Nodejs getting all needed scripts and stylesheets route from a page request(without browsers)

Question

I'm wondering if it is possible for me to get all the scripts and stylesheets routes I need from a html file.( not using browsers at all. )

for example:

I run index.html , and the website need ./style.css and also need ./script.js and lot of images.

---

How could I know what index.html need the routes of style.css and script.js and images.

another question: Does the browser works like rendering HTML file while it need some files, the browser request to server and wait for response? So is there a way I could use nodejs to analog the action and not using browsers?

Answer 1

You have to specify the location of your static content in your app.js file from where server will serve the files to browser when index.html will render.

for example if i have following directory structure

ROOT
 |
  ---public
 |     |
 |      ----css
 |     |     |
 |     |      ----style.css
 |     |     |
 |     |
 |      ----js
 |     |     |
 |     |      ----script.js
 |     |     |
 |     |
 |      ----images
 |     |
 |
 |
  ---app.js
 |
  ---index.html
 |

 Browser requested urls format are 
 css - > <link href="/css/style.css" rel="stylesheet" type="text/css" />
 js  - > <script type="text/javascript" src="/js/script.js"></script>

 after that you have to add following code in your app.js file;

app.use(express.static(__dirname + '/public'));

 if express is not installed in your system install it form cmd using command 

npm install express

 add following line of code in app.js

var express = require('express') , app = express();

Answer 2

找出解决问题的好方法是使用 phantomjs( http://phantomjs.org/ )，网络监控可能是我正在寻找的东西。