sending named Mulitpart Data wit HttpClient - c#

I try to Send Multipart Content as a Post to an WebService using c# HttpClient
My Code is the Following
public string UploadDocument(Stream stream)
{
MultipartContent multipart = new MultipartContent();
//multipart.Add(new StreamContent(stream), "importFile");
StreamContent content = new StreamContent(stream);
multipart.Add(content);
Client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
HttpResponseMessage response = Client.PostAsync("URL to webservice", multipart).Result;
string result = response.Content.ReadAsStringAsync().Result;
//Wirft eine Exception wenn der StatusCode nicht 200 (OK) ist.
response.EnsureSuccessStatusCode();
return result;
}
I Need the Stream Content to be named, how can i do it.
I found a Java Way to do it, but in C# I don't find a Parameter "Name"
Here the Java Example (see "importFile")
httpRequest = new HttpPostwebviewer/?action=import");
// set headers
httpRequest.setHeader(HEADER_NAME_ACCEPT, "application/json");
// prepare upload entity
MultipartEntityBuilder meBuilder = MultipartEntityBuilder.create();
meBuilder.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);
meBuilder.setCharset(Consts.UTF_8);
meBuilder.addPart("importFile", contentBody);
// build upload entity
HttpEntity uploadEntity = meBuilder.build();
// set properties to http request
((HttpPost) httpRequest).setEntity(uploadEntity);

You can use System.Net.HttpWebRequest to do the same, Below is the code that I am using:
public static void UploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
{
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
url = handlerLocation + url;
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
foreach (string key in nvc.Keys)
{
rs.Write(boundarybytes, 0, boundarybytes.Length);
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
rs.Write(formitembytes, 0, formitembytes.Length);
}
rs.Write(boundarybytes, 0, boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
string header = string.Format(headerTemplate, paramName, file, contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();
byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
rs.Write(trailer, 0, trailer.Length);
rs.Close();
WebResponse wresp = null;
try
{
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
}
catch (Exception ex)
{
//WriteTrace(ex.Message);
if (wresp != null)
{
wresp.Close();
wresp = null;
}
}
finally
{
wr = null;
}
}

Related

Http Post file to server from Windows Forms

I'm making a Windows Application that need to make a HTTP post to my Server. In my .NET MVC application i use this code:
Html.BeginForm("Index","UploadData",
FormMethod.Post,
new { enctype = "multipart/form-data", #class = "contact-form", id = "frmUploadData" })
My question is how do i post a File from my computer to the server from a windows forms application?
I have tried to use, this method, but i am not sure how to call it correctly:
public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
{
Console.WriteLine(string.Format("Uploading {0} to {1}", file, url));
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
foreach (string key in nvc.Keys)
{
rs.Write(boundarybytes, 0, boundarybytes.Length);
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
rs.Write(formitembytes, 0, formitembytes.Length);
}
rs.Write(boundarybytes, 0, boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
string header = string.Format(headerTemplate, paramName, file, contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();
byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
rs.Write(trailer, 0, trailer.Length);
rs.Close();
WebResponse wresp = null;
try
{
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
Console.WriteLine(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
}
catch (Exception ex)
{
Console.WriteLine("Error uploading file", ex);
if (wresp != null)
{
wresp.Close();
wresp = null;
}
}
finally
{
wr = null;
}
}
Not sure if i am calling it correctly, i get a 404 error when i try to call this method:
public static void OnChanged(object sender, FileSystemEventArgs e)
{
NameValueCollection nvc = new NameValueCollection();
nvc.Add("id", "frmUploadData");
HttpUploadFile("http://classicraider.com/UploadData",
#e.FullPath, "file", "lua", nvc);
Console.WriteLine("File: " + e.FullPath + " " + e.ChangeType);
}

