File Upload in API. Dotnet core Web controller or WPF Windows App

Question

User1064314092 posted

I am trying to post File in API. in the Dotnet core Web controller. and Successfully upload Files in dot net core. but i am Tired in asp.net. in from data

API Controller

public class TestDocumentModel
{
    public int Id { get; set; }
    public string Name { get; set; }
    public IFormFile formFiles { get; set; }
}

API Controller

public async Task<IActionResult> UploadModelFiles([FromForm]TestDocumentModel models)
    {

        var filePath = Path.Combine(@"Custome File Save Path");

        if (models.formFiles != null)
        {
            long sizeOfFile = models.formFiles.Length;
            if (filePath.Length > 0)
            {
                if (!Directory.Exists(filePath))
                {
                    Directory.CreateDirectory(filePath);
                
                   if (Directory.Exists(filePath))
                   {
                      using (var fileStream = new FileStream(Path.Combine(filePath, models.formFiles.FileName), FileMode.Create))
                      {
                        await models.formFiles.CopyToAsync(fileStream);
                      }
                  }
              }
            }
            return Ok(new { sizeOfFile, filePath });
        }
        else
        {
            long fileSize = 0;
            filePath = "File note Found";
            return Ok(new { fileSize, filePath });
        }
    }

Web Controller or WPF Windows Application Class

obj = new TestModelClass();

if (dlg != null)
{
using (var httpClient = new HttpClient())
{
string fileName = dlg.SafeFileName;
using (var content = new MultipartFormDataContent())
{
obj.Id = int.Parse(tb_id.Text);
obj.Name = tb_name.Text;
FileStream fileStream = File.OpenRead(dlg.FileName);
HttpContent fileStreamContent = new StreamContent(fileStream);

content.Add(new StringContent(Convert.ToString(obj.Id)), "Id");
content.Add(new StringContent(Convert.ToString(obj.Name)), "Name");
fileStreamContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data") { Name = "formFiles", FileName = fileName };
fileStreamContent.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");

content.Add(fileStreamContent);

using (var response = await httpClient.PostAsync("https://localhost:44359/weatherforecast/UploadModelFiles", content))
{
string apiResponse = await response.Content.ReadAsStringAsync();
}
}
}
}

File Post in Api With postman 

https://i.stack.imgur.com/kEKpR.png

File Post in Api With WPF or Web app

https://i.stack.imgur.com/Rkyhp.png

Answer 1

User-474980206 posted

you are doing a multipart form upload. each form field and file is a separate part (with boundary and mime-header). StringContent() is for text data (text files or text fields). you need to use one of the binary content types. say StreamContent() or ByteArrayContent() 

Answer 2

User1064314092 posted

I already try this way.

    byte[] data;
    using (var br = new BinaryReader(dilog.FileName.OpenReadStream())) 
    {
        data = br.ReadBytes((int)dilog.FileName.OpenReadStream().Length);
    }
   ByteArrayContent bytes = new ByteArrayContent(data);

////This Line
   content.Add(data,"formFiles",dilog.FileName);

Please can you provide the source code or link ? thanks

Answer 3

User665608656 posted

Hi rezaulkhan111,

Your current problem is not file conversion, but the mismatch between the name of the parameters your API accepts and the one you pass.

You need to change the receiving parameter name of the current API Controller method to be consistent with the name parameter value of MultipartFormDataContent.

API Controller:

public async Task<IActionResult> UploadModelFiles([FromForm]TestDocumentModel formFiles)
{
 ....
}

   byte[] data;
    using (var br = new BinaryReader(dilog.FileName.OpenReadStream())) 
    {
        data = br.ReadBytes((int)dilog.FileName.OpenReadStream().Length);
    }
   ByteArrayContent bytes = new ByteArrayContent(data);

////This Line
   content.Add(data,"formFiles",dilog.FileName);

Best Regards,

YongQing.

Answer 4

User1064314092 posted

Thanks For Yor Replay. and now I am Fix the problem