Hjælp til en manuel file post

jeg lavede for et par aar siden denne her kode (den er lidt speciel fordi
den uploader 2 filer, men du kan se lidt af teknikken):

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

public class FileUpload
{
    private const string BOUNDARY = "ArneArne";
    public static void Main(string[] args)
    {
        //upload("http://localhost/upload2.php", "C:\\z.c", "C:\\z.exe");
        upload("http://localhost/upload.php", "C:\\z.c", "C:\\z.exe");
    }
    public static void upload(string url, string textfile, string binfile)
    {
        HttpWebRequest req = (HttpWebRequest)WebRequest.Create(url);
        req.Method = "POST";
        req.ContentType = "multipart/form-data, boundary=" + BOUNDARY;
        StreamWriter post = new StreamWriter(req.GetRequestStream());
        post.WriteLine("--" + BOUNDARY);
        //post.WriteLine("Content-disposition: attachment; name=\"filename[1]\"; filename=\"" + textfile + "\"");
        post.WriteLine("Content-disposition: attachment; name=\"textfile\"; filename=\"" + textfile + "\"");
        post.WriteLine("Content-type: text/plain");
        post.WriteLine("");
        StreamReader txtf = new StreamReader(new FileStream(textfile, FileMode.Open));
        string line;
        while((line = txtf.ReadLine()) != null)
        {
            post.WriteLine(line);
        }
        txtf.Close();
        post.WriteLine("--" + BOUNDARY);
        //post.WriteLine("Content-disposition: attachment; name=\"filename[2]\"; filename=\"" + binfile + "\"");
        post.WriteLine("Content-disposition: attachment; name=\"binfile\"; filename=\"" + binfile + "\"");
        post.WriteLine("Content-type: application/octet-stream");
        post.WriteLine("Content-Transfer-Encoding: binary");
        post.WriteLine("");
        post.Flush();
        Stream binf = new FileStream(binfile, FileMode.Open);
        int c;
        while((c = binf.ReadByte()) >= 0) {
            post.BaseStream.WriteByte((byte)c);
        }
        binf.Close();
        post.BaseStream.Flush();
        post.Close();
        HttpWebResponse resp = (HttpWebResponse)req.GetResponse();
        Console.WriteLine(resp.StatusCode);
        resp.Close();
    }
}

den poster fint til siden men det virker ikke helt og jeg kan ikke gennemskue hvad der er galt.
Koden skal gøre det samme som denne form:

<form action="http://localhost/test.php" method="post"  enctype="multipart/form-data">
<input name="uploadfile" type="file" />
<button type="submit">send</button>
</form>

Denne form virker perfekt, kan man ikke få koden til at gøre det samme ?

Dette er helt plain html, hvis jeg laver en aspx side med den samme form, skal der stå runat="server" og så laver den om action så den ikke peger på det rigtige sted. Lige nu fatter jeg hat !?! :-(

hvordan ser din kode ud og hvilken fejl faar du ?

Lige nu ser min kode således ud

public static void upload(string url, string textfile)
    {
        HttpWebRequest req = (HttpWebRequest)WebRequest.Create(url);
        req.Method = "POST";
        req.ContentType = "multipart/form-data, boundary=" + BOUNDARY;
        StreamWriter post = new StreamWriter(req.GetRequestStream());
        post.WriteLine("--" + BOUNDARY);
        //post.WriteLine("Content-disposition: attachment; name=\"filename[1]\"; filename=\"" + textfile + "\"");
        post.WriteLine("Content-disposition: attachment; name=\"uploadfile\"; filename=\"" + textfile + "\"");
        post.WriteLine("Content-type: text/plain");
        post.WriteLine("");
        StreamReader txtf = new StreamReader(new FileStream(textfile, FileMode.Open));
        string line;
        while ((line = txtf.ReadLine()) != null)
        {
            post.WriteLine(line);
        }
        txtf.Close();
        post.BaseStream.Flush();
        post.Close();
        HttpWebResponse resp = (HttpWebResponse)req.GetResponse();
        Console.WriteLine(resp.StatusCode);
        resp.Close();
    }

jeg kan se at min php side modtager noget, men der sker ingen ting. Når jeg kalder min php side fra formen

<form action="http://localhost/test.php" method="post"  enctype="multipart/form-data">
<input name="uploadfile" type="file" />
<button type="submit">send</button>
</form>

kommer jeg hen på phpsiden og kan se at den skriver ok filen er behandlet, når jeg bruger din kode bliver den kommer jeg aldrig hen på php siden.

Jeg har lige tilrettet lidt kode og testet at det virker.

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

public class FileUpload
{
    private const string BOUNDARY = "ArneArne";
    public static void Main(string[] args)
    {
        upload("http://localhost/upload.php", "C:\\z.c");
    }
    public static void upload(string url, string textfile)
    {
        HttpWebRequest req = (HttpWebRequest)WebRequest.Create(url);
        req.Method = "POST";
        req.ContentType = "multipart/form-data, boundary=" + BOUNDARY;
        StreamWriter post = new StreamWriter(req.GetRequestStream());
        post.WriteLine("--" + BOUNDARY);
        post.WriteLine("Content-disposition: form-data; name=\"f\"; filename=\"" + textfile + "\"");
        post.WriteLine("Content-type: text/plain");
        post.WriteLine("");
        StreamReader txtf = new StreamReader(new FileStream(textfile, FileMode.Open));
        string line;
        while ((line = txtf.ReadLine()) != null)
        {
            post.WriteLine(line);
        }
        txtf.Close();
        post.BaseStream.Flush();
        post.WriteLine("--" + BOUNDARY + "--");
        post.Close();
        HttpWebResponse resp = (HttpWebResponse)req.GetResponse();
        Console.WriteLine(resp.StatusCode);
        resp.Close();
    }
}

<?php
$uploaddir = $_SERVER['DOCUMENT_ROOT'] . "/upload/";
$uploadfile = $uploaddir . basename($_FILES['f']['name']);
move_uploaded_file($_FILES['f']['tmp_name'], $uploadfile);
?>

yes det virker :-)

Hvis jeg nu skal have et hiddenfield med, hvordan gør jeg så?

Smid lige et svar så du kan få point

svar

studer foelgende 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();
    }
}

takker