我如何使用 Python gRPC 处理流式消息

Question

有问题的示例在不回复特定消息的情况下启动并阅读消息。后者是我想要实现的目标。

这是我目前的情况：

import grpc
...

channel = grpc.insecure_channel(conn_str)
try:
    grpc.channel_ready_future(channel).result(timeout=5)
except grpc.FutureTimeoutError:
    sys.exit('Error connecting to server')
else:
    stub = MyService_pb2_grpc.MyServiceStub(channel)
    print('Connected to gRPC server.')
    this_is_just_read_maybe(stub)


def this_is_just_read_maybe(stub):
    responses = stub.MyEventStream(stream())
    for response in responses:
        print(f'Received message: {response}')
        if response.something:
            # okay, now what? how do i send a message here?

def stream():
    yield my_start_stream_msg
    # this is fine, i receive this server-side
    # but i can't check for incoming messages here

存根上似乎没有read()或write()，一切似乎都是用迭代器实现的。

如何从 this_is_just_read_maybe(stub) 发送消息？这是正确的方法吗？

我的原型是双向流：

service MyService {
  rpc MyEventStream (stream StreamingMessage) returns (stream StreamingMessage) {}
}

Answer 1

您尝试做的事情是完全可能的，并且可能涉及编写您自己的请求 iterator 对象，该对象可以在响应到达时给出响应，而不是使用简单的生成器作为您的请求迭代器。也许像

class MySmarterRequestIterator(object):

    def __init__(self):
        self._lock = threading.Lock()
        self._responses_so_far = []

    def __iter__(self):
        return self

    def _next(self):
        # some logic that depends upon what responses have been seen
        # before returning the next request message
        return <your message value>

    def __next__(self):  # Python 3
        return self._next()

    def next(self):  # Python 2
        return self._next()

    def add_response(self, response):
        with self._lock:
            self._responses.append(response)

你然后像这样使用

my_smarter_request_iterator = MySmarterRequestIterator()
responses = stub.MyEventStream(my_smarter_request_iterator)
for response in responses:
    my_smarter_request_iterator.add_response(response)

。在你的 _next 实现中可能会有锁定和阻塞来处理 gRPC Python 询问你的对象它想要发送的下一个请求和你的响应（实际上）的情况 "wait, hold on, I don't know what request I want to send until after I've seen how the next response turned out" .

Answer 2

除了编写自定义迭代器，您还可以使用阻塞队列为客户端存根实现类似发送和接收的行为：

import queue
...

send_queue = queue.SimpleQueue()  # or Queue if using Python before 3.7
my_event_stream = stub.MyEventStream(iter(send_queue.get, None))

# send
send_queue.push(StreamingMessage())

# receive
response = next(my_event_stream)  # type: StreamingMessage

这利用了 iter 的哨兵形式，它将常规函数转换为迭代器，迭代器在达到哨兵值时停止（在本例中为 None）。

我如何使用 Python gRPC 处理流式消息

How do i handle streaming messages with Python gRPC

python

stream

grpc