FileWebRequest not returning what it should

Question

I am trying to send a file from my machine to the browser, so that it can be downloaded, in a .NET application I am working on. I am using the code in the this SO answer, but instead of using an HttpWebRequest I am using a FileWebRequest because I am accessing the file on my local machine. The request looks like this: FileWebRequest fileReq = (FileWebRequest)WebRequest.Create(@"file:///C:/Tmp/new.html"); and when I copy the url file:///C:/Tmp/new.html into the browser, it gives me the correct file. But when I use fileReq.ContentLength in my code it always returns 0, which leads me to believe that the file is not being read for some reason. Can anyone tell me what's going on here?

EDIT: Here's my code, like I said exactly like from the other SO question, but I used FileWebRequest instead of HttpWebRequest.

        Stream stream = null;
        int bytesToRead = 10000;
        byte[] buffer = new Byte[bytesToRead];
        try
        {               
            FileWebRequest fileReq = (FileWebRequest)WebRequest.Create(@"file:///C:/Tmp/new.html");
            FileWebResponse fileResp = (FileWebResponse)fileReq.GetResponse();

            if (fileReq.ContentLength > 0)
            {
                fileResp.ContentLength = fileReq.ContentLength;
                stream = fileResp.GetResponseStream();
                var resp = HttpContext.Current.Response;
                resp.ContentType = "application/octet-stream";
                resp.AddHeader("Content-dsiposition", "attachment; filename=" + url);
                resp.AddHeader("Content-Length", fileResp.ContentLength.ToString());

                int length;
                do
                {
                    if (resp.IsClientConnected)
                    {
                        length = stream.Read(buffer, 0, bytesToRead);
                        resp.OutputStream.Write(buffer, 0, length);
                        resp.Flush();
                        buffer = new Byte[bytesToRead];
                    }
                    else
                    {
                        length = -1;
                    }
                } while (length > 0);
            }
        }
        catch (Exception ex)
        {
            FileLabel.Text = ex.Message.ToString();
        }
        finally
        {
            if (stream != null)
            {
                stream.Close();
            }
        }

Why don't you just open the file and feed it into the response? Why "open" it with a request/response pair? Have you considered Response.TransmitFile? — Alexander, Jun 12 '13 at 19:35
@Alexander this sounds like what I want to do, could you elaborate on how to do this? — azrosen92, Jun 12 '13 at 19:42

score 0 · Answer 1 · answered Jun 12 '13 at 19:31

Try WebClient class.

 public static void Main (string[] args)
{
    if (args == null || args.Length == 0)
    {
        throw new ApplicationException ("Specify the URI of the resource to retrieve.");
    }
    WebClient client = new WebClient ();

    // Add a user agent header in case the 
    // requested URI contains a query.

    client.Headers.Add ("user-agent", "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.2; .NET CLR 1.0.3705;)");

    Stream data = client.OpenRead (args[0]);
    StreamReader reader = new StreamReader (data);
    string s = reader.ReadToEnd ();
    Console.WriteLine (s);
    data.Close ();
    reader.Close ();

tutorial here

score 0 · Answer 2 · answered Jun 12 '13 at 19:56

0

Start here: http://msdn.microsoft.com/en-us/library/12s31dhy(v=vs.110).aspx This method just accepts the file path and does the rest for you.

answered Jun 12 '13 at 19:56

Alexander

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how do I get the HttpResponse object? – azrosen92 Jun 12 '13 at 19:58
Like you do in your own code. Or just use the Page's Response. – Alexander Jun 12 '13 at 20:04
ok, that puts the contents of the file on that actual page. I want the user to be able to download the actual file, but I feel like I'm getting close. – azrosen92 Jun 12 '13 at 20:07

FileWebRequest not returning what it should

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