От: | Аноним | ||
Дата: | 22.02.05 16:05 | ||
Оценка: |
XmlSerializer xs = new XmlSerializer(responce.GetType());
System.IO.Stream mStream = new System.IO.MemoryStream();
xs.Serialize(mStream, responce);
XmlDocument xmlResponce = new XmlDocument();
try
{
xmlResponce.Load(mStream);
}
catch (XmlException ex)
{
throw new SoapException(ex.Message, XmlQualifiedName.Empty);
}