Root Element Falta al deserializar una respuesta XML usando XmlSerializer

Question

Tengo la siguiente respuesta XML. Estoy tratando de deserializar usando XmlSerializer. Cuando elimino la llamada al serializador XML, no arrojo errores. Cada vez que uso XmlSerializer obtengo una excepción de . ¿Qué me estoy perdiendo?

Excepción es:

System.Xml.XmlException: Root element is missing. 
    at System.Xml.XmlTextReaderImpl.Throw(Exception e) 
    at System.Xml.XmlTextReaderImpl.ParseDocumentContent() 
    at System.Xml.XmlTextReaderImpl.Read() 
    at System.Xml.XmlTextReader.Read() 
    at System.Xml.XmlReader.MoveToContent() 
    at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderSubmitReportResponse.Read5_NeweggAPIResponse() 

en documento XML es:

<?xml version="1.0" encoding="utf-8"?> 
<NeweggAPIResponse> 
    <IsSuccess>true</IsSuccess> 
    <OperationType>OrderListReportResponse</OperationType> 
    <SellerID>myID</SellerID> 
    <ResponseBody> 
     <ResponseList> 
      <ResponseInfo> 
       <RequestId>XXXXXXXX</RequestId> 
       <RequestType>ORDER_LIST_REPORT</RequestType> 
       <RequestDate>07/26/2012 09:27:06</RequestDate> 
       <RequestStatus>SUBMITTED</RequestStatus> 
      </ResponseInfo> 
     </ResponseList> 
    </ResponseBody> 
</NeweggAPIResponse> 

Mi llamado a XmlSerializer es:

XmlSerializer serializer = new XmlSerializer(typeof(SubmitReportResponse)); 
reportReq = serializer.Deserialize(respStream) as SubmitReportResponse; 

clase SubmitReportResponse es:

public enum RequestStatus 
{ 
    ALL, 
    SUBMITTED, 
    IN_PROGRESS, 
    FINISHED, 
    CANCELLED 
} 

/// <summary> 
/// TODO: Update summary. 
/// </summary> 
[XmlRoot("NeweggAPIResponse")] 
public class SubmitReportResponse 
{ 
    public string IsSuccess { get; set; } 
    public string OperationType { get; set; } 
    public string SellerID { get; set; } 
    public ReportResponseBody ResponseBody { get; set; } 

    public SubmitReportResponse() 
    { 
     ResponseBody = new ReportResponseBody(); 
    } 
} 

public class ReportResponseBody 
{ 
    public string Memo { get; set; } 
    public ReportResponseList[] ResponseList { get; set; } 



    public ReportResponseBody() 
    { 

     ResponseList = new ReportResponseList[0]; 
    } 
} 

public class ReportResponseList 
{ 
    public ResponseInfo[] ResponseInfo { get; set; } 

    public ReportResponseList() 
    { 
     ResponseInfo = new ResponseInfo[0]; 
    } 

} 

public class ResponseInfo 
{ 
    public string RequestId { get; set; } 
    public string RequestType { get; set; } 
    public string RequestDate { get; set; } 
    public RequestStatus RequestStatus { get; set; } 

    public ResponseInfo() 
    { 
     RequestStatus = new RequestStatus(); 
    } 
} 

EDIT:

Código Solicitud:

  HttpWebRequest request = WebRequest.Create(endpoint) as HttpWebRequest; 
      request.Proxy = null; 
      request.Method = "POST"; 
      //Specify the xml/Json content types that are acceptable. 
      request.ContentType = "application/xml"; 
      request.Accept = "application/xml"; 

      //Attach authorization information 
      request.Headers.Add("Authorization", apikey); 
      request.Headers.Add("Secretkey", secretkey); 

      GetOrderListRequest requestObj = new GetOrderListRequest(); 
      requestObj.OperationType = OperationType.OrderListReportRequest; 
      requestObj.RequestBody = new OrderListRequestBody(); 
      requestObj.RequestBody.OrderReportCriteria = new OrderReportCriteria(); 
      requestObj.RequestBody.OrderReportCriteria.Status = 3; 
      requestObj.RequestBody.OrderReportCriteria.KeywordsType = 0; 
      requestObj.RequestBody.OrderReportCriteria.OrderDateFrom = "2012-01-01"; 
      requestObj.RequestBody.OrderReportCriteria.OrderDateTo = "2012-07-26"; 
      requestObj.RequestBody.OrderReportCriteria.OrderDownloaded = "false"; 

      string requestBody = SerializeToString(requestObj); 

      byte[] byteStr = Encoding.UTF8.GetBytes(requestBody); 
      request.ContentLength = byteStr.Length; 

      using (Stream stream = request.GetRequestStream()) 
      { 
       stream.Write(byteStr, 0, byteStr.Length); 
      } 

      //Parse the response 
      using (HttpWebResponse response = request.GetResponse() as HttpWebResponse) 
      { 
       //Business error 
       if (response.StatusCode != HttpStatusCode.OK) 
       { 
        Console.WriteLine(string.Format("Error: response status code is{0}, at time:{1}", response.StatusCode, DateTime.Now.ToString())); 

        return; 
       } 
       else if (response.StatusCode == HttpStatusCode.OK)//Success 
       { 
        using (Stream respStream = response.GetResponseStream()) 
        { 
         StreamReader readerOK = new StreamReader(respStream); 
         //Console.WriteLine(String.Format("Result:{0}", DateTime.Now.ToString())); 
         Console.WriteLine(String.Format("{0}", readerOK.ReadToEnd())); 

         XmlSerializer serializer = new XmlSerializer(typeof(SubmitReportResponse)); 
         reportReq = serializer.Deserialize(respStream) as SubmitReportResponse; 
        } 
       } 
      } 


    public string SerializeToObj(object obj) 
    { 

     XmlSerializerNamespaces ns = new XmlSerializerNamespaces(); 
     ns.Add("", ""); 
     XmlWriterSettings settings = new XmlWriterSettings(); 
     settings.OmitXmlDeclaration = true; 
     settings.Encoding = new UTF8Encoding(false); 
     settings.Indent = true; 
     XmlSerializer xs = new XmlSerializer(obj.GetType()); 
     MemoryStream ms = new MemoryStream(); 

     // xs.Serialize(ms, obj,ns); 


     XmlWriter writer = XmlWriter.Create(ms, settings); 
     xs.Serialize(writer, obj, ns); 

     return Encoding.UTF8.GetString(ms.ToArray()); 
    } 

Resolución:

Parece que llamar Console.WriteLine(String.Format("{0}", readerOK.ReadToEnd())); hace que la corriente para ser consumido y no estará disponible para la serialización. Eliminar esta línea me permitió serializar correctamente el XML en mis clases.

Answer 1

Llamar a Console.WriteLine(String.Format("{0}", readerOK.ReadToEnd())); hace que la secuencia se consuma y no esté disponible para la serialización.

Eliminar esta línea me permitió serializar correctamente el XML en mis clases.