HTTP POST file upload in C#
Sending files via HTTP POST is a common task in web applications. In C#, you can use HttpWebRequest
classes and MultipartFormDataContent
objects to achieve this functionality.
Create HttpWebRequest object
First, create a HttpWebRequest
object and specify the URI of the web server. Then, set the Method
attribute to "POST" and the ContentType
attribute to the appropriate content type of the file being sent.
HttpWebRequest req = WebRequest.Create(uri) as HttpWebRequest; req.Method = "POST"; req.ContentType = file.ContentType;
File ready for upload
To send a file, create a MultipartFormDataContent
object and include it in the request. MultipartFormDataContent
Allows you to send both form data and binary data in one request.
using (var formData = new MultipartFormDataContent()) { formData.Add(new StringContent(file.Name), "file"); formData.Add(new StreamContent(file.Content), file.Name, file.Name); }
Send request
Finally, use GetResponse()
or GetResponseAsync()
to send the request to the web server.
HttpWebResponse response = null; try { response = req.GetResponse() as HttpWebResponse; } catch (Exception e) { // 處理異常 }
Alternatives for .NET 4.0 and below
For .NET 4.0 and below, you can use the Microsoft.Net.Http
package from NuGet to simplify the file upload process:
using System.Net.Http; using System.Net.Http.Headers; private async Task<System.IO.Stream> Upload(string actionUrl, string paramString, Stream paramFileStream, byte[] paramFileBytes) { using (var client = new HttpClient()) { using (var formData = new MultipartFormDataContent()) { formData.Add(new StringContent(paramString), "param1"); formData.Add(new StreamContent(paramFileStream), "file1"); formData.Add(new ByteArrayContent(paramFileBytes), "file2"); var response = await client.PostAsync(actionUrl, formData); if (!response.IsSuccessStatusCode) { return null; } return await response.Content.ReadAsStreamAsync(); } } }
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