Question

我不确定如何对存储在列表中的XML文件进行深层复制。我在这里阅读了一些帖子，有些人使用了Serialize＆amp;一些使用Icloneable。哪种方法最好？

目前我只阅读了XML文件并将其存储在列表中。

TextAsset XMLFile;
public GameObject spawnAlert;

List<AlertInfo> m_AlertInfo = new List<AlertInfo>();

public class AlertInfo
{
    public string strId     { get; set; }
    public string strName   { get; set; }
    public Vector3 vectSpawnPos = new Vector3();
}

void Start()
{
    TextAsset textXML = (TextAsset)Resources.Load("Alert", typeof(TextAsset));
    XmlDocument xml = new XmlDocument();
    xml.LoadXml(textXML.text);


    XmlNode root = xml.FirstChild;

    for( int x = 0; x < root.ChildNodes.Count; x++)
    {
        AlertInfo alertI = new AlertInfo();

        alertI.strId = root.ChildNodes[x].Attributes[0].Value;
        alertI.strName = root.ChildNodes[x].Attributes[1].Value;

        alertI.vectSpawnPos.x = float.Parse( root.ChildNodes[x].Attributes[2].Value);
        alertI.vectSpawnPos.y = float.Parse( root.ChildNodes[x].Attributes[3].Value);
        alertI.vectSpawnPos.z = float.Parse( root.ChildNodes[x].Attributes[4].Value);

        m_AlertInfo.Add( alertI );
    }

<ListOfAlerts>
    <AlertInfo id = "1"     Name = "Unsecured Aircraft"                                     pos_X = "0" pos_Y = "0" pos_Z = "0" />
    <AlertInfo id = "2"     Name = "Suspecious Baggage Found at Lobby"                          pos_X = "0" pos_Y = "0" pos_Z = "0" />
    <AlertInfo id = "3"     Name = "Disruptive person at Departure Hall"                                pos_X = "0" pos_Y = "0" pos_Z = "0" />
    <AlertInfo id = "4"     Name = "Suspicious person loitering around at Arrival Hall"             pos_X = "0"pos_Y = "0"  pos_Z = "0" />
</ListOfAlerts>

Answer 1

如何使用LINQ to XML？

XDocument xDoc = XDocument.Parse("textXML.text");

var myList = xDoc.Descendants("AlertInfo").Select(x => new AlertInfo()
        {
            strId = (string) x.Attribute("id"),
            strName = (string) x.Attribute("Name"),
            vectSpawnPos = new Vector3
            {
                x = (float) x.Attribute("pos_X"),
                y = (float) x.Attribute("pos_Y"),
                z = (float) x.Attribute("pos_Z")
            }
        }).ToList();

要使用此代码，您需要进行一些更改。使用vectSpawnPos自动实现的属性，例如：

public Vector3 vectSpawnPos { get; set; }

您的Vector3课程应如下所示：

public class Vector3
{
    public float x { get; set; }

    public float y { get; set; }

    public float z { get; set; }
}

列表中XML文件的深层复制

1 个答案: