GCP PubSub：Python 中的同步拉取订阅者？

Question

全部，

我正在尝试学习如何使用 GCP PubSub，并且可以通过 CLI 命令对其进行测试（创建主题、订阅、发布到主题、从订阅中拉取等），但是当我跳转到 python 时（v 2.7，当前公司标准）我正在努力以同步方式拉取消息。

我已经查看了这个网址，它告诉你去睡觉和“While True”，但我无法想象有人在现实世界中这样做，对吗？ https://cloud.google.com/pubsub/docs/quickstart-client-libraries#pubsub-subscribe-python

这个网址告诉您可以使用 future.result()，我尝试过，但它不会像您想象的那样阻塞会话/线程： http: //google-cloud-python.readthedocs.io/en/latest/ pubsub/订阅者/index.html

有人还有其他想法吗？这是我的函数，它几乎直接来自示例之一：

def sample_receive_messages(subscription_name="my-sub", project=None):
"""Receives messages from a pull subscription."""
if not project:
    credentials, project = google.auth.default()  

subscriber = psub.SubscriberClient()
subscription_path = subscriber.subscription_path(
    project, subscription_name)

def callback(message):
    # print('Received message: {}'.format(message))
    message.ack()
    print('<message>' + str(message.data) + '</message>')

subscription = subscriber.subscribe(subscription_path)

future = subscription.open(callback)

myResult = future.result()

subscription.close()

print("done")

我最后的目标是有一个进程经常醒来，抓取消息并确认它们，将消息写入文件，然后结束。

到目前为止，该进程读取消息并将其打印出来（很棒），但它然后坐啊坐啊坐，最后错误地出现了一些乱码：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "pubSubTools.py", line 50, in sample_receive_messages
    myResult = future.result()
  File "/usr/local/lib/python2.7/site-packages/google/cloud/pubsub_v1/futures.py", line 98, in result
    raise err
grpc._channel._Rendezvous: <_Rendezvous of RPC that terminated with (StatusCode.UNAVAILABLE, The service was unable to fulfill your request. Please try again. [code=8a75])>

Answer 1

库文档中对此进行了描述：

from google.cloud import SubscriberClient

pubsub_client = SubscriberClient()
subscription_path = 'projects/{project}/subscriptions/{subscription}'.format(project=project_name, subscription=subscription_name)
pull_response= pubsub_client.pull(subscription=subscription_path, max_messages=10)

for msg in pull_response.received_messages:
    message = msg.message.data.decode('utf-8')
    # do your thing
    pubsub_client.acknowledge(subscription_path, [msg.ack_id])