Check if AngularJS module is bootstrapped

It's not good practice to access scope from outside the app, so it's not enabled in well-built production applications. If you need to access/apply scope then there's something strange/unsupported about your use case.

However, the right way to check whether an element has been bootstrapped is the way the Angular library does it which is to load up the element and check for an injector. So you'd want angular.element(document.querySelector('#mainDiv')).injector(); which makes your code:

function onEndRequest(sender, args) {    var element = angular.element(document.querySelector('#mainDiv'));    //This will be truthy if initialized and falsey otherwise.    var isInitialized = element.injector();    if (!isInitialized) {        angular.bootstrap(element, ['defaultApp']);    }    // Can't get at scope, and you shouldn't be doing so anyway}

Can you tell us why you need to apply the scope?

You could simply check for the scope of mainDiv, if angular.element(document.querySelector('#mainDiv')).scope() is not undefined then that means angular has been not initialized yet.

You code will be like below.

CODE

function onEndRequest(sender, args) {    //below flag will be undefined if app has not bootsrap by angular.    var doesAppInitialized = angular.element(document.querySelector('#mainDiv')).scope();    if (angular.isUndefined(doesAppInitialized)) //if it is not         angular.bootstrap($('#mainDiv'), ['defaultApp']);    var rootscope = angular.element('#mainDiv').scope();    if (rootscope) {        rootscope.$apply(); //I don't know why you are applying a scope.this may cause an issue    }}

Update

After angular 1.3+ release in later Aug 2015, there it added performance related improvement by disabling debugging information by disabling debug info. So normally we should enable debuginfo option to false to have good performance improvement on Production environment. I don't wanted to write too much about it as its already covered by @AdamMcCormick answer, which is really cool.