我有这个简单的python脚本,它连接到ZMQ提要并吐出一些数据:
#!/usr/bin/env python2
import zlib
import zmq
import simplejson
def main():
context = zmq.Context()
subscriber = context.socket(zmq.SUB)
# Connect to the first publicly available relay.
subscriber.connect('tcp://relay-us-east-1.eve-emdr.com:8050')
# Disable filtering.
subscriber.setsockopt(zmq.SUBSCRIBE, "")
while True:
# Receive raw market JSON strings.
market_json = zlib.decompress(subscriber.recv())
# Un-serialize the JSON data to a Python dict.
market_data = simplejson.loads(market_json)
# Dump typeID
results = rowsets = market_data.get('rowsets')[0];
print results['typeID']
if __name__ == '__main__':
main()
这在我的家庭服务器上运行。有时,我的家庭服务器失去了与互联网的连接,这是一个住宅连接的诅咒。然而,当网络辍学并重新开启时,脚本会停止。有没有办法重新初始化连接?我还是蟒蛇新手,正确方向上的一点很精彩。 =)
答案 0 :(得分:2)
不确定这是否仍然相关,但这里有:
使用超时(示例here,here和here)。在ZMQ< 3.0它看起来像这样(未经测试):
#!/usr/bin/env python2
import zlib
import zmq
import simplejson
def main():
context = zmq.Context()
while True:
subscriber = context.socket(zmq.SUB)
# Connect to the first publicly available relay.
subscriber.connect('tcp://relay-us-east-1.eve-emdr.com:8050')
# Disable filtering.
subscriber.setsockopt(zmq.SUBSCRIBE, "")
this_call_blocks_until_timeout = recv_or_timeout(subscriber, 60000)
print 'Timeout'
subscriber.close()
def recv_or_timeout(subscriber, timeout_ms)
poller = zmq.Poller()
poller.register(subscriber, zmq.POLLIN)
while True:
socket = dict(self._poller.poll(stimeout_ms))
if socket.get(subscriber) == zmq.POLLIN:
# Receive raw market JSON strings.
market_json = zlib.decompress(subscriber.recv())
# Un-serialize the JSON data to a Python dict.
market_data = simplejson.loads(market_json)
# Dump typeID
results = rowsets = market_data.get('rowsets')[0];
print results['typeID']
else:
# Timeout!
return
if __name__ == '__main__':
main()
ZMQ> 3.0允许您设置套接字RCVTIMEO
选项,这将导致其recv()
引发超时错误,而不需要Poller
对象。