I am working on mean application,i have a requirement to load my html,css from another url like("/myApp") rather then root, so how can get this? i mean i want to start my app form url "localhost:3003/myApp", bydefault it is running on "localhost:3003/"
My configuration file is
var express = require('express'),
routes = require('./routes'),
fs = require('fs'),
bodyParser = require('body-parser'),
router = express.Router();
var port = 80;
var app = module.exports = express();
// Configuration
//app.use(router);
app.use(bodyParser());
//app.use(express.methodOverride());
app.use(express.static(__dirname + '/dev'));
//app.use(express.compress());
app.all('*', function (req, res, next) {
res.header("Access-Control-Allow-Origin", "*");
res.header("Access-Control-Allow-Headers", "Content-Type, Authorization, Content-Length,X-Requested-With");
res.header('Access-Control-Allow-Methods', 'GET,PUT,POST,DELETE,OPTIONS');
next();
})
router.get('/', routes.index);
router.get('/template/:name', routes.partials);
// JSON API
router.get('/api/name', routes.name);
// redirect all others to the index (HTML5 history)
router.get('*', routes.index);
app.use('/', router);
app.listen(3003, function(){
console.log("Express server listening on port mode");
});
process.on('uncaughtException', function (err) {
console.log((new Date).toUTCString() + ' uncaughtException:', err.message)
console.log(err); //process.setMaxListeners(0);
// process.setMaxListeners(0);
process.exit(1)
})
