ASP.NET WebApi: how to perform a multipart post with file upload using WebApi HttpClient

I have a WebApi service handling an upload from a simple form, like this one:

    <form action="/api/workitems" enctype="multipart/form-data" method="post">
        <input type="hidden" name="type" value="ExtractText" />
        <input type="file" name="FileForUpload" />
        <input type="submit" value="Run test" />

However, I can’t figure out how to simulate the same post using the HttpClient API. The FormUrlEncodedContent bit is simple enough, but how do I add the file contents with the name to the post?


Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

After much trial and error, here’s code that actually works:

using (var client = new HttpClient())
    using (var content = new MultipartFormDataContent())
        var values = new[]
            new KeyValuePair<string, string>("Foo", "Bar"),
            new KeyValuePair<string, string>("More", "Less"),

        foreach (var keyValuePair in values)
            content.Add(new StringContent(keyValuePair.Value), keyValuePair.Key);

        var fileContent = new ByteArrayContent(System.IO.File.ReadAllBytes(fileName));
        fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
            FileName = "Foo.txt"

        var requestUri = "/api/action";
        var result = client.PostAsync(requestUri, content).Result;

Method 2

Thank you @Michael Tepper for your answer.

I had to post attachments to MailGun (email provider) and I had to modify it slightly so it would accept my attachments.

var fileContent = new ByteArrayContent(System.IO.File.ReadAllBytes(fileName));
fileContent.Headers.ContentDisposition = 
        new ContentDispositionHeaderValue("form-data") //<- 'form-data' instead of 'attachment'
    Name = "attachment", // <- included line...
    FileName = "Foo.txt",

Here for future reference. Thanks.

Method 3

You need to look for various subclasses of HttpContent.

You create a multiform http content and add various parts to it. In your case you have a byte array content and form url encoded along the lines of:

HttpClient c = new HttpClient();
var fileContent = new ByteArrayContent(new byte[100]);
fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
                                                FileName = "myFilename.txt"

var formData = new FormUrlEncodedContent(new[]
                                                new KeyValuePair<string, string>("name", "ali"),
                                                new KeyValuePair<string, string>("title", "ostad")

MultipartContent content = new MultipartContent();
c.PostAsync(myUrl, content);

All methods was sourced from or, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

0 0 votes
Article Rating
Notify of

Inline Feedbacks
View all comments
Would love your thoughts, please comment.x