如何构建Node/Angular/Socket.io项目？

Question

我正在开发一个使用AngularJS和Socket.io的项目.我发现了这个非常好的集成示例.

这是项目结构:

app.js                  --> app config
bower.json              --> for bower
package.json            --> for npm
public/                 --> all of the files to be used in on the client side
  css/                  --> css files
    app.css             --> default stylesheet
  img/                  --> image files
  js/                   --> javascript files
    app.js              --> declare top-level app module
    controllers.js      --> application controllers
    directives.js       --> custom angular directives
    filters.js          --> custom angular filters
    services.js         --> custom angular services
  bower_components/
    angular/            --> angular.js
    angular-socket-io/  --> socket.io adapter for angular
routes/
  index.js              --> route for serving HTML pages and partials
  socket.js             --> serve content over a socket
  api.js                --> serve JSON to our AngularJS client
views/
  index.jade            --> main page for app
  layout.jade           --> doctype, title, head boilerplate
  partials/             --> angular view partials (partial jade templates)
    partial1.jade
    partial2.jade

在app.js中:

var express = require('express'),
    routes = require('./routes'),
    api = require('./routes/api'),
    socket = require('./routes/socket');

...

// serve index and view partials
app.get('/', routes.index);
app.get('/partials/:name', routes.partials);

// JSON API
app.get('/api/name', api.name);

// redirect all others to the index (HTML5 history)
app.get('*', routes.index);

// Socket.io Communication
io.sockets.on('connection', require('./routes/socket'));

现在,虽然通常我只是将服务器逻辑放在app.js其中似乎逻辑在这里划分api.js,socket.js并且index.js- 我非常喜欢这个.

但是,让我们说,在socket.js我需要使用所定义的东西api.js.我应该添加var api = require('./api');的socket.js？

Answer 1

我最终创建了模块/对象，将所有内容导入 app.js 并通过相互引用传递它们（需要时）。

  var mashup = require('./routes/mashupModule'),
  socket = require('./routes/socketModule'),
  browser = require('./routes/browserModule');

  socket.init(server, browser, mashup);
  browser.init(socket, mashup);

不确定这是否是在代码中进行某种分离的最佳方式。我习惯了 Java，而在 JS 中它通常是一个大源文件，这很糟糕。