AngularJS cannot seem to take a variable duration for a $timeout

Question

I am trying to implement a function within the controller which will be incharge of showing various notification to the user.

the problem is that I want the duration to be a function parameter, and that doesn't seem to work.

How come?.

How can I fix this?.

    $scope.layout.showNotification = function(msg, duration){
            $scope.layout.notification.message = msg;
            $scope.layout.notification.visible = true;

            if(!duration || duration === null)
                return

            $timeout(function(){
                $scope.layout.notification.visible = false;
                $scope.layout.notification.message = "";
            }, duration);
   }

Are you sure your duration is an integer ? Maybe you should just parseInt() it ? — LoremIpsum, Sep 17 '13 at 12:14
What do you mean by 'that doesn't seem to work' ? Do you have an error in the console ? Have you traced the duration value or added a breakpoint ? — LoremIpsum, Sep 17 '13 at 12:20

score 1 · Answer 1 · answered Sep 17 '13 at 12:20

1

Try This

$scope.notification = {
    message : '',
    visible: true
};
$scope.showNotification = function(msg, duration) {
        $scope.notification.message = msg;
        $scope.notification.visible = true;

        if(!duration || duration === null)
            return

        $timeout(function(){
            $scope.notification.visible = false;
            $scope.notification.message = "";
        }, duration);
};

$scope.showNotification('MSH',5000);

DEMO

answered Sep 17 '13 at 12:20

Nitish Kumar

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you can modify now according to your condition – Nitish Kumar Sep 17 '13 at 12:22
Can you please explain what is the difference between my implementation and yours, because I don't see any difference, thus I cannot use it to solve my problem. Thank for your time and effort – Oleg Belousov Sep 18 '13 at 09:56
then whats your problem ? – Nitish Kumar Sep 18 '13 at 10:14

AngularJS cannot seem to take a variable duration for a $timeout

1 Answers1