我写了一个这样的异步程序。一个永远的运行循环同时启动 4 个事件。每个事件都会运行该rpc
服务。在nameko
服务中,我使用time.sleep(10)
.
我很困惑为什么服务每秒钟就完成一次10
。我认为服务应该同时完成。我怎样才能让工作同时完成?
def start_loop(loop):
asyncio.set_event_loop(loop)
loop.run_forever()
async def job(x):
try:
with ClusterRpcProxy(CONFIG) as rpc:
res = rpc.helloworldService.helloworld(x)
print(res)
except Exception as e:
print(f"{e}")
async def do_sleep(x, queue):
try:
await job(x)
queue.put("ok")
except Exception as e:
print(f"{e}")
def consumer():
asyncio.run_coroutine_threadsafe(do_sleep('10', queue), new_loop)
asyncio.run_coroutine_threadsafe(do_sleep('11', queue), new_loop)
asyncio.run_coroutine_threadsafe(do_sleep('12', queue), new_loop)
asyncio.run_coroutine_threadsafe(do_sleep('13', queue), new_loop)
if __name__ == '__main__':
print(time.ctime())
new_loop = asyncio.new_event_loop()
loop_thread = Thread(target=start_loop, args=(new_loop,))
loop_thread.setDaemon(True)
loop_thread.start()
CONFIG = {'AMQP_URI': "amqp://guest:guest@localhost"}
queue = Queue()
sema = asyncio.Semaphore(2)
consumer_thread = Thread(target=consumer)
consumer_thread.setDaemon(True)
consumer_thread.start()
while True:
msg = queue.get()
print("current:", time.ctime())
nameko
rpc
服务是:
class HelloWorld:
name = 'helloworldService'
@rpc
def helloworld(self,str):
time.sleep(10)
return 'hello_'+str
输出如下:</p>
hello_10
current: Sat Jan 26 13:04:57 2019
hello_11
current: Sat Jan 26 13:05:07 2019
hello_12
current: Sat Jan 26 13:05:17 2019
hello_13
current: Sat Jan 26 13:05:28 2019