Post'e fil + data til ekstern server

Upload eksempel:

using System;
using System.IO;
using System.Net;

public class DataAndFileUpload
{
    private const string BOUNDARY = "ArneArne";
    public static void Main(string[] args)
    {
        upload("http://localhost:8080/test/xuploaddo.jsp", "test", "C:\\last.cpp");
    }
    public static void upload(string url, string description, string filename)
    {
        ServicePointManager.Expect100Continue = false; // to avoid .NET - Tomcat 5 bug
        HttpWebRequest req = (HttpWebRequest)WebRequest.Create(url);
        req.Method = "POST";
        req.KeepAlive = false;
        req.ContentType = "multipart/form-data; boundary=" + BOUNDARY;
        StreamWriter post = new StreamWriter(req.GetRequestStream());
        post.WriteLine("--" + BOUNDARY);
        post.WriteLine("Content-Disposition: form-data; name=\"beskrivelse\"");
        post.WriteLine("");
        post.WriteLine(description);
        post.WriteLine("--" + BOUNDARY);
        post.WriteLine("Content-Disposition: form-data; name=\"fil\"; filename=\"" + filename + "\"");
        //post.WriteLine("Content-Type: text/plain");
        post.WriteLine("Content-type: application/octet-stream");
        post.WriteLine("Content-Transfer-Encoding: binary");
        post.WriteLine("");
        post.Flush();
        Stream binf = new FileStream(filename, FileMode.Open);
        int c;
        while((c = binf.ReadByte()) >= 0) {
            post.BaseStream.WriteByte((byte)c);
        }
        binf.Close();
        post.BaseStream.Flush();
        post.WriteLine("--" + BOUNDARY + "--");
        post.Close();
        HttpWebResponse resp = (HttpWebResponse)req.GetResponse();
        Console.WriteLine(resp.StatusCode);
        resp.Close();
    }
}

Tak for dit svar, det prøver jeg lige.
Det KAN egentlig bare være den første linie jeg mangler, da det er en Tomcat server :-)

ja, nu er der lige pludselig hul igennem. Det var kun nogle få ting jeg manglede.
Tak for det.

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