CODEWITHSUNDEEP
c#httppostasp.net-coremultipartform-data
Deivydas Voroneckis
Deivydas Voroneckis

Reputation: 2513

How to post form-data IFormFile with HttpClient?

I have backend endpoint Task<ActionResult> Post(IFormFile csvFile) and I need to call this endpoint from HttpClient. Currently I am getting Unsupported media type error. Here is my code:

var filePath = Path.Combine("IntegrationTests", "file.csv");
var gg = File.ReadAllBytes(filePath);
var byteArrayContent = new ByteArrayContent(gg);
var postResponse = await _client.PostAsync("offers", new MultipartFormDataContent
{
    {byteArrayContent }
});

Upvotes: 31

Views: 51037

Answers (6)

Anubis77
Anubis77

Reputation: 142

Please see the following working code with .NET 5.0 Environment.

You send the file as a byte[] and receive it as a IFormFile in the API.

//api controller receiver

[HttpPost("SendBackupFiles")]
public IActionResult SendBackupFiles(IFormFile file)
{
    var filePath = Path.GetTempFileName();

    using (var stream = System.IO.File.Create(filePath)) 
        file.CopyToAsync(stream);
}



//client code sender example, not optimized lol.

const string filePath = @"C:\temp\hallo.csv";

using (var httpClient = new HttpClient())
{
    var form = new MultipartFormDataContent();
               
    byte[] fileData = File.ReadAllBytes(filePath);

    ByteArrayContent byteContent = new ByteArrayContent(fileData);

    byteContent.Headers.ContentType = MediaTypeHeaderValue.Parse("multipart/form-data");
    
    form.Add(byteContent, "file", Path.GetFileName(filePath));

    var result = httpClient.PostAsync("http://localhost:5070/..yourControllerName.../SendBackupFiles", form).ConfigureAwait(false).GetAwaiter().GetResult().Content.ReadAsStringAsync().Result;
}

Upvotes: 3

Enrico
Enrico

Reputation: 3479

Post the attachment as an MultipartFormDataContent

var type = typeof(Startup);
            var stream = type.Assembly.GetManifestResourceStream(type, "Resources.New.docx");
        var fileContent = new StreamContent(stream);
var data = new MultipartFormDataContent
            {
                { fileContent, "file", "New.docx" }
            };

        var response = await _client.PostAsync("upload", multipartContent);

Source: https://medium.com/@woeterman_94/c-webapi-upload-files-to-a-controller-e5ccf288b0ca

Upvotes: 2

Danilo Popovikj
Danilo Popovikj

Reputation: 157

This worked for me as a generic

public static Task<HttpResponseMessage> PostFormDataAsync<T>(this HttpClient httpClient, string url, string token, T data)
    {
        var content = new MultipartFormDataContent();

        foreach (var prop in data.GetType().GetProperties())
        {
            var value = prop.GetValue(data);
            if (value is FormFile)
            {
                var file = value as FormFile;
                content.Add(new StreamContent(file.OpenReadStream()), prop.Name, file.FileName);
                content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data") { Name = prop.Name, FileName = file.FileName };
            }
            else
            {
                content.Add(new StringContent(JsonConvert.SerializeObject(value)), prop.Name);
            }
        }

        if (!string.IsNullOrWhiteSpace(token))
            httpClient.DefaultRequestHeaders.Authorization = new AuthenticationHeaderValue("Bearer", token);
        return httpClient.PostAsync(url, content);
    }

Upvotes: 6

Deivydas Voroneckis
Deivydas Voroneckis

Reputation: 2513

Solved by using this code:

        const string fileName = "csvFile.csv";
        var filePath = Path.Combine("IntegrationTests", fileName);
        var bytes = File.ReadAllBytes(filePath);
        var form = new MultipartFormDataContent();
        var content = new StreamContent(new MemoryStream(bytes));
        form.Add(content, "csvFile");
        content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
        {
            Name = "csvFile",
            FileName = fileName
        };
        content.Headers.Remove("Content-Type");
        content.Headers.Add("Content-Type", "application/octet-stream; boundary=----WebKitFormBoundaryMRxYYlVt8KWT8TU3");
        form.Add(content);

        //Act
        var postResponse = await _sellerClient.PostAsync("items/upload", form);

Upvotes: 4

Alexander
Alexander

Reputation: 9642

You need to specify parameter name in MultipartFormDataContent collection matching action parameter name (csvFile) and a random file name

var multipartContent = new MultipartFormDataContent();
multipartContent.Add(byteArrayContent, "csvFile", "filename");
var postResponse = await _client.PostAsync("offers", multipartContent);

or equivalent

var postResponse = await _client.PostAsync("offers", new MultipartFormDataContent {
    { byteArrayContent, "csvFile", "filename" }
});

Upvotes: 31

Moien Tajik
Moien Tajik

Reputation: 2321

Use this snippet:

const string url = "https://localhost:5001/api/Upload";
const string filePath = @"C:\Path\To\File.png";

using (var httpClient = new HttpClient())
{
    using (var form = new MultipartFormDataContent())
    {
        using (var fs = File.OpenRead(filePath))
        {
            using (var streamContent = new StreamContent(fs))
            {
                using (var fileContent = new ByteArrayContent(await streamContent.ReadAsByteArrayAsync()))
                {
                    fileContent.Headers.ContentType = MediaTypeHeaderValue.Parse("multipart/form-data");

                    // "file" parameter name should be the same as the server side input parameter name
                    form.Add(fileContent, "file", Path.GetFileName(filePath));
                    HttpResponseMessage response = await httpClient.PostAsync(url, form);
                }
            }
        }
    }
}

Upvotes: 13

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