How send a PDF by POST HttpWebRequest in C#

(i'm french, sorry for my english)
I don't find / understand how send a simple PDF file by post request at a webService (REST protocol).
I tried some examples but it's doesn't word. And when i use a , it's work, but i want do this in the code behind only !
My question is the title : How send this PDF file?
The url : https://test.website.fr/Website/api/transactions/" + sVal + "/contrat
PDF file : questions.pdf
Thanks.
Iris Touchard
I found the solution :
//Identificate separator
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
//Encoding
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
//Creation and specification of the request
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create("https://test.website.fr/Website/api/transactions/" + sVal + "/contrat"); //sVal is id for the webService
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
string sAuthorization = "login:password";//AUTHENTIFICATION BEGIN
byte[] toEncodeAsBytes = System.Text.ASCIIEncoding.ASCII.GetBytes(sAuthorization);
string returnValue = System.Convert.ToBase64String(toEncodeAsBytes);
wr.Headers.Add("Authorization: Basic " + returnValue); //AUTHENTIFICATION END
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}"; //For the POST's format
//Writting of the file
rs.Write(boundarybytes, 0, boundarybytes.Length);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(Server.MapPath("questions.pdf"));
rs.Write(formitembytes, 0, formitembytes.Length);
rs.Write(boundarybytes, 0, boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
string header = string.Format(headerTemplate, "file", "questions.pdf", contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(Server.MapPath("questions.pdf"), FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();
byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
rs.Write(trailer, 0, trailer.Length);
rs.Close();
rs = null;
WebResponse wresp = null;
try
{
//Get the response
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
string responseData = reader2.ReadToEnd();
}
catch (Exception ex)
{
string s = ex.Message;
}
finally
{
if (wresp != null)
{
wresp.Close();
wresp = null;
}
wr = null;
}
This should be what you are looking for:
http://msdn.microsoft.com/en-us/library/debx8sh9.aspx
Here's a quick code example:
byte[] pdfFile = File.ReadAllBytes("pdf file path here");
WebRequest request = WebRequest.Create("https://test.site.fr/Testfile");
request.Method = "POST";
request.ContentLength = pdfFile.Length;
request.ContentType = "application/pdf";
Stream stream = request.GetRequestStream();
stream.Write(pdfFile, 0, pdfFile.Length);
stream.Close();
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
StreamReader reader = new StreamReader(response.GetResponseStream());
Console.WriteLine(reader.ReadToEnd());
reader.Close();

Streamsend API with C#

I cannot seem to get this code to work with Streamsends API. I have been all over the internet and have found very little info pertaining to what I am trying to do. I am trying to upload a file to their server. Can anyone look at this and maybe help me out here?
I keep getting a 500 error from Streamsends servers which they say is within the Streamsend application itself. What is funny is if I take the line near the bottom that is commented out
(//request.ContentType = "multipart/form-data";)
and uncomment it I get a 422 error which is saying invalid data. It seems like there is something wrong with the boundary.
I am fairly new to C# and this is the first time I have worked with Streamsends API so any help would be greatly appreciated.
m_Method = "POST"
m_Action = "uploads"
private string StreamSendResponse(string Path, string FileName)
{
string sReturn = String.Empty;
string sLocation = String.Empty;
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
byte[] boundarybytes = Encoding.ASCII.GetBytes(Environment.NewLine + "--" + boundary + Environment.NewLine);
Uri addy = new Uri(m_URI + m_Action);
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(addy);
request.ContentType = "multipart/form-data; boundary=" + boundary;
request.Method = m_Method;
request.KeepAlive = true;
request.Headers.Add("Authorization", "Basic " + Merv.base64Encode(m_ID + ":" + m_Key));
Stream rs = request.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n{1}";
NameValueCollection nvc = new NameValueCollection();
foreach (string key in nvc.Keys)
{
rs.Write(boundarybytes, 0, boundarybytes.Length);
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
rs.Write(formitembytes, 0, formitembytes.Length);
}
rs.Write(boundarybytes, 0, boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n";
string header = string.Format(headerTemplate, "data", FileName);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(Path + FileName, FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();
byte[] trailer = System.Text.Encoding.ASCII.GetBytes(Environment.NewLine + "--" + boundary + "--" + Environment.NewLine);
rs.Write(trailer, 0, trailer.Length);
rs.Close();
request.Accept = "application/xml";
//request.ContentType = "multipart/form-data";
try
{
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
StreamReader srResponse = new StreamReader(response.GetResponseStream());
sLocation = response.Headers["Location"];
sReturn = srResponse.ReadToEnd().Trim();
}
catch (WebException exc)
{
StreamReader srResponse = new StreamReader(response.GetResponseStream());
sReturn = srResponse.ReadToEnd().Trim();
sLocation = exc.Message;
}
if (sReturn == String.Empty)
{
sReturn = sLocation;
}
return sReturn;
}

