Showing an URL in browser on onShow method

Classic Servoy
1

Hi,
I want to create an onShow method that does this: 5 seconds after showing the form it loads an URL and closes the application.
Is there a way I make so that
application.showURL(URL,‘_self’)
application.exit()
is reached only after a number of seconds only?
application.updateUI() does not work in this case…

2

I think the easiest way to do this is to use html. Place an htmlarea on your form with a global as dataprovider. And then in your onShow-method do:

globals.myGlobal = "<html><head><meta http-equiv='refresh' content='5;url=http://www.servoy.com/' /></head></html>";

The htmlarea can have size 0,0 so you don’t see it on the form.

3

Thank you Joas for your reply!
It’s a very good and useful thing to know.
But what i want is that the solution closes after www.servoy.com opens. Is that possible?

globals.myGlobal = "<html><head><meta http-equiv='refresh' content='5;url=http://www.servoy.com/' /></head></html>";
application.exit()

…does not work in this case, because it does not show the final form any more because of application.exit() on the onShow method…

DB

4

You can use the scheduler plugin to perform tasks later.
See sample code for plugins.scheduler.addJob(…).