例如,我有以下类:
public abstract class Device
{
}
public class WindowsDevice: Device
{
}
public class AndroidDevice: Device
{
}
现在我想将WindowsDevice和AndroidDevice序列化/反序列化为XML:
public static string Serialize(object o, Type[] additionalTypes = null)
{
var serializer = new XmlSerializer(o.GetType(), additionalTypes);
using (var stringWriter = new StringWriterWithEncoding(Encoding.UTF8))
{
serializer.Serialize(stringWriter, o);
return stringWriter.ToString();
}
}
这将产生以下输出:
<?xml version="1.0" encoding="utf-8"?>
<WindowsDevice xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
</WindowsDevice>
但是现在我无法反序列化这个,因为在我的应用程序中我不知道XML是WindowsDevice还是AndroidDevice,所以我必须反序列化为typeof(Device)。但是我会得到一个例外,“WindowsDevice”在XML中是意外的。
我尝试了XmlInclude和extraTypes但没有成功。
我不明白的是,如果我有以下样本类:
public class SampleClass
{
public List<Device> Devices {get;set}
}
如果我序列化SampleClass并使用XmlInclude或extraTypes,我会得到我想要的东西:
<Devices>
<Device xsi:type="WindowsDevice"></Device>
</Devices>
但是我没有那个课,我没有设备列表。我只想序列化/反序列化WindowsDevice和AndroidDevice但是反序列化我不知道它是AndroidDevice还是WindowsDevice所以我必须使用typeof(Device)并且想要获得正确的子类AndroidDevice或WindowsDevice,所以代替:
<WindowsDevice></WindowsDevice>
我希望:
<Device xsi:type="WindowsDevice"></Device>
如何做到这一点?
答案 0 :(得分:1)
您的问题是您在序列化和反序列化过程中构建XmlSerializer
不一致。在这两种情况下,您都需要使用相同的Type
参数构造它,特别是基类型typeof(Device)
。因此,我建议您将现有的完全通用的序列化方法替换为Device
的特定序列化方法:
public static class DeviceExtensions
{
public static string SerializeDevice<TDevice>(this TDevice o) where TDevice : Device
{
// Ensure that [XmlInclude(typeof(TDevice))] is present on Device.
// (Included for clarity -- actually XmlSerializer will make a similar check.)
if (!typeof(Device).GetCustomAttributes<XmlIncludeAttribute>().Any(a => a.Type == o.GetType()))
{
throw new InvalidOperationException("Unknown device type " + o.GetType());
}
var serializer = new XmlSerializer(typeof(Device)); // Serialize as the base class
using (var stringWriter = new StringWriterWithEncoding(Encoding.UTF8))
{
serializer.Serialize(stringWriter, o);
return stringWriter.ToString();
}
}
public static Device DeserializeDevice(this string xml)
{
var serial = new XmlSerializer(typeof(Device));
using (var reader = new StringReader(xml))
{
return (Device)serial.Deserialize(reader);
}
}
}
然后,将[XmlInclude(typeof(TDevice))]
应用于Device
以获取所有可能的子类型:
[XmlInclude(typeof(WindowsDevice))]
[XmlInclude(typeof(AndroidDevice))]
public abstract class Device
{
}
然后,两种类型的设备现在都可以成功序列化和反序列化,同时保留其类型,因为XmlSerializer
将包含"xsi:type"
属性以明确指出类型:
<Device xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:type="WindowsDevice" />
或者
<Device xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:type="AndroidDevice" />
示例fiddle。
<强>更新强>
所以问题是,我用typeof(WindowsDevice)而不是typeof(Device)序列化了?
是
如果我必须使用typeof(WindowsDevice),那么解决方案的任何想法都会有效吗?因为我有数百个课程,并且不想使用数百个不同的XmlSerializer初始化......
这更像是一个建筑问题,而不是一个如何问题。一种可能性是引入一个custom attribute,您可以将其应用于类,以指示该类的任何子类型应始终序列化为属性基类型。还需要所有适当的[XmlInclude(typeof(TDerivedType))]
属性:
[System.AttributeUsage(System.AttributeTargets.Class, AllowMultiple = false, Inherited = false)]
public class XmlBaseTypeAttribute : System.Attribute
{
}
[XmlInclude(typeof(WindowsDevice))]
[XmlInclude(typeof(AndroidDevice))]
[XmlBaseType]
public abstract class Device
{
}
然后修改通用XML序列化代码以查找为[XmlBaseType]
属性序列化的对象的类型层次结构,并(de)序列化为该类型:
public static class XmlExtensions
{
static Type GetSerializedType(this Type type)
{
var serializedType = type.BaseTypesAndSelf().Where(t => Attribute.IsDefined(t, typeof(XmlBaseTypeAttribute))).SingleOrDefault();
if (serializedType != null)
{
// Ensure that [XmlInclude(typeof(TDerived))] is present on the base type
// (Included for clarity -- actually XmlSerializer will make a similar check.)
if (!serializedType.GetCustomAttributes<XmlIncludeAttribute>().Any(a => a.Type == type))
{
throw new InvalidOperationException(string.Format("Unknown subtype {0} of type {1}", type, serializedType));
}
}
return serializedType ?? type;
}
public static string Serialize(this object o)
{
var serializer = new XmlSerializer(o.GetType().GetSerializedType());
using (var stringWriter = new StringWriterWithEncoding(Encoding.UTF8))
{
serializer.Serialize(stringWriter, o);
return stringWriter.ToString();
}
}
public static T Deserialize<T>(this string xml)
{
var serial = new XmlSerializer(typeof(T).GetSerializedType());
using (var reader = new StringReader(xml))
{
return (T)serial.Deserialize(reader);
}
}
}
当然这意味着如果您的代码尝试反序列化XML,它希望包含WindowsDevice
,它实际上可能会取回AndroidDevice
,具体取决于XML的内容。
示例fiddle #2。