CODEWITHSUNDEEP
angularjs
rkmax
rkmax

Reputation: 18125

Angular call a service on run

I have a service for handle the menu of my application I dont want call from any controller, where is the best place for call my service

my service has a register method

// sample

menuService.register({name: "Person", label: "Person", url: "/persons"});
menuService.register({name: "Company", label: "Companies", url: "/companies"});

is defined like

app.service('MenuService', ['$rootScope', function($r) { /*...*/ }

Note: my service $rootScope.$emit and is listen by a directive and depends of $rootScope and $location

Upvotes: 0

Views: 706

Answers (2)

Narek Mamikonyan
Narek Mamikonyan

Reputation: 4611

you should use .run block for that, but keep in mind you cannot inject .provider to run block 

  yourApp.run(function ($rootScope, $location) {
                // your code goes herer
        });

from docs

Run blocks are the closest thing in Angular to the main method. A run block is the code which needs to run to kickstart the application. It is executed after all of the services have been configured and the injector has been created. Run blocks typically contain code which is hard to unit-test, and for this reason should be declared in isolated modules, so that they can be ignored in the unit-tests.

Upvotes: 1

bmleite
bmleite

Reputation: 26870

You could expose the menus as a model in the MenuService.

app.service('MenuService', ['$rootScope', function($r) { 
  var svcModel = {
    menus: []
  };

  var registerMenu = function(menu) {
    svcModel.menus.push(menu);
  };

  /*...*/

  return {
    model: svcModel,
    register: registerMenu
    /*...*/
  };
}

And then access that model directly on the directive:

app.directive('menuDirective', ['MenuService', function(MenuService) {
  return {
    /*...*/
    link: function(scope, element, attrs) {
      /*...*/
      scope.menus = MenuService.model.menus;
    }
}]);

Upvotes: 0

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