问题

I am making some script for a game I'm making. The game's map is a randomly generated series of hallways, and there are different shaped hallways. For example, one hallway is straight, another turns, another turns and has a set of stairs that go down. If there is no way to do anything I mentioned above in the question, is there an alternative?

I haven't really tried anything yet for moving one node to another, because all the tutorials are outdated. For duplicating a random node, I tried something like this:

var parent_node = $Node3D
var random_index = randi() % parent_node.get_child_count()
var random_child = parent_node.get_child(random_index)
var duplicated_child = random_child.duplicate()

It gave me an error saying cannot call duplicate on a null instance.

var parent_node = $Node3D
var random_index = randi() % parent_node.get_child_count()
var random_child = parent_node.get_child(random_index)
var duplicated_child = random_child.duplicate()

答案1

得分: 0

I figured out your problem:
您已经解决了您的问题：

You are trying to get the node: $Node3D and call .get_child_count() on it before the scene has finished loading.
您在场景加载完成之前尝试获取节点：$Node3D 并对其调用 .get_child_count()。

To fix this, add @onready at the beginning of each of the lines, or put the in the _ready() function.
为了解决这个问题，在每行的开头添加 @onready，或者放在 _ready() 函数中。

Doing this would make sure the code only runs when the scene is loaded and everything is ready.
这样做可以确保代码仅在场景加载完毕并且一切准备就绪时运行。

This is the final code I would suggest:
这是我建议的最终代码：

@onready var parent_node = $Node3D

func _ready():
    var random_index = randi() % parent_node.get_child_count()
    var random_child = parent_node.get_child(random_index)
    var duplicated_child = random_child.duplicate()

You also said that you want to move the node to another parent node.
您还说您想将节点移动到另一个父节点。

I will assume you are trying to move the random node. So first lets add another variable for the new parent node:
我假设您想要移动随机节点。所以首先让我们为新的父节点添加另一个变量：

@onready var new_parent = $Node

Replace $Node with the node you want to be the new parent.
将 $Node 替换为您想要作为新父节点的节点。

Next we want to add the child to the new parent in the _ready() function:
接下来，我们希望在 _ready() 函数中将子节点添加到新的父节点中：

new_parent.add_child(duplicated_child)

You are now done unless you want to remove the node from the original parent, then you will add this after:
除非您想要从原始父节点中删除节点，否则您已经完成，然后您将在此之后添加以下内容：

parent_node.remove_child(duplicated_child)

This is now the final Code:
这现在是最终的代码：

@onready var parent_node = $Node3D
@onready var new_parent = $Node

func _ready():
    var random_index = randi() % parent_node.get_child_count()
    var random_child = parent_node.get_child(random_index)
    var duplicated_child = random_child.duplicate()
    new_parent.add_child(duplicated_child)
    parent_node.remove_child(duplicated_child)

@onready var parent_node = $Node3D

func _ready():
	var random_index = randi() % parent_node.get_child_count()
	var random_child = parent_node.get_child(random_index)
	var duplicated_child = random_child.duplicate()

@onready var new_parent = $Node

new_parent.add_child(duplicated_child)

parent_node.remove_child(duplicated_child)

@onready var parent_node = $Node3D
@onready var new_parent = $Node

func _ready():
	var random_index = randi() % parent_node.get_child_count()
	var random_child = parent_node.get_child(random_index)
	var duplicated_child = random_child.duplicate()
    new_parent.add_child(duplicated_child)
    parent_node.remove_child(duplicated_child)