node.js - 如果 socket.send() 仅使用一次，则 zmq 订阅者无法订阅已发布的消息

Question

如果 socket.send() 在发布者中使用仅一次，zmq 订阅者将无法订阅消息。

在发布者中使用以下代码时，订阅者能够订阅消息:

var zmq = require('zmq')
  , sock = zmq.socket('pub')

sock.bindSync('tcp://127.0.0.1:3000');
var message = {"test" : true};
setInterval(function(){
    sock.send(['notify_anomaly', JSON.stringify(message)]);
},1000);

但是如果在publisher代码中去掉setInterval就不行了，如下:

var zmq = require('zmq')
  , sock = zmq.socket('pub')

sock.bindSync('tcp://127.0.0.1:3000');
var message = {"test" : true};
sock.send(['notify_anomaly', JSON.stringify(message)]);

Answer 1

这是“慢连接器”问题的结果。

这是来自 guide 的引述:

There is one more important thing to know about PUB-SUB sockets: you do not know precisely when a subscriber starts to get messages. Even if you start a subscriber, wait a while, and then start the publisher, the subscriber will always miss the first messages that the publisher sends. This is because as the subscriber connects to the publisher (something that takes a small but non-zero time), the publisher may already be sending messages out.

基本上，当发布者运行时，它会与订阅者握手。由于这是异步的，发布者可能会在握手完成之前完成发送消息。在那种情况下，订阅者将错过消息。为了让订阅者收到第一条消息，发布者需要等待发送，直到确定订阅者已连接。

这里是进一步的引用:

In Chapter 2 - Sockets and Patterns we'll explain how to synchronize a publisher and subscribers so that you don't start to publish data until the subscribers really are connected and ready.

它显示了如何使用另一个套接字对使用 REQ-REP 在订阅者准备好在订阅套接字上接收信号时发出信号 here .