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Home/ Questions/Q 7839555
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T15:28:19+00:00 2026-06-02T15:28:19+00:00

I’m working on a project which requires me to work with an REST API.

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I’m working on a project which requires me to work with an REST API. The API expects a DTO as input parameter.

 [OperationContract]
        [WebInvoke(Method = "POST", UriTemplate = "campaign/create?session={sessionkey}", ResponseFormat = WebMessageFormat.Xml, BodyStyle = WebMessageBodyStyle.Bare, RequestFormat = WebMessageFormat.Xml)]
        [XmlSerializerFormat(Style = OperationFormatStyle.Document, Use = OperationFormatUse.Literal)]
        FeedbackDTO PersistCampaign(string sessionkey, CampaignDTO PaycentoCampaign);

I’m trying to send the data to this method with the following method:

public static HttpWebResponse DoHttpWebRequest(String url, String method, string data)
{
    HttpWebRequest req = WebRequest.Create(url) as HttpWebRequest;
    req.KeepAlive = false;
    req.ContentType = "application/xml";
    req.Method = method;
    if ((method.Equals("POST") || method.Equals("PUT")) && data != null)
    {
        byte[] buffer = Encoding.UTF8.GetBytes(data);
        Stream PostData = req.GetRequestStream();
        PostData.Write(buffer, 0, buffer.Length);
        PostData.Close();
    }
    return req.GetResponse() as HttpWebResponse;
}

After hours of debugging I found that the problem is caused by the DateTime proprties. I started to ‘manually’ create the XML and do property per property.

 sb.Append("<Startdate>").Append(HttpUtility.HtmlEncode(Startdate.ToString(Helper.DATE_FORMAT_STRING))).Append("</Startdate>");

I tried adding a format (ddMMyyyy) in the toString but it’s still causing a bad request. How should I add a DateTime property to the XML so ASP.NET is able to parse it into the DTO?

Stuff I tried:

  • StartDate.ToLongDateString()
  • StartDate.ToString()
  • StartDate.toString(“ddhhyyyy”)
  • StartDate.Ticks
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-02T15:28:21+00:00Added an answer on June 2, 2026 at 3:28 pm

    StartdDate.toString(“s”) did the trick.

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