void OnDocumentChanged(
object sender,
DocumentChangedEventArgs e)
{
ICollection <ElementId> idsAdded
= e.GetAddedElementIds();
int n = idsAdded.Count;
Debug.Print("{0} id{1} added.",
n, Util.PluralSuffix(n));
// This does not work, because the handler will
// be called each time a new instance is added,
// overwriting the previous ones recorded:
//_added_element_ids = e.GetAddedElementIds();
_added_element_ids.AddRange(idsAdded);
if (_place_one_single_instance_then_abort &&
0 < n)
{
// Why do we send the WM_KEYDOWN message twice?
// I tried sending it once only, and that does
// not work. Maybe the proper thing to do would
// be something like the Press.OneKey method...
//
//Press.OneKey( _revit_window.Handle,
// (char) Keys.Escape );
//
// Nope, that did not work.
//
// Answer: When you place instances with
// PromptForFamilyInstancePlacement, the previous
// one remains selected just until you drop the
// next one. The first esc key hit removes that
// selection while still allowing you to continue
// adding instances to the model. Only a second
// esc hit aborts the command.
//Press.PostMessage( _revit_window.Handle,
// (uint) Press.KEYBOARD_MSG.WM_KEYDOWN,
// (uint) Keys.Escape, 0 ); // 2018
//Press.PostMessage( _revit_window.Handle,
// (uint) Press.KEYBOARD_MSG.WM_KEYDOWN,
// (uint) Keys.Escape, 0 ); // 2018
Press.PostMessage(_revit_window,
(uint)Press.KEYBOARD_MSG.WM_KEYDOWN,
(uint)Keys.Escape, 0); // 2019
Press.PostMessage(_revit_window,
(uint)Press.KEYBOARD_MSG.WM_KEYDOWN,
(uint)Keys.Escape, 0); // 2019
}
}
