python - Python zmq SUB 套接字未接收到 MQL5 Zmq PUB 套接字

Question

我正在尝试在 MQL5 中设置一个 PUB 套接字，在 Python 中设置一个 SUB 套接字来接收消息。

我在 MQL5 中有这个:

#include <Zmq/Zmq.mqh>

Context context("helloworld");
Socket socket(context,ZMQ_PUB);

string BROKER;

int OnInit()
{
   if (socket.bind("tcp://*:5556"))
   {
       Print("Error");
   }
   else
       Print("Bound");
   BROKER = AccountInfoString(ACCOUNT_COMPANY);
   return(INIT_SUCCEEDED);
}

void OnTick()
{  
   MqlTick last_tick; 
   string str;
   if(SymbolInfoTick(Symbol(),last_tick)) 
   { 
      StringConcatenate(str, BROKER, ",", Symbol(), ",", last_tick.time_msc, ",", last_tick.ask, ",", last_tick.bid, ",", last_tick.last, ",", last_tick.volume); 
   } 
   else 
      str = "FAIL";

   Print(str);
   ZmqMsg reply(str);
   socket.send(reply);
}

在 Python 中:

import zmq
import random
import sys
import time

context = zmq.Context()
socket = context.socket(zmq.SUB)

ports = [5556]
for port in ports:
    print(port)
    socket.connect("tcp://localhost:{}".format(port))
socket.setsockopt_string(zmq.SUBSCRIBE, '')

print('connected')

f = open('metatrader-1.csv', 'a')
while True:
    msg = socket.recv()
    print(msg)
    f.write(str(msg) + '\n')

这个问题是，这似乎在 Python 端没有收到任何东西，recv 调用只是永远阻塞。 OnTick 方法在 MT 中触发，因为可以看到指纹。

我怎样才能让它工作？

请注意，如果我切换到 REP/REQ 对，它会起作用。

MQL5:

#include <Zmq/Zmq.mqh>


Context context("helloworld");
Socket socket(context,ZMQ_REQ);

string BROKER;

int OnInit()
{
   if (socket.connect("tcp://localhost:5555"))
   {
      Print("Error");
   }
   else
      Print("Bound");
   BROKER = AccountInfoString(ACCOUNT_COMPANY);
   return(INIT_SUCCEEDED);
}

void OnTick()
{  
   MqlTick last_tick; 
   string str;
   if(SymbolInfoTick(Symbol(),last_tick)) 
   { 
      StringConcatenate(str, BROKER, ",", Symbol(), ",", last_tick.time_msc, ",", last_tick.ask, ",", last_tick.bid, ",", last_tick.last, ",", last_tick.volume); 
   } 
   else 
      str = "FAIL";

   Print(str);
   ZmqMsg reply(str);
   socket.send(reply);
   socket.recv(reply);
}

python :

import zmq
import random
import sys
import time

context = zmq.Context()
socket = context.socket(zmq.REP)

ports = [5555]
for port in ports:
    print(port)
    socket.bind("tcp://*:{}".format(port))
#socket.setsockopt_string(zmq.SUBSCRIBE, '')

print('connected')

f = open('metatrader-1.csv', 'a')
while True:
    msg = socket.recv()
    socket.send_string('ack')
    print(msg)
    f.write(str(msg) + '\n')

但这有一些缺点，所以如果可以的话我宁愿不使用它。

Answer 1

您的套接字选项必须放在套接字连接之前，因此您的代码将是:

import zmq
import random
import sys
import time

context = zmq.Context()
socket = context.socket(zmq.SUB)

ports = [5556]
socket.setsockopt(zmq.SUBSCRIBE, b"")  # Note.

for port in ports:
    print(port)
    socket.connect("tcp://localhost:{}".format(port))

print('connected')

f = open('metatrader-1.csv', 'a')
while True:
    msg = socket.recv()
    print(msg)
    f.write(str(msg) + '\n')

此外，这部分:

if (socket.bind("tcp://*:5556"))
{
   Print("Error");
}
else
   Print("Bound");

应该是反过来的。当套接字成功绑定(bind)时，这实际上会打印错误。