python gRPC 需要上下文参数

Question

我刚刚开始使用 gRPC，并在阅读入门指南后尝试实现一个简单的 python 服务。但是当我调用我的客户端调用时，python 要求提供上下文参数。为什么我的代码需要提供示例中不需要的上下文对象？

附注我开始尝试创建自己的具体上下文子类，但不确定应该如何实现它。我已经添加了我的开始，但如果可能的话，我真的很感激一个可以使用的例子

谢谢！

原型文件

syntax = "proto2";

package parsefile;

service ParseFile {
  rpc SendFile (File) returns (Empty) {}
}

message File {

  message MetaData {
    optional string file_name = 1;
    optional string file_path = 2 [default = '.'];
    optional string mime_type = 3 [default = 'application/pdf'];
  }

  message Source {
    optional string title = 1;
    optional int32 id = 2;
  }

  optional MetaData document = 1;
  optional Source supplier = 2;
}

message Empty {
}

服务器

from concurrent import futures
import time

import grpc

import parsefile_pb2_grpc
import parsefile_pb2

_ONE_DAY_IN_SECONDS = 60 * 60 * 24

class ParseFileServicer(parsefile_pb2_grpc.ParseFileServicer):

    def SendFile(self, request, context):
        supplier = request.supplier.title
        file_name = request.document.file_name
        print('Received {} from {}'.format(file_name, supplier))
        return parsefile_pb2.Empty()

def serve():
    server = grpc.server(futures.ThreadPoolExecutor(max_workers=2))
    parsefile_pb2_grpc.add_ParseFileServicer_to_server(
        ParseFileServicer, server)
    server.add_insecure_port('[::]:50051')
    server.start()
    try:
        while True:
            time.sleep(_ONE_DAY_IN_SECONDS)
    except KeyboardInterrupt:
        server.stop(0)

if __name__ == '__main__':
    serve()

客户

import grpc

import parsefile_pb2_grpc
import parsefile_pb2

def get_file_info():
    return parsefile_pb2.File(
        document = parsefile_pb2.File.MetaData(
            file_name = 'example.txt'
        ),
        supplier = parsefile_pb2.File.Source(
            title = 'Example Supplier'
        )
    )

def run():
    channel = grpc.insecure_channel('localhost:50051')
    stub = parsefile_pb2_grpc.ParseFileStub(channel)
    context = RequestContext()
    print('object created')
    response = stub.SendFile(get_file_info())
    print('File info sent to server')

if __name__ == '__main__':
    run()

错误追踪

Traceback (most recent call last):   File "parse_client.py", line 60, in <module>
    run()   File "parse_client.py", line 56, in run
    response = stub.SendFile(get_file_info(), 2)   File "/Users/davidbowe/.virtualenvs/post/lib/python3.6/site-packages/grpc/_channel.py", line 507, in __call__
    return _end_unary_response_blocking(state, call, False, deadline)   File "/Users/davidbowe/.virtualenvs/post/lib/python3.6/site-packages/grpc/_channel.py", line 455, in _end_unary_response_bl ocking
    raise _Rendezvous(state, None, None, deadline) grpc._channel._Rendezvous: <_Rendezvous of RPC that terminated with (StatusCode.UNKNOWN, Exception calling application: SendF ile() missing 1 required positional argument: 'context')>

Answer 1

在您的服务器代码中，您缺少用于实例化服务程序对象的括号。你写了...

parsefile_pb2_grpc.add_ParseFileServicer_to_server(ParseFileServicer, server)

应该是：

parsefile_pb2_grpc.add_ParseFileServicer_to_server(ParseFileServicer(), server)

Answer 2

您不需要创建

context

参数，它是由grpc自动创建的。

Answer 3

我遇到这个错误是因为我添加了一个额外的方法。我刚刚将该方法移到了类之外。