AngularJS: How to execute a controller function AFTER completion of AJAX call in a service?

Question

Here's my code in the service.

this.loginUser = function(checkUser) {
    Parse.User.logIn(checkUser.username, checkUser.password, {
        success: function(user) {
            $rootScope.$apply(function (){
                $rootScope.currentUser = user;
            });
        }
    });
};

Here's my code in the controller:

$scope.logIn = function(){
    authenticationService.loginUser($scope.checkUser);
        console.log($scope.currentUser)
};

So, what I want to do is, execute some code AFTER the completion of AJAX call, whose success function sets the value of $scope.currentUser, which, I can use for some conditional logic (like redirecting etc) The success function is correctly setting the value, but the console.log should be executed AFTER the execution of authenticationService.loginUser() function.

Answer 1

You need to return a promise using $q and act on that.

For instance in your service:

this.loginUser = function(checkUser) {
    var deferred = $q.defer();
    Parse.User.logIn(checkUser.username, checkUser.password, {
        success: function(user) {
            $rootScope.$apply(function (){
                $rootScope.currentUser = user;
            });
            deferred.resolve();
        }
    });
    return deferred.promise;
};

Then in your controller act on the success:

$scope.logIn = function(){
    authenticationService.loginUser($scope.checkUser).then(function() {
        console.log($rootScope.currentUser));
    });
};

Answer 2

Try using $rootScope.$broadcast in your service then listen for it in your controller:

Service

Parse.User.logIn(checkUser.username, checkUser.password, {
    success: function(user) {
        $rootScope.$apply(function (){
            $rootScope.currentUser = user;
            $rootScope.$broadcast('user.online');
        });
    }
});

Controller

$scope.$on('user.online',function(){
    [ DO STUFF HERE ]
});

This isn't the best way to do this though @comradburk's use of $q is probably a better way.