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Hello,

in my application I use PresentDialog()/DismissDialog()/SwitchToDialog() and effects.

Additional I implemented a container for my GUI content. Now I have noticed that only by the fade-in effect the effect is visible. If I use DismissDialog() or SwitchToDialog() the screen is changing every time, but only the first change has a fade out effect. My goal would be to change the screens with a simultaneously fade in /fade out effect.

The method is currently implemented in the Application::ContentContainer class like this:

// The value doesn't change - nothing to do.
if ( pure Content == value )
      return;

if ( pure Content != null )
  Remove( pure Content );

// Remember the property's new value.
pure Content = value;



 if(OutletSystem^ == Application::SystemState.Standby)
 {
     this.SwitchToDialog( value, DesignComp::MyFadeInOut, DesignComp::MyFadeInOut, null, null, null, null, null, null, null, true );
 }
 else
 {
     this.SwitchToDialog( value, DesignComp::MyFadeInOut, DesignComp::MyFadeInOut, null, null, null, null, null, null, null, true );

 }

Do you have any idea why only the first fade-out effect is working?

1 Answer

0 votes
by

Hi Maxi,

is there a reason for the value true passed in the last argument of SwitchToDialog()? Usually you need it only if you want the switch operation to be perfomed simultanously with other operations (e.g. a separate DismissDialog()). SwitchToDialog() should itself coordinate the effects of the both affected dialogs (the new one and the old one). Try to use false here.

If the problem still persists, can you send us (support@embedded-wizard.de) your project for debugging purpose? In such case please add some description how to provoke the error.

Best regards

Paul Banach

by
Hi Paul,

i used true because i thought if the fade-in and fade-out process should start at the same time, it is the right one. Unfortunately if I change it to false it won't work either.

I will send you a project with the described behavior.

Thanks and best regards

Maxi

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