Upload file with HttpWebRequest

Here is my code:
private void UploadFilesToRemoteUrl(string url, string[] files, string logpath, NameValueCollection nvc)
{
long length = 0;
string boundary = "----------------------------" +
DateTime.Now.Ticks.ToString("x");
HttpWebRequest httpWebRequest2 = (HttpWebRequest)WebRequest.Create(url);
httpWebRequest2.ContentType = "multipart/form-data; boundary=" + boundary;
httpWebRequest2.Method = "POST";
httpWebRequest2.KeepAlive = true;
httpWebRequest2.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream memStream = new System.IO.MemoryStream();
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
foreach(string key in nvc.Keys)
{
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
memStream.Write(formitembytes, 0, formitembytes.Length);
}
memStream.Write(boundarybytes,0,boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\";filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";
for(int i=0;i<files.Length;i++)
{
string header = string.Format(headerTemplate,"file"+i,files[i]);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
memStream.Write(headerbytes,0,headerbytes.Length);
FileStream fileStream = new FileStream(files[i], FileMode.Open,
FileAccess.Read);
byte[] buffer = new byte[1024];
int bytesRead = 0;
while ( (bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0 )
{
memStream.Write(buffer, 0, bytesRead);
}
memStream.Write(boundarybytes,0,boundarybytes.Length);
fileStream.Close();
}
httpWebRequest2.ContentLength = memStream.Length;
Stream requestStream = httpWebRequest2.GetRequestStream();
memStream.Position = 0;
byte[] tempBuffer = new byte[memStream.Length];
memStream.Read(tempBuffer,0,tempBuffer.Length);
memStream.Close();
requestStream.Write(tempBuffer,0,tempBuffer.Length );
requestStream.Close();
WebResponse webResponse2 = httpWebRequest2.GetResponse();
Stream stream2 = webResponse2.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
MessageBox.Show(reader2.ReadToEnd());
webResponse2.Close();
httpWebRequest2 = null;
webResponse2 = null;
}
And here is how I call it:
string[] filenames = new string[] { #"C:\Users\John\Desktop\ex.txt" };
NameValueCollection nvc = new NameValueCollection();
nvc.Add("cmd", "new");
nvc.Add("sTitle", "bugX");
nvc.Add("token", "someToken");
UploadFilesToRemoteUrl("https://myUrl.fogbugz.com/api.asp", filenames, "", nvc);
And everything is working fine, except the file isn't uploaded.
I also tried this code:
http://stackoverflow.com/questions/566462/upload-files-with-httpwebrequest-multipart-form-data
and the same thing happens. How to resolve this?
Here is the response from the server:
<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<response>
<case ixBug=\"123486\" operations=\"edit,assign,resolve,email,remind\"></case>
</response>
you're just sending a stream to the server, it's impossible to know what happens at the server for us if you don't provide any code for us from the server side. I guess you've forgotten to receive and create the file on the server side.
if this is not the problem I would suggest you get the invaluable tool fiddler and start catching your httprequests and see what they contain.

HttpWebRequest C# uploading a file

using C#, I'm trying to integrate my web store w/ an email marketing client. I want to upload a comma delimited file of subscribers once a night. They say to get this to work, it has to be a form posts: multipart/form-data, but I'm not using a form. I'm able to connect to their servers but I keep getting back a Data can't be blank. Can anyone help me to get this working?
public static string Create()
{
string authInfo = "username" + ":" + "password";
string root = AppDomain.CurrentDomain.BaseDirectory;
string file = root + "Folder\\work.txt";
FileInfo fi = new FileInfo(file);
int fileLength = (int)fi.Length;
FileStream rdr = new FileStream(file, FileMode.Open);
HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(url);
httpWebRequest.Method = "POST";
httpWebRequest.ContentType = "application/x-www-form-urlencoded";
httpWebRequest.Accept = "application/xml";
authInfo = Convert.ToBase64String(Encoding.Default.GetBytes(authInfo));
httpWebRequest.Headers["Authorization"] = "Basic " + authInfo;
byte[] requestBytes = new byte[fileLength];
int bytesRead = 0;
httpWebRequest.ContentLength = requestBytes.Length;
using (Stream requestStream = httpWebRequest.GetRequestStream())
{
while ((bytesRead = rdr.Read(requestBytes, 0, requestBytes.Length)) != 0)
{
requestStream.Write(requestBytes, 0, bytesRead);
requestStream.Close();
}
}
//READ RESPONSE FROM STREAM
string responseData;
using (StreamReader responseStream = new StreamReader(httpWebRequest.GetResponse().GetResponseStream()))
{
responseData = responseStream.ReadToEnd();
responseStream.Close();
}
return responseData;
}
I had got the same problem and this following code answered perfectly at this problem :
//Identificate separator
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
//Encoding
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
//Creation and specification of the request
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url); //sVal is id for the webService
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
string sAuthorization = "login:password";//AUTHENTIFICATION BEGIN
byte[] toEncodeAsBytes = System.Text.ASCIIEncoding.ASCII.GetBytes(sAuthorization);
string returnValue = System.Convert.ToBase64String(toEncodeAsBytes);
wr.Headers.Add("Authorization: Basic " + returnValue); //AUTHENTIFICATION END
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}"; //For the POST's format
//Writting of the file
rs.Write(boundarybytes, 0, boundarybytes.Length);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(Server.MapPath("questions.pdf"));
rs.Write(formitembytes, 0, formitembytes.Length);
rs.Write(boundarybytes, 0, boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
string header = string.Format(headerTemplate, "file", "questions.pdf", contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(Server.MapPath("questions.pdf"), FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();
byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
rs.Write(trailer, 0, trailer.Length);
rs.Close();
rs = null;
WebResponse wresp = null;
try
{
//Get the response
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
string responseData = reader2.ReadToEnd();
}
catch (Exception ex)
{
string s = ex.Message;
}
finally
{
if (wresp != null)
{
wresp.Close();
wresp = null;
}
wr = null;
}

Categories

Resources