Reputation: 5082
I'm trying to write a simple zmq system to replace an http client/server. My client times out when the server is off/unavailable, but does not retry or stop. What am I missing?
zmq_client.rb (modified version of Han Holl's lazy pirate client from zeromq guide)
require 'rubygems'
require 'zmq'
context = ZMQ::Context.new
socket = context.socket(ZMQ::REQ)
socket.connect('tcp://localhost:5559')
retries = 2
timeout = 10
retries.times do |tries|
message = "Hello #{tries}"
raise("Send: #{message} failed") unless socket.send(message)
puts "Sending string [#{message}]"
if ZMQ.select( [socket], nil, nil, timeout)
message = socket.recv
puts "Received reply [#{message}]"
break
else
puts "timeout"
end
end
socket.close
zmq_broker.rb (modified version of Oleg Sidorov's code found on zeromq guide)
require 'rubygems'
require 'ffi-rzmq'
context = ZMQ::Context.new
frontend = context.socket(ZMQ::ROUTER)
frontend.bind('tcp://*:5559')
poller = ZMQ::Poller.new
poller.register(frontend, ZMQ::POLLIN)
loop do
poller.poll(:blocking)
poller.readables.each do |socket|
if socket === frontend
loop do
socket.recv_string(message = '')
more = socket.more_parts?
puts "#{message}#{more}"
socket.send_string(message, more ? ZMQ::SNDMORE : 0)
break unless more
end
end
end
end
Upvotes: 1
Views: 1085
Reputation: 1948
You should get an error Send: #{message} failed
as soon as you try to send
again after the first timeout, because your 2nd send
will happen directly after the 1st send
, and the REQ socket enforces that each send
must go after (successful, not timeout-ed) recv
.
In the lazy pirate pattern, you may need to send several requests before getting a reply. Solution suggested in the 0MQ Guide is to close and reopen the REQ socket after an error. Your client doesn't close/reopen the REQ socket.
You may find helpful the "Lazy Pirate client in Ruby" example from the Guide.
Upvotes: 1