post multipart/form-data in c# HttpClient 4.5
15,354
This method works for me. You can use form data and file
public async Task<bool> Upload(FileUploadRequest model)
{
var httpClientHandler = new HttpClientHandler()
{
Proxy = new WebProxy("proxyAddress", "proxyPort")
{
Credentials = CredentialCache.DefaultCredentials
},
PreAuthenticate = true,
UseDefaultCredentials = true
};
var fileContent = new StreamContent(model.File.OpenReadStream())
{
Headers =
{
ContentLength = model.File.Length,
ContentType = new MediaTypeHeaderValue(model.File.ContentType)
}
};
var formDataContent = new MultipartFormDataContent();
formDataContent.Add(fileContent, "File", model.File.FileName); // file
formDataContent.Add(new StringContent("Test Full Name"), "FullName"); // form input
using (var client = new HttpClient(handler: httpClientHandler, disposeHandler: true))
{
client.DefaultRequestHeaders.Add("Authorization", "Bearer " + tokenString);
using (var res = await client.PostAsync("http://filestorageurl", formDataContent))
{
return res.IsSuccessStatusCode;
}
}
}
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Author by
Malik Kashmiri
Updated on June 04, 2022Comments
-
Malik Kashmiri almost 2 years
Problem
I am trying to post API to send data to API which calls my internal API service to send that data to other API i service. Entity contains property with files . this send only file to the other derive but the NameSender property not send with the file.
Entity
public class Email { public string NameSender{ get; set; } public List<IFormFile> Files { get; set; } }
Api
[Consumes("multipart/form-data")] [HttpPost] public IActionResult SendEmail([FromForm]Entity entity) { try { string Servicesfuri = this.serviceContext.CodePackageActivationContext.ApplicationName + "/" + this.configSettings.SendNotificationServiceName; string proxyUrl = $"http://localhost:{this.configSettings.ReverseProxyPort}/{Servicesfuri.Replace("fabric:/", "")}/api/values/Send"; //attachments var requestContent = new MultipartFormDataContent(); foreach (var item in entity.Files) { StreamContent streamContent = new StreamContent(item.OpenReadStream()); var fileContent = new ByteArrayContent(streamContent.ReadAsByteArrayAsync().Result); requestContent.Add(fileContent, item.Name, item.FileName); } HttpResponseMessage response = this.httpClient.PostAsync(proxyUrl, requestContent).Result; if (response.StatusCode != System.Net.HttpStatusCode.OK) { return this.StatusCode((int)response.StatusCode); } return this.Ok(response.Content.ReadAsStringAsync().Result); } catch (Exception e) { throw e; } }