我有一个与XML序列化和实现IXmlSerializable的派生类兼容的基类。C#Xml-使用IXmlSerializable序列化派生类
在这个例子中,基类确实实现IXmlSerializable的:
using System.Diagnostics;
using System.Text;
using System.Xml;
using System.Xml.Schema;
using System.Xml.Serialization;
namespace XmlSerializationDerived
{
public class Foo
{
public int fooProp;
public XmlSchema GetSchema()
{
return null;
}
public void ReadXml(XmlReader reader)
{
fooProp = int.Parse (reader.ReadElementString ("fooProp"));
}
public void WriteXml(XmlWriter writer)
{
writer.WriteElementString ("fooProp", fooProp.ToString());
}
}
public class Bar : Foo, IXmlSerializable
{
public new void ReadXml(XmlReader reader)
{
base.ReadXml (reader);
}
public new void WriteXml(XmlWriter writer)
{
base.WriteXml (writer);
}
static void Main(string[] args)
{
StringBuilder sb = new StringBuilder();
XmlWriter writer = XmlWriter.Create (sb);
Bar bar = new Bar();
bar.fooProp = 42;
XmlSerializer serializer = new XmlSerializer (typeof (Bar));
serializer.Serialize (writer, bar);
Debug.WriteLine (sb.ToString());
}
}
}
这将产生以下输出:
<?xml version="1.0" encoding="utf-16"?><Bar><fooProp>42</fooProp></Bar>
然而,我想使用一个基类,它确实不实现IXmlSerializable。这可以防止使用base.Read/WriteXml
。结果将是:
<?xml version="1.0" encoding="utf-16"?><Bar />
有什么办法仍然可以得到理想的结果吗?
来自Java和它的死容易序列化,我有点难过听到这一点。我实际上创建了基类,我很幸运地可以访问它,实现IXmlSerializable。它有效,但它有点痛,因为我正在编写不应该存在的代码。 – mafu 2009-02-09 17:00:15
除非有人想出一个新的方法,这是正式的答案。 – mafu 2009-02-10 14:54:03