Question

到目前为止我调试了它，列表中有值..但是它没有将它们写到文件中。

我不确定为什么它是空的。

GameObjects类包含所有字段。

GameObjectData仅用于列表。

然后ChestPlate继承自GameObjects。这样做的原因是，我正在制作游戏，GameObjects中的所有值都与ChestPlate相关。

代码如下：

[Serializable]
public class GameObjects
{
    //Defines name within XML file:
    [XmlElement("Item_ID")]
    public int Item_ID { get; set; }
    [XmlElement("Item_Name")]
    public string Item_Name = "bob";
    [XmlElement("Item_type")]
    public string Item_type = "GameObject";
    [XmlElement("Item_Level")]
    public int Item_Level = 5;
    [XmlElement("Item_description")]
    public string Item_description = "best description evar";

    public GameObjects(int id, string name, string type, int level, string description)
    {
        this.Item_ID = id;
        this.Item_Name = name;
        this.Item_type = type;
        this.Item_Level = level;
        this.Item_description = description;
    }
}

[Serializable]
[XmlInclude(typeof(GameObjects))]
public class GameObjectData
{
    [XmlArrayItem(typeof(GameObjects))]
    public List<GameObjects> GameList { get; set;}
}

public class ChestPlate : GameObjects
{
    [XmlElement("Armour_Rating")]
    int Armour_Rating = 5;

    public ChestPlate(int Armour_id, string Armour_name, string Armour_type, int Armour_level, string Armour_description)
        : base(Armour_id, Armour_name, Armour_type, Armour_level, Armour_description)
    {
        this.Item_ID = Armour_id;
        this.Item_Name = Armour_name;
        this.Item_type = Armour_type;
        this.Item_Level = Armour_level;
        this.Item_description = Armour_description;       
    }

    public  void SerializeToXML(List<GameObjects> responsedata)
    {        
        GameObjectData f = new GameObjectData();
        f.GameList = new List<GameObjects>();
        f.GameList.Add(new GameObjects { Item_ID = 1234, Item_Name = "OMG", Item_type =  "CHESTPLATE", Item_Level = 5, Item_description = "omg" });

        XmlSerializer serializer = new XmlSerializer(typeof(GameObjectData));
        TextWriter textWriter = new StreamWriter(@"C:\Test.xml");

        serializer.Serialize(textWriter, f);

        Console.WriteLine(f);           
    }

class Program
{
    static void Main(string[] args)
    {
        GameObjects a = new GameObjects();
        ChestPlate b = new ChestPlate();
        List<GameObjects> d = new List<GameObjects>();
        b.SerializeToXML(d);
        Console.ReadLine();
     }
 }

Answer 1

我已经运行了你的代码，似乎正确地序列化你的数据，测试文件是用基类设置的默认参数创建的。

让我担心的是实现我确信你还有更多的工作要做，但是你需要正确处理对象，确保继承类的属性和序列化以及父类是否正确完成否则你将无法完成正确地反序列化等。我建议你在网上寻找.net中xml序列化的最佳实践。网上有无数的例子，但请看下面列出的几个。

要检查的其他方法可能是好的，如果你真的需要使用XML，json就足够了，甚至是二进制序列化。

stackoverflow上有几个社区wiki条目，包括有关最佳实践的问题。

请在下面输入的代码输出下面的完整源代码。请注意，我将保存xml文件的位置更改为“。\ test.xml”，通常是bin文件夹和调试文件夹，应用程序不会有任何写入权限等问题。

另请注意，在下面的xml中，您已经丢失了序列化ChestPlate对象的事实。

<?xml version="1.0" encoding="utf-8"?>
<GameObjectData xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <GameList>
    <GameObjects>
      <Item_Name>OMG</Item_Name>
      <Item_type>CHESTPLATE</Item_type>
      <Item_Level>5</Item_Level>
      <Item_description>omg</Item_description>
      <Item_ID>1234</Item_ID>
    </GameObjects>
  </GameList>
</GameObjectData>

用于生成XML的代码库

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;

namespace ConsoleApplication12
{
    class Program
    {
        static void Main(string[] args)
        {
            GameObjects a = new GameObjects();
            ChestPlate b = new ChestPlate();
            List<GameObjects> d = new List<GameObjects>();
            b.SerializeToXML(d);
            Console.ReadLine();
        }

    }

    [Serializable]
    public class GameObjects
    {
        //Defines name within XML file:
        [XmlElement("Item_ID")]
        public int Item_ID { get; set; }
        [XmlElement("Item_Name")]
        public string Item_Name = "bob";
        [XmlElement("Item_type")]
        public string Item_type = "GameObject";
        [XmlElement("Item_Level")]
        public int Item_Level = 5;
        [XmlElement("Item_description")]
        public string Item_description = "best description evar";

        public GameObjects(int id, string name, string type, int level, string description)
        {
            this.Item_ID = id;
            this.Item_Name = name;
            this.Item_type = type;
            this.Item_Level = level;
            this.Item_description = description;
        }

        public GameObjects()
        {

        }

    }

    [Serializable]
    [XmlInclude(typeof(GameObjects))]
    public class GameObjectData
    {
        [XmlArrayItem(typeof(GameObjects))]
        public List<GameObjects> GameList { get; set; }


    }

    public class ChestPlate : GameObjects
    {
        [XmlElement("Armour_Rating")]
        int Armour_Rating = 5;

        public ChestPlate(int Armour_id, string Armour_name, string Armour_type, int Armour_level, string Armour_description)
            : base(Armour_id, Armour_name, Armour_type, Armour_level, Armour_description)
        {
            this.Item_ID = Armour_id;
            this.Item_Name = Armour_name;
            this.Item_type = Armour_type;
            this.Item_Level = Armour_level;
            this.Item_description = Armour_description;
        }

        public ChestPlate()
        {


        }


        public void SerializeToXML(List<GameObjects> responsedata)
        {
            GameObjectData f = new GameObjectData();
            f.GameList = new List<GameObjects>();
            f.GameList.Add(new GameObjects { Item_ID = 1234, Item_Name = "OMG", Item_type = "CHESTPLATE", Item_Level = 5, Item_description = "omg" });


            XmlSerializer serializer = new XmlSerializer(typeof(GameObjectData));
            TextWriter textWriter = new StreamWriter(@".\Test.xml");

            serializer.Serialize(textWriter, f);

            Console.WriteLine(f);
        }
    }
}

XML：写入文件是空的？

1 个答案: