Sending Files using HTTP POST in c# [closed]

Answer 1

Using .NET 4.5 (or .NET 4.0 by adding the Microsoft.Net.Http package from NuGet) there is an easier way to simulate form requests. Here is an example:

private System.IO.Stream Upload(string actionUrl, string paramString, Stream paramFileStream, byte [] paramFileBytes)
{
    HttpContent stringContent = new StringContent(paramString);
    HttpContent fileStreamContent = new StreamContent(paramFileStream);
    HttpContent bytesContent = new ByteArrayContent(paramFileBytes);
    using (var client = new HttpClient())
    using (var formData = new MultipartFormDataContent())
    {
        formData.Add(stringContent, "param1", "param1");
        formData.Add(fileStreamContent, "file1", "file1");
        formData.Add(bytesContent, "file2", "file2");
        var response = client.PostAsync(actionUrl, formData).Result;
        if (!response.IsSuccessStatusCode)
        {
            return null;
        }
        return response.Content.ReadAsStreamAsync().Result;
    }
}

Answer 2

try this

string fileToUpload = @"c:\user\test.txt";
string url = "http://example.com/upload";
using (var client = new WebClient())
{
byte[] result = client.UploadFile(url, fileToUpload);
string responseAsString = Encoding.Default.GetString(result);
}

Answer 3

You can try the following code:

    public void PostMultipleFiles(string url, string[] files)
{
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(url);
    httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
    httpWebRequest.Method = "POST";
    httpWebRequest.KeepAlive = true;
    httpWebRequest.Credentials = System.Net.CredentialCache.DefaultCredentials;
    Stream memStream = new System.IO.MemoryStream();
    byte[] boundarybytes =System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary     +"\r\n");
    string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition:  form-data; name=\"{0}\";\r\n\r\n{1}";
    string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";
    memStream.Write(boundarybytes, 0, boundarybytes.Length);
    for (int i = 0; i < files.Length; i++)
    {
        string header = string.Format(headerTemplate, "file" + i, files[i]);
        //string header = string.Format(headerTemplate, "uplTheFile", files[i]);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        memStream.Write(headerbytes, 0, headerbytes.Length);
        FileStream fileStream = new FileStream(files[i], FileMode.Open,
        FileAccess.Read);
        byte[] buffer = new byte[1024];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            memStream.Write(buffer, 0, bytesRead);
        }
        memStream.Write(boundarybytes, 0, boundarybytes.Length);
        fileStream.Close();
    }
    httpWebRequest.ContentLength = memStream.Length;
    Stream requestStream = httpWebRequest.GetRequestStream();
    memStream.Position = 0;
    byte[] tempBuffer = new byte[memStream.Length];
    memStream.Read(tempBuffer, 0, tempBuffer.Length);
    memStream.Close();
    requestStream.Write(tempBuffer, 0, tempBuffer.Length);
    requestStream.Close();
    try
    {
        WebResponse webResponse = httpWebRequest.GetResponse();
        Stream stream = webResponse.GetResponseStream();
        StreamReader reader = new StreamReader(stream);
        string var = reader.ReadToEnd();

    }
    catch (Exception ex)
    {
        response.InnerHtml = ex.Message;
    }
    httpWebRequest = null;
}