Why doesn't this code show the collectible counter UI, then hide it after the timer ends?

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1
:information_source: Attention Topic was automatically imported from the old Question2Answer platform.
:bust_in_silhouette: Asked By DJSlimeball

Sorry if this is REALLY stupid, I’m not good.

extends Area2D



var timer = Timer.new()

func _on_[COIN]_body_entered(body: Node) -> void:
	queue_free()
	get_parent().get_parent().get_node("Node2D/[COIN]Counter/UI").show()

if get_parent().get_parent().get_node("Node2D/[COIN]Counter/UI").show():
	timer.connect("timeout",self,"do_this")
	timer.wait_time = 3
	timer.one_shot = true
	add_child(timer)
	timer.start()

func do_this():
	get_parent().get_parent().get_node("Node2D/[COIN]Counter/UI").hide()

[COIN]

is just a placeholder to make this more readable, if that’s possible.

2
:bust_in_silhouette: Reply From: exuin

Queue free deletes the node and its children, including the timer, so it never times out.

How would I go about rearranging this or changing it to make it only delete the collectible? Thanks.

P.S. Sorry for the late response.

DJSlimeball | 2022-12-04 09:22

hide the coin instead of deleting it, and then free it when the timer times out

exuin | 2022-12-04 16:44

The other comment has been hidden, cuz’ I actually figured THE WHOLE THING out:

    extends Area2D



var timer = Timer.new()

func _on_GeluLumenLapis_body_entered(body: Node) -> void:
	hide()
	$CollisionShape2D.set_deferred("disabled", true)
	get_parent().get_parent().get_node("Node2D/LapisCounter/UI").show()
	timer.connect("timeout",self,"do_this")
	timer.wait_time = 3
	timer.one_shot = true
	add_child(timer)
	timer.start()

func do_this():
	get_parent().get_parent().get_node("Node2D/LapisCounter/UI").hide()
	queue_free()

Thanks for the help, though.

EDIT: Better code that doesn’t break when touching a second collectible.

DJSlimeball | 2022-12-06 02:20