JAVASCRIPT

How to return $http.post() object with factory function

return $http.put(url + obj.id + '/', obj).
            then(function onSuccess(response) {
                loaderHide();
                angular.extend(obj, response);
                errors = {};
            }, function onError(response) {
                loaderHide();
                handleErrors(response, status, errors, not_show_error_toast);
            });

How to return $http.post() object with factory function

save: function(url, obj, errors, not_show_error_toast) {            
    var defer = $.Deferred();

    not_show_error_toast = typeof not_show_error_toast !== 'undefined' ? not_show_error_toast : false;

    if (angular.isDefined(obj.id)) {
        loaderShow(); // NOTE: let create manage the loader start if it needs to
        $http.put(url + obj.id + '/', obj).
            then(function onSuccess(response) {
                loaderHide();
                // HACK: because .then is used instead of .success, you probably want the response.data, please review this
                angular.extend(obj, response.data);

                defer.resolve(obj);
            }, function onError(response) {
                loaderHide();
                handleErrors(response, status, errors, not_show_error_toast);

                defer.reject(response);
            });
    } else {
        // If create is a deferral, it should be safe to return it directly
        return this.create(url, obj, errors, not_show_error_toast);
    }

    return defer.promise();
},