An HttpWebRequest has the properties ContentLength and ContentType, but how do you actually set the content of the request?
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here is a simple tutorial: http://msdn.microsoft.com/en-us/library/system.net.httpwebrequest.contentlength.aspx – Zameer Ansari May 13 '14 at 15:20
5 Answers
The following should get you started
byte[] buffer = ...request data as bytes
var webReq = (HttpWebRequest) WebRequest.Create("http://127.0.0.1/target");
webReq.Method = "REQUIRED METHOD";
webReq.ContentType = "REQUIRED CONTENT TYPE";
webReq.ContentLength = buffer.Length;
var reqStream = webReq.GetRequestStream();
reqStream.Write(buffer, 0, buffer.Length);
reqStream.Close();
var webResp = (HttpWebResponse) webReq.GetResponse();

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4Though it may be obvious to some, make sure to set the Content Type BEFORE writing to the request stream. – bsayegh Feb 26 '14 at 20:13
.NET 4.5 (or .NET 4.0 by adding the Microsoft.Net.Http package from NuGet) provides a lot of additional flexibility in setting the request content. Here is an example:
private System.IO.Stream Upload(string actionUrl, string paramString, Stream paramFileStream, byte [] paramFileBytes)
{
HttpContent stringContent = new StringContent(paramString);
HttpContent fileStreamContent = new StreamContent(paramFileStream);
HttpContent bytesContent = new ByteArrayContent(paramFileBytes);
using (var client = new HttpClient())
using (var formData = new MultipartFormDataContent())
{
formData.Add(stringContent, "param1", "param1");
formData.Add(fileStreamContent, "file1", "file1");
formData.Add(bytesContent, "file2", "file2");
var response = client.PostAsync(actionUrl, formData).Result;
if (!response.IsSuccessStatusCode)
{
return null;
}
return response.Content.ReadAsStreamAsync().Result;
}
}

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Hi, do you have an example on how to call this function? should I create the Web request ( request) first , set the content type (request.ContentType = "application/x-www-form-urlencoded"), method = Post, content length, then do a Stream (e.g. Stream dataStream = WebRequest.GetRequestStream();) And after that pass it to the parameter paramFileStream ? like this? Stream uploadstream = Upload(actionUrl, paramString, dataStream, byteArray); ? – Alexei May 10 '17 at 12:51
Here's a different option for posting info without messing with Bytes and Streams. I personally find it easier to follow, read, and debug.
// Convert Object to JSON
var requestMessage = JsonConvert.SerializeObject(requestObject);
var content = new StringContent(requestMessage, Encoding.UTF8, "application/json");
// Create the Client
var client = new HttpClient();
client.DefaultRequestHeaders.Add(AuthKey, AuthValue);
// Post the JSON
var responseMessage = client.PostAsync(requestEndPoint, content).Result;
var stringResult = responseMessage.Content.ReadAsStringAsync().Result;
// Convert JSON back to the Object
var responseObject = JsonConvert.DeserializeObject<ResponseObject>(stringResult);

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1Oh you are using Microsoft.Net.Http, which is new and a better answer. – Frank Schwieterman Jul 17 '15 at 23:12
HttpWebRequest's RequestStream is where the action is at - rough code...
//build the request object
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(http://someapi.com/);
//write the input data (aka post) to a byte array
byte[] requestBytes = new ASCIIEncoding().GetBytes(inputData);
//get the request stream to write the post to
Stream requestStream = request.GetRequestStream();
//write the post to the request stream
requestStream.Write(requestBytes, 0, requestBytes.Length);
If you're sending extended chars, use UTF8Encoding, make sure you set the right content-type/charset header too.

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need to set the method on the request as well - and usually also the contenttype – Robert Levy Apr 03 '11 at 04:27
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yeah sure Robert - I left out *a lot of code* like actually sending the request too :) It's good for the OP and brevity to complete the dots. I marked it as a community answer if you want to make it more complete. – stephbu Apr 04 '11 at 07:33
HttpWebRequest.GetRequestStream()
gets the request Stream. After you have set the headers, use GetRequestStream()
and write the content to the stream.
This post explains how to transmit files using HttpWebRequest
, which should provide a good example of how to send content.
But, basically the format would be
var stream = request.GetRequestStream();
stream.Write( stuff );
stream.Close();
var response = request.GetResponse();

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