single page application deep linking with login page

Question

My team is going to build a single-page-application for our future project. At the moment, I have a problem with designing the app with login page. There are 2 approaches:

Answer 1

I decided to go with approach 2: The app has only 1 page and the login page will be a view in the app which is switched back and forth using javascript.. I found out that it's not difficult to do and I can still use normal cookie-based authentication with session on server, redirect users to default page (main page) after successful login, and so on. Here is a sample code how I do it with angularjs.

Routing:

var App = angular.module('App', ["ui.state"]);

App.config(function ($stateProvider, $routeProvider) {
   $stateProvider.
   .state('login', {
        url: "/login?returnUrl",
        templateUrl: '/Home/Login',
        controller:"LoginController"
    })
    .state('main', {
        url: "/main",
        abstract:true,
        templateUrl: '/Home/Main',
        controller: "MainController"
     })
})
.run(function ($rootScope, $state, $stateParams, $location) {
     $rootScope.$on('$stateChangeError', function(event, toState, toParams, fromState, fromParams, error){ 
        if (error.status == 401) {
            $state.transitionTo("login", { returnUrl: $location.url() });
        }
     })
});

The point here is when there is a route change error with status 401 (from the server) indicating that the user is not logged in, I will transition to login state with the return url.

After the user successfully logging in using ajax, I will transition the state back to my main view. Something like this:

$http.post("/Login", JSON.stringify({UserName:username,Password:password}))
     .success(function (data, status, headers, config) {
        var returnUrl = $stateParams.returnUrl ? $stateParams.returnUrl : "mydefaulturl";
        $location.url(returnUrl);
 })

With this approach, now I'm able to create deep-link to jump to a specific state in my app with login page and return url.