Here's one way to accomplish the first part of your question. As to the
second part, I'll leave that to someone else because it's late :)
private FavoritesFrm _FavoritesFrm;
....
private void button1_Click(o bject sender, System.EventArg s e)
{
if (_FavoritesForm == null)
{
_FavoritesForm = new FavoritesFrm();
_FavoritesForm. Closed += new EventHandler(_F avoritesForm_Cl osed);
}
_FavoritesForm. WindowState = FormWindowState .Normal; // bring it from
the taskbar if necessary
_FavoritesForm. Show();
}
private void _FavoritesForm_ Closed(object sender, EventArgs e)
{
_FavoritesForm = null;
}
HTH,
ShaneB
"VMI" <vo******@yahoo .com> wrote in message
news:%2******** ********@tk2msf tngp13.phx.gbl. ..
Through my MDI application, I open a FormX that is opened outside of the
application (not an MDI child). How can I check if the Form's open so I
don't load two instances of the same window? Is it possible that, within
the MDI container, a Form can have TopMost equal to true? For some reason,
TopMost property only works when the window's opened outside the MDI
container.
Thanks.