How do I always serve the same file with express? - node.js

Question

Is there any way I can always serve the same file?
So, if they go to website.com/ajsdflkasjd it still serves the same file as website.com/asdnw
I would like to do this using express with node.
The file I have is a static html file, not a jade file.
By the way, the reason I'm wanting to do this, in case you were wondering, is I have an angularjs app that handles all the routing for me. So, all I need to do is serve that one page, and it will take care of the rest.
Thanks in advance!

Answer 1

new answer
const app= require('express')()
// static file serve
app.use(express.static(__dirname))
// not found in static files, so default to index.html
app.use((req, res) => res.sendFile(`${__dirname}/index.html`))
app.listen(3000)
old answer
var express = require('express');
var bodyParser = require('body-parser')
var path = require('path')
var app = express();
// url encoding
app.use(bodyParser.urlencoded({extended:false}));
// gzip
// redirect all html requests to `index.html`
app.use(function (req, res, next) {
if (path.extname(req.path).length > 0) {
// normal static file request
next();
}
else {
// should force return `index.html` for angular.js
req.url = '/index.html';
next();
}
});
// static file serve
app.use(express.static(__dirname))
app.listen(3000)

Answer 2

Below is what I'm using express with angularjs in my project. It will always send index.html unless the browser requests resource files (images, css, js, etc.) which contains extname.
var express = require('express');
var app = express();
app.configure(function () {
// url encoding
app.use(express.urlencoded());
// gzip
app.use(express.compress());
// redirect all html requests to `index.html`
app.use(function (req, res, next) {
if (path.extname(req.path).length > 0) {
// normal static file request
next();
}
else {
// should force return `index.html` for angular.js
req.url = '/index.html';
next();
}
});
// static file serve
app.use(express.static(__dirname));
});

Answer 3

Basic configuration for Express 4 is:
var express = require('express');
express()
.get(/.*/, function(req, res) {
res.sendFile('index.html', {
root: __dirname
});
})
.listen(8080);
Working example
Those snippets with GZip, BodyParser etc. are pretty cool, but I think over-complicated if you want to just test your single page app. Of course you can add all this "production stuff" when it starts to be needed.
Read more:
sending files
routing

Answer 4

Here a simple implementation with ExpressJs to create a virtual host and whenever return the index.html
var express = require('express');
var app = express();
var vhost = require('vhost');
// Function to create virtualhost
function createVhost(domain,rootDirectory){
var exp = express();
exp.use(express.static(rootDirectory));
exp.get(/.*/,function(req,res){
res.sendFile('index.html',{root:rootDirectory});
})
app.use(vhost(domain,exp));
}
// Virtual Host to create
createVhost('example.com','/home/[user]/[www]/[example.com]');
createVhost('othersite.com','/home/[user]/[www]/[othersite.com]');
// Start Server
app.listen(80,function(){
console.log('Node server on port 80');
});
Remember:
Add the domains in the "/etc/host" (in linux)
127.0.0.1 example.com
127.0.0.1 othersite.com
And run in the terminal the "app.js" with "sudo" for port 80
~/home/[server]$ sudo node app.js

Answer 5

You can do this in both angular as well as node side.
In Node side you can do something like this:
res.sendfile('<ur html file path');
In Angular if you using ui-router you can make use of
$urlRouterProvider.otherwise('/otherwise');
and this otherwise state needs to be defined as well
$stateProvider
.state("otherwise", { url : '/urPage'...})
If you using ngRoute, you can do
$routeProvider.otherwise({redirectTo: '/urPage'});
UPDATE
Since your routers are not configured to show a default urPage, in the server you can have something as:
var app = express.createServer();
app.get('/urPage',function(req,res){
res.sendfile('<ur html page>');
});