我有网络应用程序。该应用程序具有将一些对象数据推送到 redis
channel 的端点。
另一个端点处理 websocket
连接,其中数据从 channel 中获取并通过 ws
发送到客户端。
当我通过 ws 连接时,消息仅获取第一个连接的客户端。
如何使用多个客户端从 redis
channel 读取消息而不创建新订阅?
Websocket 处理程序。
在这里,我订阅了 channel ,将其保存到应用程序 (init_tram_channel
)。然后在我收听 channel 和发送消息的地方运行作业(run_tram_listening
)。
@routes.get('/tram-state-ws/{tram_id}')
async def tram_ws(request: web.Request):
ws = web.WebSocketResponse()
await ws.prepare(request)
tram_id = int(request.match_info['tram_id'])
channel_name = f'tram_{tram_id}'
await init_tram_channel(channel_name, request.app)
tram_job = await run_tram_listening(
request=request,
ws=ws,
channel=request.app['tram_producers'][channel_name]
)
request.app['websockets'].add(ws)
try:
async for msg in ws:
if msg.type == aiohttp.WSMsgType.TEXT:
if msg.data == 'close':
await ws.close()
break
if msg.type == aiohttp.WSMsgType.ERROR:
logging.error(f'ws connection was closed with exception {ws.exception()}')
else:
await asyncio.sleep(0.005)
except asyncio.CancelledError:
pass
finally:
await tram_job.close()
request.app['websockets'].discard(ws)
return ws
订阅并保存 channel 。
每个 channel 都与唯一对象相关,为了不创建与同一对象相关的多个 channel ,我只将一个 channel 保存到应用程序。
app['tram_producers']
是字典。
async def init_tram_channel(
channel_name: str,
app: web.Application
):
if channel_name not in app['tram_producers']:
channel, = await app['redis'].subscribe(channel_name)
app['tram_producers'][channel_name] = channel
运行 coro 以收听 channel 。 我通过 aiojobs 运行它:
async def run_tram_listening(
request: web.Request,
ws: web.WebSocketResponse,
channel: Channel
):
"""
:return: aiojobs._job.Job object
"""
listen_redis_job = await spawn(
request,
_read_tram_subscription(
ws,
channel
)
)
return listen_redis_job
我在 Coro 收听和发送消息:
async def _read_tram_subscription(
ws: web.WebSocketResponse,
channel: Channel
):
try:
async for msg in channel.iter():
tram_data = msg.decode()
await ws.send_json(tram_data)
except asyncio.CancelledError:
pass
except Exception as e:
logging.error(msg=e, exc_info=e)
最佳答案
在一些 aioredis github issue 中发现了以下代码(我已将其用于我的任务)。
class TramProducer:
def __init__(self, channel: aioredis.Channel):
self._future = None
self._channel = channel
def __aiter__(self):
return self
def __anext__(self):
return asyncio.shield(self._get_message())
async def _get_message(self):
if self._future:
return await self._future
self._future = asyncio.get_event_loop().create_future()
message = await self._channel.get_json()
future, self._future = self._future, None
future.set_result(message)
return message
那么,它是如何工作的呢? TramProducer 包装了我们获取消息的方式。
正如@Messa所说
message is received from one Redis subscription only once.
因此只有 TramProducer 的一个客户端正在从 Redis 中检索消息,而其他客户端正在等待从 channel 接收消息后设置的 future 结果。
如果 self._future
被初始化,这意味着有人正在等待来自 redis 的消息,所以我们将等待 self._future
结果。
TramProducer 用法(我从我的问题中举了一个例子):
async def _read_tram_subscription(
ws: web.WebSocketResponse,
tram_producer: TramProducer
):
try:
async for msg in tram_producer:
await ws.send_json(msg)
except asyncio.CancelledError:
pass
except Exception as e:
logging.error(msg=e, exc_info=e)
TramProducer 初始化:
async def init_tram_channel(
channel_name: str,
app: web.Application
):
if channel_name not in app['tram_producers']:
channel, = await app['redis'].subscribe(channel_name)
app['tram_producers'][channel_name] = TramProducer(channel)
我认为这可能对某些人有帮助。
完整项目在这里 https://gitlab.com/tram-emulator/tram-server
关于python - 如何用aioredis pub/sub实现单生产者多消费者,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54159292/