如何解決以下easyrtc問題？

Question

我已經寫了下面的代碼：

var http = require("http");    // http server core module 
var express = require("express");   // web framework external module 
var io  = require("socket.io");   // web socket external module 
var easyrtc = require("easyrtc");   // EasyRTC external module 
var app = express(); 

var server = http.createServer(app).listen(app.get('port')); 

io= io.listen(server,{"log level":1}); 
var rtc = easyrtc.listen(server, io); 

這是給下面的錯誤：

$node server info - EasyRTC: Starting EasyRTC Server (v1.0.10) on Node (v0.10.26) [TypeError: Object # has no method 'get'] TypeError: Object # has no method 'get' at async.waterfall.pub.socketServer.sockets.on.easyrtcid (/home/ritzy1/Downloads/downloaded codes/VEDIO/change2/testexpandwb/node_modules/easyrtc/lib/easyrtc_default_event_listeners.js:1472:29) at fn (/home/ritzy1/Downloads/downloaded codes/VEDIO/change2/testexpandwb/node_modules/easyrtc/node_modules/async/lib/async.js:582:34) at Object._onImmediate (/home/ritzy1/Downloads/downloaded codes/VEDIO/change2/testexpandwb/node_modules/easyrtc/node_modules/async/lib/async.js:498:34) at processImmediate [as _immediateCallback] (timers.js:330:15)

Ηow我能解決這個問題？

Answer 1

你必須在listen（）函數中給出一個特定的端口號。 更改以下行
var server = http.createServer(app).listen(app.get('port'));
與
var server = http.createServer(app).listen(8080);
注意：8080是一個端口號，您可以指定任何其他目前未使用的端口號。

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