hi All

Basic stuff but i just noticed my script does not open the popup.

There is some dataloogging and when finished the last line in the script is mean to call the popup to display 

HMIRuntime.UI.SysFct.OpenScreenInPopup("Popup Name","Screen name",true,"caption",0,0,false);

Instead of popup i get an error instead:

Object member: UI not supported in the current scope.

I get the gist that i cannot call UI functions from scripts but how can i call it in correct way?

Please help

 HIi

I am not calling it from the button - the script triggers from the scheduler on the change of other PLC tag.

Regards

rdoliveira

Hello,

Are you calling the script through a button event, from a screen?

Where does the script call that contains the popup call come from?

good luck.

 Any idea how to open it then?

My idea is now to write at the end of my script a value 1 to the internal tag which controls the visibility of the popup. But how to i embed the popup to the screen? Simple copy and paste of the popup does not work. Is there any wrapper structure i should use and then place the popup inside?

Regards

rdoliveira

ok it is not allowed by scheduler.