Shulou(Shulou.com)06/03 Report--

Editor to share with you how to use the function to convert the scene tree into XML. I hope you will get something after reading this article. Let's discuss it together.

A few days ago, I encountered some GameObject that were not destroyed as expected in the runtime scenario tree.

You need to print out the scene tree structure for comparison, so write this simple function.

Public static void ExchangeHierarchy2XmlNode (Transform transParant, System.Xml.XmlNode node,System.Xml.XmlDocument doc)

{

Int indexCount = transParant.transform.childCount

String name = transParant.name

If (string.IsNullOrEmpty (name))

{

Name = "Untitiled"

}

Name=name.Replace ('(','_'); name=name.Replace (')','_')

Name = name.Replace ('','_')

Name = "_" + name

System.Xml.XmlElement curentElement = doc.CreateElement (name)

Node.AppendChild (curentElement)

Foreach (Component comp in transParant.gameObject.GetComponents (typeof (MonoBehaviour)

{

System.Xml.XmlAttribute newAttr = doc.CreateAttribute (comp.GetType () .Name)

NewAttr.Value = ""

CurentElement.Attributes.Append (newAttr)

}

For (int I = 0; I < indexCount; iTunes +)

{

Transform transChild = transParant.transform.GetChild (I)

ExchangeHierarchy2XmlNode (transChild, curentElement, doc)

}

}

After reading this article, I believe you have a certain understanding of "how to use the function to convert the scene tree into XML". If you want to know more about it, welcome to follow the industry information channel, thank you for reading!

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