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Attention | Topic was automatically imported from the old Question2Answer platform. |
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Asked By | alexzheng |
When the VisibilityNotifier2D is removed from parent(not by queue_free), it also emits _screen_exited, How can I determine whether it is emitted by move out of screen or removed from the scene?
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Reply From: | volzhs |
I think you can determine that with is_inside_tree() method.
func _on_screen_exited():
if is_inside_tree():
# it's out of screen
else:
# it's removed from tree
Have you tried this in 3.1?
After call get_parent().remove_child(self).
the _on_screen_exited will be called and the is_inside_tree() still return true.
alexzheng | 2019-05-04 06:54
i was just guessing.
there could be workaround but it looks not clean.
func on_screen_exited():
yield(get_tree(), "idle_frame")
if is_inside_tree():
print("node is out of screen")
else:
print("it's removed from tree")
tested it and worked as expected but hm…
volzhs | 2019-05-04 14:07
yes,I knew this workaround.
I used another method call_deferred also works.
As you said, it looks not clean, I’m looking for a straightforward way.
alexzheng | 2019-05-05 03:13