Angular JS: $location injection undefined

Question

I'm trying to configure my route in such a way to redirect to the login page if user is not logged. This is my code:

angular.module('witBiz.services', []);
angular.module('witBiz.controllers', ['witBiz.services']);
var witBizModule = angular.module('witBiz', ['ngRoute' , 'witBiz.controllers', 'witBiz.services'   ]);

witBizModule.config(['$routeProvider', function($routeProvider) {
    $routeProvider.when('/login', {templateUrl: 'resources/views/login.html', controller: 'LoginController'});
    $routeProvider.when('/index', {templateUrl: 'resources/views/index.html', controller: 'IndexController'});
    $routeProvider.otherwise({redirectTo: 'resources/views/index.html'});
}])
    .run(['$rootScope',  'checkLogin', function($rootScope, checkLogin ,$routeProvider ) {
        $rootScope.$on('$routeChangeSuccess', function () {
           if (checkLogin())
                $location.url("/index");
        })
    }])
    .factory('checkLogin', function(){
        return function() {
            //perform real logic here
            return true;
        }
    })

where basically I declare modules and services. Now the problem is $location is not defined so I get error. I tried to inject $location as dependency like this but I got undefined injection (injpt):

.run(['$rootScope',  'checkLogin', '$location', function($rootScope, checkLogin ,$location) {
        $rootScope.$on('$routeChangeSuccess', function () {
           if (checkLogin())
                $location.url("/ciao");
        })
    }])

So I'm wondering how I can use the builtin $location services inside my "run" method...why can I inject it as a dependency as $rootScope or checkLogin?!

I'm using angular 1.2.0 rc2.

Thanks

bekite · Accepted Answer · 2013-10-30 00:14:33Z

2

Here is a working example http://plnkr.co/edit/gf5LhTjqhz4MoZfVcqIt?p=preview

Couldn't reproduce the error with $location not working inside .run

answered Oct 30, 2013 at 0:14

bekite

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2 Comments

Simo Mafuxwana Over a year ago

I have been struggling with understanding app.run your example helped

demaniak Over a year ago

This feels kludgy, but it worked, so after two days of struggle - I will accept it as a working answer. Thanks!

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