You can fix this by making stack1 a function

let stack1() = ...

and then calling it at its use site

...Content=stack1()

This way, each new window created by the button gets its own new controls, rather than trying to reuse the ones from the prior window.

By on 3/14/2009 7:54 PM ()

Thank-you!!!

By on 3/15/2009 4:27 AM ()
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