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Angularjs [$rootScope:inprog] inprogress error

I am getting angularjs [$rootScope:inprog] error.

Error: [$rootScope:inprog] http://errors.angularjs.org/1.2.7/$rootScope/inprog?p0=%24digest.

this is the function calling

 Members.get({}, function (response) { // success
   $scope.family_mem = response.data;    
  }, function (error) { // ajax loading error

    Data.errorMsg(); // display error notification
  });

in console i am getting results by php controller function.but not updating $scope.family_mem instead going to error part. this is the directive

myApp.directive('mySelect', function() {
  return{
    restrict: 'A',
    link: function(scope, element){
      $(element).select2();
    }
  };
});
like image 261
Nisham Mahsin Avatar asked Mar 29 '14 16:03

Nisham Mahsin


2 Answers

Usually this means that you defined $rootScope.$apply somewhere manually inside of the another angular code which has lifecycle already. This should not be happen in common cases as angular tracks the lifecycle itself. The one common case where it is needed is when you need to update scope from non-angular code (like jquery or old-fashioned js stuff). So please check if you have this somewhere. In case you really need it's better to use safe apply (the common code snippet):

angular.module('main', []).service('scopeService', function() {
     return {
         safeApply: function ($scope, fn) {
             var phase = $scope.$root.$$phase;
             if (phase == '$apply' || phase == '$digest') {
                 if (fn && typeof fn === 'function') {
                     fn();
                 }
             } else {
                 $scope.$apply(fn);
             }
         },
     };
});

Then you can inject this service and make necessary calling by:

scopeService.safeApply($rootScope, function() {
    // you code here to apply the changes to the scope
});
like image 195
Alexander Kalinovski Avatar answered Oct 11 '22 05:10

Alexander Kalinovski


[$rootScope:inprog] inprogress error in my case:
Case 1: It means you're executing two actions at a time.

Example:

goView();
hidePopOver(); //Will trigger error

Use $timeout to make sure two actions (function) are not running at the same time.

goView();
$timeout(function () {
    hidePopOver();
}, 300);

Case 2: an action has not executed completely.
Use $timeout to make sure first action has executed.

$timeout(function () {
    $(element.target).trigger('click');
}, 300);
like image 7
Rain Avatar answered Oct 11 '22 03:10

Rain