What event should I handle to execute code when WinForms application switches from Run mode to Design mode.

Question

I am running a Windows form application and I need to execute a piece of code when I switch to design mode. I have a handler for the OnEnterDesignMode debugger event and this gets hit if I am debugging the application and then switch to design mode. However, this does not get hit if I initially start without debugging and then switch to design mode. What event do I need to handle in order that certain code is executed when switching from Run mode to Design mode?

Answer 1

Are you referring to release mode and debug mode?

If so you can wrap your code as follows:

#if DEBUG
            //Execute debug mode code
#else
            //execute release mode code
#endif

Answer 2

Not sure about the event, but there is a DesignMode bool property on a windows form, probably inherited from Control, that returns true if the form is open in the designer. Beware though, as DesignMode returns false in a constructor even when in design mode. So always use it in something like the load event, and not in the constructor.

Answer 3

Try Component.DesignMode in your control. See MSDN

VB.net Example

 Private Sub txtSmartDate_TextChanged(ByVal sender As Object, ByVal e As System.EventArgs) Handles txtSmartDate.TextChanged
    If Not Me.DesignMode Then
        If _ValueInitialised And Not _SuppressEventCode Then
            '   Apply the changes to the property value, now the text box has been updated.
            Call SetPropertyValues()
            Call ApplyDateFormating()
        End If
    End If
End Sub