CODEWITHSUNDEEP
c#asp.netasp.net-web-apidotnet-httpclient
Michael Teper
Michael Teper

Reputation: 4651

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" />
    </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?

Upvotes: 68

Views: 77716

Answers (3)

Aliostad
Aliostad

Reputation: 81660

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);

Upvotes: 11

ThiagoPXP
ThiagoPXP

Reputation: 5462

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.

Upvotes: 14

Michael Teper
Michael Teper

Reputation: 4651

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;
    }
}

Upvotes: 136

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