我试图解析这个枚举:
public enum ResponseFormat
{
XML,
JSON,
PDF,
}
与 ksoap2 一起进入 wcf 服务。我尝试使用Marshal
但我不确定它是否正确。
我有一个名为 Format 的类:
public class Format implements Marshal
{
public enum ResponseFormat
{
XML,
JSON,
PDF,;
}
public Object readInstance(XmlPullParser xpp, String string, String string1, PropertyInfo pi)
{
try
{
return ResponseFormat.valueOf(xpp.nextText());
}
catch(Exception e)
{
return null;
}
}
public void writeInstance(XmlSerializer xs, Object o)
{
try
{
xs.text(((ResponseFormat)o).name());
}
catch (Exception e)
{
}
}
public void register(SoapSerializationEnvelope sse)
{
sse.addMapping(sse.xsd, "ResponseFormat", ResponseFormat.class, new Format());
}
}
然后我像这样调用 wcf 服务:
try
{
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
request.addProperty("UserSessionToken", guid);
request.addProperty("Reference", "Nicolas");
request.addProperty("ResponseFormat", Format.ResponseFormat.JSON);
request.addProperty("Surname", "Tyler");
request.addProperty("Firstname", "Nicolas");
request.addProperty("IDNumber", "1234567890123");
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.addMapping("http://schemas.datacontract.org/2004/07/SearchWorksAPI", "ResponseFormat", Format.ResponseFormat.JSON.getClass(), new Format());
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
androidHttpTransport.call(SOAP_ACTION + METHOD_NAME, envelope);
SoapObject response = (SoapObject)envelope.getResponse();
return response.toString();
}
catch (Exception e)
{
return null;
}
我得到了例外:
There was an error reflecting type 'ResponseObjects.Person[]'.; at System.Xml.Serialization.XmlReflectionImporter.ImportTypeMapping(TypeModel model, String ns, ImportContext context, String dataType, XmlAttributes a, Boolean repeats, Boolean openModel, RecursionLimiter limiter)
在我看来,序列化有问题。关于这里有什么问题的任何想法?