2017-06-14 80 views
2

我想在php文件上運行nodejs。easyRTC連接到socket.io服務器時出錯

<script src="htps://xxx.com:8445/socket.io/socket.io.js"></script> 
<script src="htps://xxx.com:8445/easyrtc/easyrtc.js" type="text/javascript"></script> 

Server.js:

// Load required modules 
var https = require("https");  // https server core module 
var fs  = require("fs");  // file system core module 
var express = require("express"); // web framework external module 
var io  = require("socket.io"); // web socket external module 
var easyrtc = require("../"); // EasyRTC external module 

// Setup and configure Express http server. Expect a subfolder called "static" to be the web root. 
var httpApp = express(); 
httpApp.use(express.static(__dirname + ":8445")); 

// Start Express https server on port 8445 
var webServer = https.createServer(
{ 
    key: fs.readFileSync("/etc/apache2/ssl/xxx.com.key"), 
    cert: fs.readFileSync("/etc/apache2/ssl/xxx.com.crt") 
}, 
httpApp).listen(8445); 

// Start Socket.io so it attaches itself to Express server 
var socketServer = io.listen(webServer, {"log level":1}); 

// Start EasyRTC server 
var rtc = easyrtc.listen(httpApp, socketServer); 

錯誤: enter image description here

工作的NodeJS,但我得到一個錯誤這樣的。如果我沒有使用PHP,它可以工作。

+0

爲什麼你在錯誤中隱藏白色?端口8445是否顯示在那裏? – itsundefined

+0

沒有域名和擴展名 – Nevermore

+0

@ slowdeath007我的意思是google.com/socket.io .....................我不知道爲什麼端口不返回 – Nevermore

回答

2

對於任何來自谷歌與這個問題,easyRTC假設http服務器和ws服務器是在同一個網址/端口。爲了解決這個問題,在包含easyRTC後需要運行方法easyrtc.setSocketUrl(" https://thedomain.com.tr:8445 ");

+0

非常感謝你:) – Nevermore

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