I have a custom widget and it needs to launch a MessageDialog and in order for me to put that message dialog on top of the window my widget is in then I need access to the parent gtk.window. Is there a way to get the parent GTK window? Thanks
For GTK4, the gtk_widget_get_toplevel()
method on GtkWidget
has been deprecated. Instead, you can use the gtk_widget_get_root()
method or the Widget:root
property, which returns a GtkRoot
. That GtkRoot
can then be cast to a GtkApplicationWindow()
or GtkWindow()
.
Here's an example in C
GtkRoot *root = gtk_widget_get_root (GTK_WIDGET (widget));
GtkWindow *window = GTK_WINDOW (root);
Here's an example in Rust
let window: Window = widget
.root()
.unwrap()
.downcast::<Window>()
.unwrap();
The GTK docs suggest:
GtkWidget *toplevel = gtk_widget_get_toplevel (widget);
if (gtk_widget_is_toplevel (toplevel))
{
/* Perform action on toplevel. */
}
get_toplevel will return the topmost widget you're inside, whether or not it's a window, thus the is_toplevel check. Yeah something is mis-named since the code above does a "get_toplevel()" then an immediate "is_toplevel()" (most likely, get_toplevel() should be called something else).
In pygtk, you can get the toplevel like toplevel = mywidget.get_toplevel()
then feed toplevel
directly to gtk.MessageDialog()
Though gtk_widget_get_toplevel should work, you may also give a try to the code below. It should get the parent gtk window for the given widget and print it's title.
GdkWindow *gtk_window = gtk_widget_get_parent_window(widget);
GtkWindow *parent = NULL;
gdk_window_get_user_data(gtk_window, (gpointer *)&parent);
g_print("%s\n", gtk_window_get_title(parent));
hope this helps, regards
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