Send file+parameters in post request

Question

I\'m using this code to send parameters to a webpage and getting correct response from it.

System.Net.WebClient oWeb = new System.Net.WebClient();
oWeb.Proxy         


        

Answer 1

Use WebClient.QueryString to pass name/value pairs associated with the request.

NameValueCollection parameters = new NameValueCollection();
parameters.Add("value1", "123");
parameters.Add("value2", "xyz");
oWeb.QueryString = parameters;
var responseBytes = oWeb.UploadFile("http://website.com/file.php", "path to file");
string response = Encoding.ASCII.GetString(responseBytes);

Answer 2

    public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
        Stream rs = wr.GetRequestStream();
        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            rs.Write(boundarybytes, 0, boundarybytes.Length);
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);
        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, file, contentType);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);
        FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
        byte[] buffer = new byte[4096];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            rs.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();
        byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
        rs.Write(trailer, 0, trailer.Length);
        rs.Close();
        WebResponse wresp = null;
        try
        {
            wresp = wr.GetResponse();
            Stream stream2 = wresp.GetResponseStream();
            StreamReader reader2 = new StreamReader(stream2);
            result = reader2.ReadToEnd();
        }
        catch (Exception ex)
        {
            System.Windows.MessageBox.Show("Error occurred while converting file", "Error!");
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }

Copied from SO but can't remember it's link. And this is how it will be used

        NameValueCollection nvc = new NameValueCollection();
        nvc.Add("parm1", "value1");
        nvc.Add("parm2", "value2");
        nvc.Add("parm3", "value3");
        HttpUploadFile("http://www.example.com/upload.php",@filepath, "file", "text/html", nvc);

Here @filepath is the path of your file you want to upload: c:\file_to_upload.doc And file is the name of the filed used in php as $_Files['file']