將socket.io事件邏輯分離到模塊中

Question

我已經看到了一些引用此問題的問題，但無法使其正常工作。

我有一個應用程序運行快速API API路由以及套接字實時應用程序。

我想保持我的server.js文件「乾淨」，並將所有與套接字相關的事件放在一個單獨的文件中。我也希望能夠從我的路線文件中激發一個事件。

我server.js loks這樣的：

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 
}; 

我想有包括socketEvents.js文件：

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

//all socket events here 

}); 

我怎樣才能做到這一點？

Answer 1

您可以測試以下模式：

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;