Angularjs resolve 404 views

Question

In angularjs is it possible to handle & resolve 404 Views?

I want to achieve conditional view rendering depending on my client, I have a view something like this:

<div ng-include="'assets/stackoverflow/breadcrumb.html'"></div>

so if suppose the above returns 404, it should try this:

<div ng-include="'assets/default/breadcrumb.html'"></div>

This approach may be wrong, but I just need a way to handle views 404, I have an interceptor which actually catches 404 requests but how do I resolve it and send it again?

app.factory('RequestResponseInterceptor', ['$q', '$rootScope', '$location',
    function($q, $rootScope, $location) {

        var loadingCount = 0;
        return {
            responseError: function(response) {

                if (response.status === 404) {
                    //Change URL and request again
                }
                return $q.reject(response);
            }
        };
    }
]);

http://stackoverflow.com/questions/20836374/how-to-catch-angular-ng-include-error — Kalhan.Toress, Feb 16 '15 at 09:47
The above link shows how I can catch error(404) but if it occurs how can I resend with new url? — Niraj Chauhan, Feb 16 '15 at 09:58

score -1 · Answer 1 · answered Feb 16 '15 at 12:55

You should use routing.

Configure your app like

.config([ '$routeProvider', function($routeProvider) {
$routeProvider.when('/', {
    templateUrl : 'assets/stackoverflow/breadcrumb.html',
    controller : 'ctrl1'
}).when('/s404', {
    templateUrl : 'assets/default/breadcrumb.html',
    controller : 'ctrl2'
}).otherwise({
    redirectTo : '/'
});

Replace ng-include with ng-view:

<ng-view></ng-view>

Set the location path in interceptor (and remember to inject $location) :

if (response.status === 404) {
   $location.path( "/s404" );
}

Angularjs resolve 404 views

1 Answers1