无法连接以在openzmq

如何解决无法连接以在openzmq

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上进行预测并将结果发送到服务器。我有以下代码。

Client-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)

Client-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)).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，我收到以下错误消息

    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,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

解决方法

一些罪过，让我们一次又一次地揭穿：

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:代码块创建了多达Context()个实例，这些实例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类的另一个新实例），从而将原始对象引用呈现给 SUB 类型的套接字实例是一个孤立实例，但所有关联资源的分配均未终止

c）一个新重新分配的 footage_socket 套接字，现在带有 PUB 类型socket的引用如实例所示，-instance确实无法处理 .send() 方法，如while True:代码块开头的脚本所示，并且抛出了（未处理的）异常以上。

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解决方案？

消除这些概念性错误-避免使预期的SUB和新的PUB的名称冲突，并避免重复使代码（无限地）生成新的和新的Context()-实例（这些操作在资源和延迟方面都是昂贵的），并且它的无数socket(PUB)-实例和无数.connect()-s以及代码应按预期工作。

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