关于Color 的Xml序列化

[laorer]

public Color LeftColor{
get;
set;
}

这样序列化后， LeftColor 在xml文件中为 null，
网上找了一下，说Color直接序列化是不行的，需要转换成 String 或者 int ，但我不清楚如果在序列化的过程中转换,
下面链接谈了一些，但没具体谈到什么， 是不是要实现XmlSerializable 接口，http://www.cnblogs.com/surfsky/archive/2009/02/19/673620.html

[icehawk]

int icolor = ColorTranslator.ToWin32(acolor).ToString();
Color color = ColorTranslator.FromWin32(icolor);

[北京的雾霾天]

1：如果类没有添加可序列化的属性则添加上：[Serializable]
2:如果类本身不能被正确的序列化，则继承ISerializable接口，自己来定义类的序列化，比如：

[Serializable]
public class ClassA : ISerializable
{
	private string str_;
	private object m_aa;
	private int num_;
	private DateTime overdueTime;

	public ClassA()
	{
	}
	protected ClassA(SerializationInfo info, StreamingContext context)
	{
		this.str_ = info.GetString("str");
		this.num_ = info.GetInt32("num_");
	}
	public ClassA(string myStr, int myNum)
	{
		str_ = myStr;
		num_ = myNum;
		this.overdueTime = DateTime.Now;
	}
	public string Str
	{
		get { return str_; }
		set
		{
			str_ = value;
		}
	}
	public int Num
	{
		get { return num_; }
		set
		{
			this.num_ = value;
		}
	}

	public DateTime OverdueTime
	{
		get { return overdueTime; }
		set { overdueTime = value; }
	}
	public object AA
	{
		get
			{
				return this.m_aa;
		}
		set {
			this.m_aa = value;
		}
	}
	public void GetObjectData(SerializationInfo info, StreamingContext context)
	{
		info.AddValue("str", str_);
		info.AddValue("num_", num_);
	}
}
public class Program1
{
	static void Main(string[] args)
	{
		ClassA ca = new ClassA("Hello ", 10);
		XmlSerializer xSerial = new XmlSerializer(typeof(ClassA));

		//StreamReader reader = new StreamReader(@"D:\temp\po.xml ");
		//object obj = xSerial.Deserialize(reader);

		//StreamWriter write = new StreamWriter(@"D:\temp\po.xml ");
		//xSerial.Serialize(write, ca);//为什么串行化不了值 

		//or use BinaryFormatter
		IFormatter formatter = new BinaryFormatter();
		Stream stream = new FileStream(@"D:\temp\po.xml", FileMode.Open, FileAccess.Read, FileShare.None);
		//formatter.Serialize(stream, ca);
		object obj = formatter.Deserialize(stream);
		stream.Close();
	}
}

[laorer]

谢谢，我试试看

[laorer]

不好意思，还是没清楚，如果我 把 private object m_aa;

改成 private Color m_color； 并将其放入序列化中，该如何做？

[北京的雾霾天]

呵呵，那就这样了：

[Serializable]
public class ClassA : ISerializable
{
	private string str_;
	private Color m_aa;
	private int num_;
	private DateTime overdueTime;

	public ClassA()
	{
	}
	protected ClassA(SerializationInfo info, StreamingContext context)
	{
		this.str_ = info.GetString("str");
		this.num_ = info.GetInt32("num_");
		this.m_aa.A = info.GetInt32("colorA");
		this.m_aa.R = info.GetInt32("colorR");
		this.m_aa.G = info.GetInt32("colorG");
		this.m_aa.B = info.GetInt32("colorB");

		//this.m_aa=Color.FromArgb(info.GetInt32("color"); 或者
	}
	public ClassA(string myStr, int myNum)
	{
		str_ = myStr;
		num_ = myNum;
		this.overdueTime = DateTime.Now;
	}
	public string Str
	{
		get { return str_; }
		set
		{
			str_ = value;
		}
	}
	public int Num
	{
		get { return num_; }
		set
		{
			this.num_ = value;
		}
	}

	public DateTime OverdueTime
	{
		get { return overdueTime; }
		set { overdueTime = value; }
	}
	public Color AA
	{
		get{
			return this.m_aa;
		}
		set {
			this.m_aa = value;
		}
	}
	public void GetObjectData(SerializationInfo info, StreamingContext context)
	{
		info.AddValue("str", str_);
		info.AddValue("num_", num_);
		info.AddValue("colorA", this.m_aa.A);
		info.AddValue("colorR", this.m_aa.R);
		info.AddValue("colorG", this.m_aa.G);
		info.AddValue("colorB", this.m_aa.B);
		//info.AddValue("color", this.m_aa.ToArgb()); 或者
	}
}
public class Program1
{
	static void Main(string[] args)
	{
		ClassA ca = new ClassA("Hello ", 10);
		XmlSerializer xSerial = new XmlSerializer(typeof(ClassA));

		//StreamReader reader = new StreamReader(@"D:\temp\po.xml ");
		//object obj = xSerial.Deserialize(reader);

		//StreamWriter write = new StreamWriter(@"D:\temp\po.xml ");
		//xSerial.Serialize(write, ca);//为什么串行化不了值 

		//or use BinaryFormatter
		IFormatter formatter = new BinaryFormatter();
		Stream stream = new FileStream(@"D:\temp\po.xml", FileMode.Open, FileAccess.Read, FileShare.None);
		//formatter.Serialize(stream, ca);
		object obj = formatter.Deserialize(stream);
		stream.Close();
	}
}

[laorer]

谢谢，不知道为什么，我转化成 xml 文件时，似乎不还是不能得到结果，现在考虑另一种方式了，建两面个属性，一个是 Color (类型是Color，前面加上[XmlIgnore]), 另一个是 HtmlColor（类型是String, 前面加上[XmlElement("Color")] )，
只是如果有多个 属性的话，需要重复多个.....