUB>-Socket
# SUB
footage_socket = context.socket( zmq.SUB )
...
PUB_TARGET = 'tcp://192.168.56.103:9999'
while True:    
    frame  = footage_socket.recv_string()                         # <SUB>.recv()-ed
    source = cv2.imdecode( np.fromstring( base64.b64decode( frame ),
                                          dtype = np.uint8
                                          ),
                           1 )
    frame  = cv2.resize( source,
                         (224,224)
                         ).astype( "float32" )
    image = img_to_array( source )
    image = image.reshape( ( 1,
                             image.shape[0],
                             image.shape[1],
                             image.shape[2]
                             )
                           )
    preds = model.predict( preprocess_input( image ) )
    ################################################################## PER-LOOP INFty-times
    ## connecting to server ###########################
    context1=zmq.Context()                           ## INSTANTIATED new Context()-instance
    footage_socket = context1.socket( zmq.PUB )      ## ASSOCIATED a new  Socket()-instance
    footage_socket.connect( PUB_TARGET )             ## .CONNECT( PUB_TARGET )
    footage_socket.send( preds )                     ## <PUB>.send()
    ################################################### LOOP AGAIN
    ###################################################      yet now the <PUB>.recv()

a) while True: -code-block创建尽可能多的 Context() -instances，具有所有分配和至少1-I/O线程，是循环运行的次数
b)循环的尾部为 footage_socket 分配了一个新对象( socket -class的另一个新实例)，将原始对象引用呈现给 SUB -type socket-instance一个孤儿，但具有所有关联的资源的分配没有终止
c)新重新分配的 footage_socket -socket，现在带有 PUB -type socket -instance的引用，实际上无法处理 .send() -method，就像在 while True:头中的脚本和-如上所示，将抛出一个(未处理的)异常。

解决方案？
消除这些概念上的错误-避免预期的 SUB和新的 PUB的名称冲突，并避免重复使代码(无穷大)生成新的和新的和新的 Context()实例(这些操作在资源和延迟方面都很昂贵)，并且其无限多个 socket(PUB) -instances和无限多个 .connect() -s及其代码应按预期工作。

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