Posting IFormFile to webAPI from c#

Question

I've got a Web API method which accepts a list of IFormFile variables within a small class structure, to upload the files to a storage account.

public class FileInputModel
{
        public int ScenarioId { get; set; }
        public IList<IFormFile> UploadFiles { get; set; }
}


[HttpPost("UploadFiles")]
public async Task<IActionResult> UploadForm([FromForm] FileInputModel files)
{
    //UploadLogic
}

This works perfectly for a https post using Postman, but i can't quite seem to figure out how to do this using a C# programme i'm writing to link directly to this api. So far I've got some code to convert a FileStreamResult variable into an IformFile to then send in a post request, but i can't figure out how to get a FileStreamResult from a file on my pc. Here's the method i have so far.

var json = JsonSerializer.Serialize(FileInputModel);
StringContent data = new StringContent(json, Encoding.UTF8, "application/json");

try
{
    using HttpClient client = new HttpClient();

    HttpResponseMessage response = await client.PostAsync(url, data);

    return response;
}

My problem isn't the receiving end, rather how to send the data in any correct form, so that [FromForm] wouldn't change anything since that function works fine as it is currently — HamHam
– HamHam, Commented Nov 10, 2021 at 11:56
Ah so depending on which front-end framework you use to make the call, make sure to add a header of Content-type: multipart/form-data; — LinkedListT
– LinkedListT, Commented Nov 10, 2021 at 15:09

HamHam · Accepted Answer · 2021-11-15 14:01:12Z

6

I was getting too caught up on the IFormFile aspect of the backend, when in reality the function was just opening the stream to then use in further functions. With this I solved it by simply using a filestream in the frontend connecting c# programme and sending the information as a MultipartFormDataContent type.

using (var client = new HttpClient())
    {
        using (var content = new MultipartFormDataContent())
        {
            var fileName = Path.GetFileName(filePath);
            var fileStream = System.IO.File.Open(filePath, FileMode.Open);
            content.Add(new StreamContent(fileStream), "file", fileName);  

            var requestUri = baseURL;
            var request = new HttpRequestMessage(HttpMethod.Post, requestUri) { Content = content };
            var result = await client.SendAsync(request);

            return;
        }
    }

answered Nov 15, 2021 at 14:01

HamHam

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Jagadisha B S Over a year ago

Can you post the API Method sample code

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