Question

我已经看到了几个引用这个问题的问题，但是无法实现这一点。

我有一个为API路由和套接字实时应用程序运行快速的应用程序。

我想保持我的server.js文件“干净”，并将所有与套接字相关的事件放在一个单独的文件中。我还希望能够从我的路径文件中激活一个事件。

我的server.js喜欢这样：

const dotenv = require('dotenv/config');
const express = require('express');
const bodyParser = require('body-parser');
const socketIO = require('socket.io');
var http = require('http');
const _ = require('lodash');
const path = require('path');
const publicPath = path.join(__dirname, '../public');


var {mongoose} = require('./db/mongoose');
var {ObjectID} = require('mongodb');
var {authenticate} = require('./middleware/authenticate');
var {socketEvents} = require('./middleware/socketEvents')(io);
var multiparty = require('connect-multiparty'),
  multipartyMiddleware = multiparty();

var app = express()
var serv = require('http').createServer(app)
var io = socketIO.listen(serv)


var PORT = process.env.PORT || '80';
app.use(bodyParser.json());
app.use(multipartyMiddleware);
app.use(express.static(publicPath));


serv.listen(PORT, () => {
  console.log('Started on port: ' + PORT);
});

module.exports = {
  app
};

我想要一个包含以下内容的socketEvents.js文件：

io.on('connection', (socket) => {
  console.log('New user connected');

//all socket events here

});

我怎样才能做到这一点？

Answer 1

您可以测试以下模式：

server.js

'use strict';
const io = require('socket.io')(3000);
const io_logic = require('./socketEvents')(io);
console.log('socket.io server started at port 3000');

socketEvents.js

function socket_events (io) {
    io.on('connection', function (socket) {
          console.log('a client has connected');
    });
}
module.exports = socket_events;

将socket.io事件逻辑分离到模块中

1 个答案: