static files with express.js

If you have this setup

/app
   /public/index.html
   /media

Then this should get what you wanted

var express = require('express');
//var server = express.createServer();
// express.createServer()  is deprecated. 
var server = express(); // better instead
server.configure(function(){
  server.use('/media', express.static(__dirname + '/media'));
  server.use(express.static(__dirname + '/public'));
});

server.listen(3000);

The trick is leaving this line as last fallback

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

As for documentation, since Express uses connect middleware, I found it easier to just look at the connect source code directly.

For example this line shows that index.html is supported https://github.com/senchalabs/connect/blob/2.3.3/lib/middleware/static.js#L140

In the newest version of express the "createServer" is deprecated. This example works for me:

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

//app.use(express.static(__dirname)); // Current directory is root
app.use(express.static(path.join(__dirname, 'public'))); //  "public" off of current is root

app.listen(80);
console.log('Listening on port 80');

res.sendFile & express.static both will work for this

var express = require('express');
var app = express();
var path = require('path');
var public = path.join(__dirname, 'public');

// viewed at http://localhost:8080
app.get('/', function(req, res) {
    res.sendFile(path.join(public, 'index.html'));
});

app.use('/', express.static(public));

app.listen(8080);

Where public is the folder in which the client side code is

As suggested by @ATOzTOA and clarified by @Vozzie, path.join takes the paths to join as arguments, the + passes a single argument to path.

const path = require('path');

const express = require('express');

const app = new express();
app.use(express.static('/media'));

app.get('/', (req, res) => {
    res.sendFile(path.resolve(__dirname, 'media/page/', 'index.html'));
});

app.listen(4000, () => {
    console.log('App listening on port 4000')
})

If you have a complicated folder structure, such as

- application
     - assets
         - images
             - profile.jpg
     - web
     - server
        - index.js

If you want to serve assets/images from index.js

app.use('/images', express.static(path.join(__dirname, '..', 'assets', 'images')))

To view from your browser

http://localhost:4000/images/profile.jpg

If you need more clarification comment, I'll elaborate.

use below inside your app.js

app.use(express.static('folderName'));

(folderName is folder which has files) - remember these assets are accessed direct through server path (i.e. http://localhost:3000/abc.png (where as abc.png is inside folderName folder)

npm install serve-index

var express    = require('express')
var serveIndex = require('serve-index')
var path = require('path')
var serveStatic = require('serve-static')
var app = express()
var port = process.env.PORT || 3000;
/**for files */
app.use(serveStatic(path.join(__dirname, 'public')));
/**for directory */
app.use('/', express.static('public'), serveIndex('public', {'icons': true}))

// Listen
app.listen(port,  function () {
  console.log('listening on port:',+ port );
})

I would add something that is on the express docs, and it's sometimes misread in tutorials or others.

app.use(mountpoint, middleware) 

mountpoint is a virtual path, it is not in the filesystem (even if it actually exists). The mountpoint for the middleware is the app.js folder.

Now

app.use('/static', express.static('public')`

will send files with path /static/hell/meow/a.js to /public/hell/meow/a.js

This is the error in my case when I provide links to HTML files.

before:

<link rel="stylesheet" href="/public/style.css">

After:

<link rel="stylesheet" href="/style.css">

I just removed the static directory path from the link and the error is gone. This solves my error one thing more don't forget to put this line where you are creating the server.

var path = require('path');
app.use(serveStatic(path.join(__dirname, 'public')));

You can make your own function to do it like this:

---

const static = (root = "") => app.get(/./, (req, res) => res.sendFile(__dirname + root + req.url));

You can also do it with express like this:

---

app.use('/', express.static(public));