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

Question

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" />
    </form>

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?

Do you mind sharing the code in the WebApi side that consumes the MultipartFormDataContent? — granadaCoder
– granadaCoder, Commented Jan 25, 2016 at 16:49
Example send parameters as multipeart dotnet 3.0 in answer stackoverflow.com/a/59308653/5333683 — D.Oleg
– D.Oleg, Commented Dec 12, 2019 at 16:33

Brijesh Mishra · Accepted Answer · 2013-06-05 07:12:08Z

136

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"
        };
        content.Add(fileContent);

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

edited Jun 5, 2013 at 7:12

Brijesh Mishra

2,7581 gold badge21 silver badges36 bronze badges

answered May 24, 2012 at 19:27

Michael Teper

4,6712 gold badges36 silver badges50 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

Jalal El-Shaer Over a year ago

Use StreamContent for large files: var fileContent = new StreamContent(File.OpenRead(fileName));

Matthew Pitts Over a year ago

Woohoo -- I had a totally different issue, but needed to post content nonetheless. I was able to modify to fit my needs. Thanks a lot!!

deerchao Over a year ago

Strangely this does not work for me. I had to add quotes around the content names, see this: stackoverflow.com/questions/15638622/…

Robert Over a year ago

FYI, there is a constant for "attachment" under System.Net.Mime.DispositionTypeNames

Rob Bird Over a year ago

I too had to add quotes around the name value: FileName = "\"TestImage.jpg\"". I am using ServiceStack on the server instead of WebAPI not sure if that is related or not.

|

ThiagoPXP · Accepted Answer · 2014-09-03 00:34:05Z

14

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",
};
multipartFormDataContent.Add(fileContent);

Here for future reference. Thanks.

answered Sep 3, 2014 at 0:34

ThiagoPXP

5,4823 gold badges34 silver badges45 bronze badges

Comments

Konrad Viltersten · Accepted Answer · 2023-03-12 07:52:39Z

11

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();
content.Add(formData);
content.Add(fileContent);
c.PostAsync(myUrl, content);

edited Mar 12, 2023 at 7:52

Konrad Viltersten

40.2k90 gold badges296 silver badges525 bronze badges

answered Apr 26, 2012 at 20:24

Aliostad

81.9k21 gold badges164 silver badges209 bronze badges

3 Comments

Michael Teper Over a year ago

That doesn't work, unfortunately, although the ContentDisposition header does seems to take me a step forward. If I use your sample code as is, I get an Unsupported Media Type exception. However, if I replace MultipartContent with MultipartFormDataContent the exception goes away. On the server, however, the value of the hidden field is missing.

Aliostad Over a year ago

@MichaelTeper I have not tested the code. I will play a bit with it and update the post, but my point was to show you a snippet of what you need to look for and not a working code.

djack109 Over a year ago

I just get StatusCode: 415, ReasonPhrase: 'Unsupported Media Type'

Collectives™ on Stack Overflow

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

3 Answers 3

7 Comments

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

7 Comments

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related