的NodeJS合併對象

Question

我工作的NodeJS創建一種能夠一次發送圖片到多個顯示器，目前我的代碼有以下語句：

var $ = require('jQuery'); 
var app = require('http').createServer(handler); 
var io = require('socket.io').listen(app, { log:false }); 
var fs = require('fs'); 

var Server = { 
    server_address : "XXX.XXX.XXX.XXX", // The server address 
    playing_now : Object // The playlist that are being synced 
}; 

var Client = { 
    clients : Object 
}; 

app.listen(1337); 
console.log("App is running on port 1337..."); 

function handler(request, response) { 
    var ip_address = null; 

    try{ ip_address = request.header('x-forwarded-for'); } 
    catch(error){ ip_address = request.connection.remoteAddress; } 

    Client.clients[ip_address] = "X"; 

    fs.readFile(__dirname + '/index.html', function (err, data) { 
     if (err) { 
      response.writeHead(500); 
      return response.end('Error loading index.html'); 
     } 

     response.writeHead(200); 
     response.end(data); 
    }); 
} 

io.sockets.on('connection', function (socket) { 
    socket.on("RequestPlaylist", function(json){ 
     Client.clients[json["Address"]] = json["PlayList"];  

     if(typeof Server.playing_now[json["PlayList"]] === 'undefined'){ 
      $.ajax({ 
       url  : "http://" + Server.server_address + "/buildPlayList.php", 
       type  : "GET", 
       dataType : "jsonp", 
       method : "GET", 
       data  : {RemoteIP : json["Address"], Name : json["PlayList"]} 
      }).done(function(playlistJSON){ 
       console.log("________________________________________"); 
       console.log("Done"); 
       console.log(Server.playing_now); 
       Server.playing_now[playlistJSON["playName"]] = playlistJSON; 
       console.log(Server.playing_now); 
       console.log("________________________________________"); 
      })  
     }else{ 
      console.log("________________________________________"); 
      console.log("Redirect"); 
      console.log(Server.playing_now); 
      console.log("________________________________________"); 
     }  
    }); 
}); 

當一個新的客戶端連接，我保存它的IP地址在Client.clients上，當我得到包含圖像的url的json時，我將它存儲在Server.playing_now上。

的問題是，當我輸出Server.playing_now，它包括Client.clients的內容：

Done 
{ [Function: Object] 'XXX.XXX.XXX.XXX': 'Default Media' } 
{ [Function: Object] 
    'XXX.XXX.XXX.XXX': 'Default Media', 
    'Default Media': 
    { '1': 
     { json: [Object], 
     duration: '5', 
     name: 'Slideshow', 
     type: 'Image', 
     sync: '1', 
     callback: 'jQuery17207063492417801172_1377555709748' }, 
    playName: 'Default Media' } } 

如果嘗試再次讀取該數據，這是結果：

Redirect 
{ [Function: Object] 
    '105.102.15.234': 'Default Media', 
    'Default Media': 
    { '1': 
     { json: [Object], 
     duration: '5', 
     name: 'Entrenamiento PL', 
     type: 'Image', 
     sync: '1', 
     callback: 'jQuery17207063492417801172_1377555709748' }, 
    playName: 'Default Media' } } 

我的代碼沒有任何合併這兩個對象，如果我註釋掉Client對象，它只返回Server.playing_now的內容。

任何想法可能發生什麼？或者是預期這種行爲？

Answer 1

您在初始化客戶端（和服務器）：

var Client = { 
    clients : Object 
}; 

您設置爲Object這是同樣的事情（和一個壞主意）。你真的想new Object()或只是{}

嘗試改變初始化代碼是這樣的：

var Server = { 
    server_address : "XXX.XXX.XXX.XXX", // The server address 
    playing_now : {} // The playlist that are being synced 
}; 

var Client = { 
    clients : {} 
};