假设我有以下课程:
public abstract class ScheduledService : ScheduledServiceBase<ScheduledService>
{
public CronInfo CronInfo;
public String ServiceName;
public ScheduledService()
{ }
}
public abstract class ScheduledServiceBase<T>
{
public ScheduledServiceBase()
{ }
public virtual void StartUp(IScheduler scheduler, ScheduledService service, Dictionary<string, object> parameters = null)
{
...
}
}
从这个基类我创建了两个继承类,如:
public AlphaService : ScheduledService
{
public String Alpha_Name;
public Int32 Alpha_Age;
public AlphaService() { }
}
public BetaService : ScheduledService
{
public String Beta_Company;
public Boolean Beta_IsOpen;
public AlphaService() { }
}
然后在我的XML中我定义了两个服务:
<ScheduledServices>
<AlphaService>
<Alpha_Name>John Jones</Alpha_Name>
<Alpha_Age>32</Alpha_Age>
<ServiceName>FirstService</ServiceName>
<CronInfo>0 0 0/2 * * ? *</CronInfo>
</AlphaService>
<BetaService>
<Beta_Company>Ajax Inc.</Beta_Company>
<Beta_IsOpen>Ajax Inc.</Beta_IsOpen>
<ServiceName>SecondService</ServiceName>
<CronInfo>0 30 0/5 * * ? *</CronInfo>
</BetaService>
</ScheduledService>
当我反序列化时,我试图将ScheduledServices分配给List<ScheduledService> ScheduledServices
,但反序列化后列表中不存在任何内容。解串器不会出错,它不会创建任何服务。
我想要做的是有效还是我需要以不同方式设置服务?
以下是我用来反序列化XML的内容:
public Boolean Load(params Type[] extraTypes)
{
deserializedObject = default(T);
XmlTextReader reader = null;
try
{
reader = new XmlTextReader(new StreamReader(_actualPath));
XmlSerializer serializer = new XmlSerializer(typeof(T), extraTypes);
deserializedObject = (T)(serializer.Deserialize(reader));
}
catch (Exception ex)
{
log.Error("Error: " + ex.Message);
log.Debug("Stack: " + ex.StackTrace);
log.Debug("InnerException: " + ex.InnerException.Message);
}
finally
{
if (reader != null) reader.Close();
}
return ((this.deserializedObject != null) && deserializedObject is T);
}
解
感谢John Arlen和他的示例以及其他帖子的链接,我可以通过创建列表来完成我想要的操作,如下所示:
[XmlArrayItem("AlphaJob", Type=typeof(AlphaJob))]
[XmlArrayItem("BetaJob", Type=typeof(BetaJob))]
public List<ScheduledService> ScheduledServices;
答案 0 :(得分:3)
XmlSerializer
类需要比XML提供的更多信息才能正常工作。
例如,它并不确切地知道&#34; AlphaService&#34; (&#34; MyNamespace.AlphaService,MyUtility&#34;?)
最简单的方法是在内存中创建一个示例对象,并调用XmlSerializer.Serialize(myObj)来了解它正在寻找的实际格式。
如果您无法控制传入的XML,则需要进一步修饰您的类(例如,使用XmlElementAttribute),或使用其他机制
编辑:标记您的课程的示例可能是:
[XmlInclude( typeof( AlphaService ) )]
[XmlInclude( typeof( BetaService ) )]
public abstract class ScheduledService : ScheduledServiceBase<ScheduledService> {...}
您可以通过将其序列化为XML并返回到具体对象来测试它:
private void SerializeAndBack()
{
var item = new ScheduledServiceHost
{
ScheduledServices = new List<ScheduledService>
{
new AlphaService {Alpha_Age = 32, Alpha_Name = "John Jones", ServiceName = "FirstService"},
new BetaService {Beta_Company = "Ajax Inc.", Beta_IsOpen = true, ServiceName = "SecondService"},
}
};
var xmlText = XmlSerializeToString( item );
var newObj = XmlDeserializeFromString( xmlText, typeof( ScheduledServiceHost ) );
}
public static string XmlSerializeToString( object objectInstance, params Type[] extraTypes )
{
var sb = new StringBuilder();
using ( TextWriter writer = new StringWriter( sb ) )
{
var serializer = new XmlSerializer( objectInstance.GetType(), extraTypes );
serializer.Serialize( writer, objectInstance );
}
return sb.ToString();
}
public static object XmlDeserializeFromString( string objectData, Type type )
{
using ( var reader = new StringReader( objectData ) )
{
return new XmlSerializer( type ).Deserialize(reader);
}
}
关于标记类的一些选项的