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I have been asked to do the following in C#:

/** * 1. Create a MultipartPostMethod * 2. Construct the web URL to connect to the SDP Server * 3. Add the filename to be attached as a parameter to the MultipartPostMethod with parameter name "filename" * 4. Execute the MultipartPostMethod * 5. Receive and process the response as required * /

I have written some code that has no errors, however, the file is not attached.

Can someone have a look at my C# code to see if I have written the code incorrectly?

Here is my code:

var client = new HttpClient(); const string weblinkUrl = "http://testserver.com/attach?"; var method = new MultipartFormDataContent(); const string fileName = "C:\file.txt"; var streamContent = new StreamContent(File.Open(fileName, FileMode.Open)); method.Add(streamContent, "filename"); var result = client.PostAsync(weblinkUrl, method); MessageBox.Show(result.Result.ToString());
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5 Answers 5

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Posting MultipartFormDataContent in C# is simple but may be confusing the first time. Here is the code that works for me when posting a .png .txt etc.

// 2. Create the url string url = "https://myurl.com/api/..."; string filename = "myFile.png"; // In my case this is the JSON that will be returned from the post string result = ""; // 1. Create a MultipartPostMethod // "NKdKd9Yk" is the boundary parameter using (var formContent = new MultipartFormDataContent("NKdKd9Yk")) { formContent.Headers.ContentType.MediaType = "multipart/form-data"; // 3. Add the filename C:\\... + fileName is the path your file Stream fileStream = System.IO.File.OpenRead("C:\\Users\\username\\Pictures\\" + fileName); formContent.Add(new StreamContent(fileStream), fileName, fileName); using (var client = new HttpClient()) { // Bearer Token header if needed client.DefaultRequestHeaders.Add("Authorization", "Bearer " + _bearerToken); client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("multipart/form-data")); try { // 4.. Execute the MultipartPostMethod var message = await client.PostAsync(url, formContent); // 5.a Receive the response result = await message.Content.ReadAsStringAsync(); } catch (Exception ex) { // Do what you want if it fails. throw ex; } } } // 5.b Process the reponse Get a usable object from the JSON that is returned MyObject myObject = JsonConvert.DeserializeObject<MyObject>(result);

In my case I need to do something with the object after it posts so I convert it to that object with JsonConvert.

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2 Comments

what is content? The same as formContent?
@dube yes, when I wrote the answer I slipped and put content instead of formContent. I corrected it. Thanks for pointing that out
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Specify the third parameter which is a fileName.

Something like this:

method.Add(streamContent, "filename", "filename.pdf");
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As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
Seems pretty clear to me.
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I debugged this the problem is here:

method.Add(streamContent, "filename");

This 'Add' doesn't actually put the file in the BODY of Multipart Content.

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I know this is an old post But to those searching for a solution, to provide a more direct answer, here's what I've found:

using System.Diagnostics; using System.Net; using System.Net.Http; using System.Threading.Tasks; using System.Web; using System.Web.Http; public class UploadController : ApiController { public async Task<HttpResponseMessage> PostFormData() { // Check if the request contains multipart/form-data. if (!Request.Content.IsMimeMultipartContent()) { throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType); } string root = HttpContext.Current.Server.MapPath("~/App_Data"); var provider = new MultipartFormDataStreamProvider(root); try { // Read the form data. await Request.Content.ReadAsMultipartAsync(provider); // This illustrates how to get the file names. foreach (MultipartFileData file in provider.FileData) { Trace.WriteLine(file.Headers.ContentDisposition.FileName); Trace.WriteLine("Server file path: " + file.LocalFileName); } return Request.CreateResponse(HttpStatusCode.OK); } catch (System.Exception e) { return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e); } } }

Here's where I found it: http://www.asp.net/web-api/overview/advanced/sending-html-form-data,-part-2

For a more Elaborate implementation: http://galratner.com/blogs/net/archive/2013/03/22/using-html-5-and-the-web-api-for-ajax-file-uploads-with-image-preview-and-a-progress-bar.aspx

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6 Comments

This reads multipart/form-data, it doesn't send it. I wonder if you read the question, which doesn't even involve ASP.NET Web API
This is server side code. OP asked for a client side code that sends the request.
@SOusedtobegood I know this comment is old, but I haven't logged in in a while. Um, the question reads C# so .NET is implied and that code is not from ASP, it's from an MVC project I was working on. O_o
@MostafaZeinali dude, when have you ever seen C# client side? You have really got to read the questions before posting comments. the question clearly states "Can someone have a look at my C# code to see if I have written the code incorrectly?". o_O
@iuppiter I have seen C# client side many times my friend. Any client C# application that wants to send/receive data to/from a server needs to write client side code. I also saw it one more time here, in the authors question, in the line: "client.PostAsync(weblinkUrl, method);" This is a client side code that is trying to send a post request to a server. Plain and simple. Your code, on the other hand, is a server side code, that receives a multipart post request and "reads" the attached file from it. You and the author could create a web app together, you do server side, he does client side.
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I know this is old, but I only want to give a clearer example. Variable image is a byte[] and filename is an string with the name of the image:

 ByteArrayContent imageContent = new ByteArrayContent(image); imageContent.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream"); MultipartFormDataContent formData = new MultipartFormDataContent(); pFormData.Add(imageContent, "image", fileName); HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, "destinatioUri"); request.Headers.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json")); request.Content = formData; HttpResponseMessage response = await _httpClient.SendAsync(request);
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