Upload file via POST to WebAPI

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Upload file via POST to WebAPI

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#1 by (6 votes)

5

I need to upload a file to WebAPI , I'm using the following code to upload

public void Enviar()
{   
    WebRequest request = WebRequest.Create(url);
    request.Method = "POST";
    byte[] byteArray = File.ReadAllBytes(fileName);
    request.ContentType = "multipart/form-data";
    request.ContentLength = byteArray.Length;

    Stream dataStream = request.GetRequestStream();
    dataStream.Write(byteArray, 0, byteArray.Length);
    dataStream.Close();
    WebResponse response = request.GetResponse();
    Console.WriteLine(((HttpWebResponse)response).StatusDescription);
    reader.Close();
    dataStream.Close();
    response.Close();
}

In WebAPI I have the following code:

public async Task<HttpResponseMessage> Post()
    {
        if (!Request.Content.IsMimeMultipartContent("form-data"))
            return new HttpResponseException(HttpStatusCode.UnsupportedMediaType);

        var streamProvider =
            new MultipartFormDataStreamProvider(".");

        await Request.Content.ReadAsMultipartAsync(streamProvider);

        var fileNames = streamProvider.BodyPartFileNames;

        foreach (var fileName in fileNames.Keys)
            Console.WriteLine(fileName + " --> " + fileNames[fileName]);

        return new HttpResponseMessage(HttpStatusCode.Created);
    }

When I run, I'm always getting

HttpStatusCode.UnsupportedMediaType

c# asp.net-web-api

asked by anonymous 16.09.2016 / 14:17

1 answer

Why are you giving this static variable error? How to only get the number after the comma

score 6 · Accepted Answer

With # , writing a text file to a Controller WebApi :

Sending:

[HttpGet]
public void Enviar()
{
    string fileName = Server.MapPath("~") + "/Files/arq.txt";

    using (HttpClient client = new HttpClient())
    using (MultipartFormDataContent content = new MultipartFormDataContent())
    using (FileStream fileStream = System.IO.File.OpenRead(fileName))
    using (StreamContent fileContent = new StreamContent(fileStream))
    {

        fileContent.Headers.ContentType = MediaTypeHeaderValue.Parse("text/plain");
        fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
        {
            FileName = "arq.txt",                   
        };
        fileContent.Headers.Add("name", "arq.txt");
        content.Add(fileContent);
        var result = client.PostAsync("http://localhost:62951/api/arquivo", content).Result;
        result.EnsureSuccessStatusCode();
    }
}

Receiving:

public Task<HttpResponseMessage> Post()
{
    HttpRequestMessage request = Request;

    if (!request.Content.IsMimeMultipartContent())
    {
        throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
    }
    string root = System.Web.HttpContext.Current.Server.MapPath("~/Data/");
    MultipartFormDataStreamProvider provider = new CustomMultipartFormDataStreamProvider(root);            
    var task = request.Content.ReadAsMultipartAsync(provider);            
    return task.ContinueWith(o =>
        {
            return new HttpResponseMessage()
            {
                Content = new StringContent("File uploaded.")
            };
        }
    );
}

To save the file with the given name implements the class:

public class CustomMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
{
    public CustomMultipartFormDataStreamProvider(string path) 
        : base(path) { }
    public override string GetLocalFileName(HttpContentHeaders headers)
    {
        return headers.ContentDisposition.FileName;
    }
}