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Hiding and showing certain popup when navigated to different views

AY2021S1
6-Contributor

Hiding and showing certain popup when navigated to different views

Hi All,

I would ask if it is possible to hide or show popup when navigating from a different view?

 

Example :

In view 1, popup 1 is shown. However, when navigating from view 2 to view 1 by a button, popup 2 will show and popup 1 will hide.

 

Thank you

1 REPLY 1

Hi @AY2021S1 ,

 

the popup is a widget of the View-> and is not generic dialog finction. So means that you can show the popup widget only when the view is displayed where the popup belogns to.

What we can do:

-create a duplicate copy of the popup for each view where it should be shown. You can use some automatic function when you navigate:

 

 

$scope.goToViewPopup(viewName, popupName){
twx.app.fn.navigate(viewName);
$scope.$applyAsync();//not required
$timeout($scope.$root.$broadcast('app.view["'+viewName+'"].wdg["'+popupName+'"].svc.showpopup'),500);//delay to be ensure that view is loaded value to be optimized after tests
}

 

 

in case that you want to use the view menu to change /select the view. Then you can use Event callback:

 

 

$scope.$on('$ionicView.afterEnter', function () { //when 2d view is intilized
console.log("ionicView.afterEnter event called !")
$scope.$applyAsync();
$timeout($scope.$root.$broadcast('app.view["TestView"].wdg["popup-1"].svc.showpopup'),500);
});

 

 

The example above is view specific. So means the code e.g. for the view "TestView" and it should be pasted in TestView.js -> this means therefore the view name could be hardcoded!

The name of the popup could be also hardcoded but also could be red from parameter name e.g. POPUPNAME:

 

 

$timeout($scope.$root.$broadcast('app.view["TestView"].wdg['+$scope.app.params['POPUPNAME']+'].svc.showpopup'),500);

 

  

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