我以前做过几次类似的事情,如果你只需要处理少量的命名空间并且你事先知道它们,这可能对你有用。创建一个简单的类继承层次结构,并为不同名称空间的不同类添加属性。请参阅以下代码示例。如果你运行这个程序,它会给出输出:
Deserialized, type=XmlSerializerExample.GpxV1, data=1
Deserialized, type=XmlSerializerExample.GpxV2, data=2
Deserialized, type=XmlSerializerExample.Gpx, data=3
这是代码:
using System;
using System.IO;
using System.Xml;
using System.Xml.Serialization;
[XmlRoot("gpx")]
public class Gpx {
[XmlElement("data")] public int Data;
}
[XmlRoot("gpx", Namespace = "http://www.topografix.com/GPX/1/0")]
public class GpxV1 : Gpx {}
[XmlRoot("gpx", Namespace = "http://www.topografix.com/GPX/1/1")]
public class GpxV2 : Gpx {}
internal class Program {
private static void Main() {
var xmlExamples = new[] {
"<gpx xmlns='http://www.topografix.com/GPX/1/0'><data>1</data></gpx>",
"<gpx xmlns='http://www.topografix.com/GPX/1/1'><data>2</data></gpx>",
"<gpx><data>3</data></gpx>",
};
var serializers = new[] {
new XmlSerializer(typeof (Gpx)),
new XmlSerializer(typeof (GpxV1)),
new XmlSerializer(typeof (GpxV2)),
};
foreach (var xml in xmlExamples) {
var textReader = new StringReader(xml);
var xmlReader = XmlReader.Create(textReader);
foreach (var serializer in serializers) {
if (serializer.CanDeserialize(xmlReader)) {
var gpx = (Gpx)serializer.Deserialize(xmlReader);
Console.WriteLine("Deserialized, type={0}, data={1}", gpx.GetType(), gpx.Data);
}
}
}
}
}