【问题标题】:Python gRPC Server Not StartingPython gRPC 服务器未启动
【发布时间】:2017-04-13 22:17:06
【问题描述】:

我已经使用 proto3 和 python 创建了一个 gRPC 服务器来对一个长时间运行的守护进程进行基本的健康检查。但是,当我启动我的应用程序时,它实际上并没有启动 gRPC 服务器。我想知道是否有人可以帮助确定它为什么不启动并为 gRPC API 提供服务

原型定义:health.proto

syntax = "proto3";

option java_multiple_files = true;
option java_package = "com.redacted.example.worker";
option java_outer_classname = "ExampleWorker";
option objc_class_prefix = "DSW";

package exampleworker;

service Worker {
  rpc Health (Ping) returns (Pong) {}
}

// The request message containing PONG
message Ping {
  string message = 1;
}

// The response message containing PONG
message Pong {
  string message = 1;
}

然后我使用以下代码生成了 python 代码:

python -m grpc_tools.protoc -I=../protos --python_out=. --grpc_python_out=. ../protos/health.proto

这会生成health_pb2.py 和health_pb2_grpc.py 文件。接下来我创建了一个服务器文件:

u"""Health server is used to create a new health monitoring GRPC server."""

from concurrent import futures
import logging
import grpc
import health_pb2
import health_pb2_grpc

# grpc related variables
grpc_host = u'[::]'
grpc_port = u'50001'
grpc_address = u'{host}:{port}'.format(host=grpc_host, port=grpc_port)
# logging related variables
logger = logging.getLogger(__name__)
logger.setLevel(logging.INFO)


class WorkerServicer(health_pb2_grpc.WorkerServicer):
    u"""Provides methods that implement functionality of health server."""

    def Health(self, request, context):
        u"""Return PONG to say the Worker is alive."""
        return health_pb2.Pong(message='PONG')


def serve_health_api():
    u"""Create and start the GRPC server."""
    server = grpc.server(futures.ThreadPoolExecutor(max_workers=10))
    health_pb2_grpc.add_WorkerServicer_to_server(WorkerServicer(), server)
    logging.info(u'adding port {grpc_address}'.format(
        grpc_address=grpc_address))
    server.add_insecure_port(grpc_address)
    server.start()

然后在我的主 run.py 文件中:

#!mac/bin/python

"""Run is the local (non-wsgi) entrypoint to the flask application."""

from subscriber.worker import Worker
from redis import StrictRedis
from examplegrpc.health_server import serve_health_api
import logging

redis_host = 'localhost'
redis_port = 6379
redis_db = 0
redis_chan = 'deployment'

if __name__ == "__main__":
    FORMAT = '%(asctime)s %(name)-12s %(levelname)-8s %(message)s'
    logging.basicConfig(format=FORMAT)
    logger = logging.getLogger(__name__)
    logger.setLevel(logging.INFO)

    logger.debug('Creating redis client')
    client = StrictRedis(host=redis_host, port=redis_port, db=redis_db)
    w = Worker(client, [redis_chan])
    try:
        logger.info('Starting Health gRPC API...')
        serve_health_api()
        logger.info('Starting worker...')
        w.run()
    except KeyboardInterrupt:
        logger.info('Exiting...')

w.run() 正确启动以从 redis 通道执行工作,但 gRPC 服务器没有启动尝试访问它

channel = grpc.insecure_channel('localhost:{port}'.format(port=grpc_port))
stub = WorkerStub(channel)
ping = examplegrpc.health_pb2.Ping(message='PING')
health = stub.Health(ping)

开始

_Rendezvous: <_Rendezvous of RPC that terminated with (StatusCode.UNAVAILABLE, Connect Failed)>

【问题讨论】:

    标签: python protocol-buffers grpc


    【解决方案1】:

    在我看来,发生的事情是您的服务器仅分配给您的 serve_health_api 函数中的本地字段,因此当该函数返回时(在启动服务器后立即),服务器被垃圾收集并关闭.

    【讨论】:

    • 就是这样!非常感谢您的回答和帮助。我真的很感激
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