Use Script Task and do post as :
public static string POST(string postData, string url, string referer, string cookie)
{
string retValue = string.Empty;
try
{
byte[] byteArray = Encoding.UTF8.GetBytes(postData);
HttpWebRequest request = (HttpWebRequest)HttpWebRequest.Create(url);
request.Method = "POST";
request.KeepAlive = true;
request.AllowAutoRedirect = false;
request.Accept = "*/*";
request.ContentType = "text/xml";
if (!string.IsNullOrEmpty(cookie))
request.Headers.Add(HttpRequestHeader.Cookie, cookie);
if (!string.IsNullOrEmpty(referer))
request.Referer = referer;
request.ContentLength = byteArray.Length;
request.UserAgent = "Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.0.5) Gecko/2008120122 Firefox/3.0.5";
Stream dataStream = request.GetRequestStream();
dataStream.Write(byteArray, 0, byteArray.Length);
dataStream.Close();
try
{
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
Stream dataResponseStream = response.GetResponseStream();
StreamReader SR = new StreamReader(dataResponseStream, Encoding.UTF8);
retValue = SR.ReadToEnd();
response.Close();
dataStream.Close();
SR.Close();
request.Abort();
}
catch
{
}
}
catch
{
}
return retValue;
}
how to call the above method :
string postdata = oReqDoc.InnerXml; //oReqDoc is the XMLDoc you need to post. If you have xml as in text form simply sent it.
string url ="http://www.abc.com/status.php php";
pass referrer and cookie as ""
now call POST as :
string strReponse = POST(postdata, url, "", "");
now you can inspect if you have got the right output xml.
you can try the following snippet too to convert string to xml:
XmlDocument doc = new XmlDocument{ XmlResolver = null };
doc.LoadXml(strReponse);
XmlNode newNode = doc.DocumentElement;
Note: You may need following namespaces:
using System.Net;
using System.IO;
using System.Web;
using System.Xml;
Thanks,
Kuthuparakkal