Http MultipartFormDataContent
Solution 1
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.
Solution 2
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
Solution 3
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.
user2985419
Updated on December 23, 2020Comments
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user2985419 over 3 years
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());