使用c#直接在类中读取XML文件

Question

我很擅长使用XML来定义C#中实际管理的对象,但必须有更好的方法.

我的游戏数据库的数据是由几个XML文件构成的,这是一个比较精细的文件,因为我用它来测试所有地方

<GameData xmlns:NSW="Napivo.StarWars">
<!-- Ships -->
<!-- AITroopAccelerator -->
<Asset Type="ShipDescription">
<Name>AITroopAccelerator</Name>
<Asset Type="SpriteAnimation">
  <Name>\Ships\AITroopAccelerator\1.png</Name>
  <Name>\Ships\AITroopAccelerator\2.png</Name>
  <Name>\Ships\AITroopAccelerator\3.png</Name>
  <Name>\Ships\AITroopAccelerator\2.png</Name>
  <Speed>150</Speed>
</Asset>

<Asset Type="WeaponGroup" Num ="0">
  <Weapon Name ="Missile" Angle ="0" X ="0" Y ="0"> </Weapon>

<Asset Type="WeaponGroup" Num ="1">
  <Weapon Name ="NegativeEnergyFlare" Angle ="180" X ="0" Y ="0"> </Weapon>
</Asset>
  <Weapon Name ="Lightning" Angle ="0" X ="0" Y ="0"> </Weapon>
</Asset>
<Asset Type="WeaponGroup" Num ="2">
  <Weapon Name ="FastLazer" Angle ="-10" X ="0" Y ="0"> </Weapon>
  <Weapon Name ="FastLazer" Angle ="+10" X ="0" Y ="0"> </Weapon>
  <Weapon Name ="Ion" Angle ="-90" X ="0" Y ="0"> </Weapon> 
</Asset>

<Asset Type="Armory">
  <Resource Type ="Energy" Ammout ="-1000" Recharge ="0.2"></Resource>
  <Resource Type ="Missile" Ammout ="-10" Recharge ="0"></Resource>
</Asset>

<HullStrength>100</HullStrength>
<ShieldStrength>100</ShieldStrength>


<ColorMask>\Ships\AITroopAccelerator\ColorMask.png</ColorMask>
<Scale>1</Scale>

这是我需要编写的代码来制作这个xml文件的正面或反面....

private void ReadShipDescription(XmlNode node)
    {
        ShipDescription sd = new ShipDescription();


        XmlNodeList NL = node.ChildNodes;

        foreach (XmlNode n in NL)
        {
            Console.WriteLine("{0} = {1}", n.Name, n.Value);

            if (n.Name == "Name")
            {
                sd.Name = n.InnerText;
            }

            if (n.Name == "Scale")
            {
                sd.Scale = float.Parse(n.InnerText);
            }

            if (n.Name == "HullStrength")
            {
                sd.HullStrength = float.Parse(n.InnerText);
            }

            if (n.Name == "ShieldStrength")
            {
                sd.ShieldStrength = float.Parse(n.InnerText);
            }

            if (n.Attributes != null)
            {
                foreach (XmlAttribute a in n.Attributes)
                {
                    if ((n.Name == "Asset") && (a.Name == "Type") && (a.Value == "SpriteAnimation"))
                    {
                        sd.HullAnimation = ReadSpriteAnimation(n);
                    }

                    if ((n.Name == "Asset") && (a.Name == "Type") && (a.Value == "WeaponGroup"))
                    {
                        sd.WeaponGroups.Add(ReadWeaponGroup(n));
                    }

                    if ((n.Name == "Asset") && (a.Name == "Type") && (a.Value == "Armory"))
                    {
                        XmlNode childnode = n.FirstChild;

                        while (childnode != null)
                        {
                            ResourceDescription rd = ReadResource(childnode);

                            if (rd.Ammount < 0)
                            {
                                rd.Ammount = float.PositiveInfinity;
                            }

                            sd.Armory.Add(rd);
                            childnode = childnode.NextSibling;
                        }
                    }

                    Console.WriteLine("{0} = {1}", a.Name, a.Value);
                }
            }

            if (n.Name == "ColorMask")
            {
                sd.ColorMask = n.InnerText;
            }
        }

        ShipDescriptions.Add(sd.Name, sd);
    }

这对我来说似乎不对.

任何人都可以发送我的地方,我可以学习如何直接从XML读取对象到类或结构？

希望我不会因为这个而得到太多的选票.

谢谢.

Answer 1

您可以使用xsd命令生成与您的XML匹配的C#类:

xsd gamedata.xml   <- creates gamedata.xsd
xsd GameData.xsd /classes

(有关详细信息,请参阅从XML生成C#类)

然后将XML直接反序列化为一组对象:http://msdn.microsoft.com/en-us/library/fa420a9y.aspx