Question

I have been struggling with this issue for almost a month now, and I read everything I could find online, with no solution. Here’s my problem: I’m implementing a client for a RESTful API service that has to send an XML file through a POST call, in vb.net. I am able to make it working when it comes to GET some data, in xml format, but when it comes to send this Xml file, I always get the “400 bad request error”.

I already figured it out that it has to be a matter of the key that has to be passed to the server (that apparently accepts only file uploading for POST, I cannot send it as a string).

Basically this call works with cURL, but I am struggling for implementing my own call in vb.net, passing the right value.

Working cURL call: (that successfully transmits the XML)

  c:>curl -u username:password -F "file=@filename.xml" -X POST http://hostname.com/URI?parameters

Not working Vb.net code: (that gives me 400 Bad Request)

Dim ss As String = "" 'server says...
Dim S As String = txb_username.Text & ":" & txb_password.Text
Dim EncodedString As String = System.Convert.ToBase64String(System.Text.Encoding.UTF8.GetBytes(S))
Dim req As HttpWebRequest = Nothing
Dim res As HttpWebResponse = Nothing

Try
    Dim xmlDoc As System.Xml.XmlDocument = New System.Xml.XmlDocument
    xmlDoc.XmlResolver = Nothing

    xmlDoc.Load("c:\path\file4.xml")

    Dim sXML As String = "file" & xmlDoc.InnerXml '<- This is where I try to put the "KEY"

    Dim url As String = "http:/host.com+URI"

    req = CType(WebRequest.Create(url), Net.HttpWebRequest) 'or Directcast ... 
    req.Method = "POST"
    req.Headers.Add("Authorization: Basic " & EncodedString)
    req.ContentType = "multipart/form-data"
    req.ContentLength = sXML.Length 
    req.Accept = "*/*"

    System.Windows.Forms.Application.DoEvents()

    Dim sw As System.IO.StreamWriter = New System.IO.StreamWriter(req.GetRequestStream)
    StatusUpdate(sXML)
    sw.Write(sXML)
    sw.Close()
    ss = "server says: "
    res = CType(req.GetResponse, HttpWebResponse)
     StatusUpdate(req.ToString)
    Catch ex As Exception
        StatusUpdate(ss & ex.Message)
    Finally

End Try

Is it because I am trying to send it as a string? (but how else can I send it as a file?)
For this I made another procedure that sends the bytes of data, but this one also gives me “400” because (I assume) I did not put the “file” key.

    Dim requestStream As Stream = Nothing
    Dim fileStream As FileStream = Nothing
    Dim uploadResponse As Net.HttpWebResponse = Nothing

    Try

        Dim uploadRequest As Net.HttpWebRequest = CType(Net.HttpWebRequest.Create(URI.Text & Uri_part2.text), Net.HttpWebRequest)

        uploadRequest.Method = Net.WebRequestMethods.Http.Post
        uploadRequest.ContentType = "text/xml; charset=utf-8"
        uploadRequest.Credentials = New NetworkCredential("user", "pass")
        uploadRequest.KeepAlive = True
        uploadRequest.UserAgent = "User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.0.10) Gecko/2009042316 Firefox/3.0.10"
        uploadRequest.Accept = ("text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8")
        uploadRequest.Headers.Add("Accept-Language: en-us,en;q=0.5")
        uploadRequest.Headers.Add("Accept-Encoding: gzip,deflate")
        uploadRequest.Headers.Add("Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7")
        uploadRequest.Headers.Add("Content-Disposition: form-data; name=""file"";")
        uploadRequest.ContentType = "application/xml; charset=utf-8"

        requestStream = uploadRequest.GetRequestStream()
        fileStream = File.Open("C:\example.xml", FileMode.Open)

        Dim a As Integer

        Dim buffer(1024) As Byte
        Dim bytesRead As Integer
        While True
            a = a + 1
            bytesRead = fileStream.Read(buffer, 0, buffer.Length)
            StatusUpdate(buffer(a))

            If bytesRead = 0 Then
                Exit While
            End If
            requestStream.Write(buffer, 0, bytesRead)

        End While

       requestStream.Close()

        uploadResponse = uploadRequest.GetResponse()
        Dim responseReader As StreamReader = New StreamReader(uploadRequest.GetResponse.GetResponseStream())
        Dim x As String = responseReader.ReadToEnd()
        responseReader.Close()
        StatusUpdate(x)


    Catch ex As UriFormatException
        StatusUpdate("UriFormatException: " & ex.Message)
    Catch ex As IOException
        StatusUpdate("IOException: " & ex.Message)
    Catch ex As Net.WebException
        StatusUpdate("Net.WebException: " & ex.Message)

    Finally

        If uploadResponse IsNot Nothing Then
            uploadResponse.Close()
        End If

        If fileStream IsNot Nothing Then
            fileStream.Close()
        End If

        If requestStream IsNot Nothing Then
            requestStream.Close()
        End If

    End Try

In any case, I tried also other 2 clients (POSTMAN and REST Console, 2 extensions for Google Chrome) and I can get it working only if I add the value “file” into the “key” field. I have to insert the specific 4 chars “file” to get it working. So, the question is: how do I add the same value in a Vb.net call? How can I translate the code of the cURL call in working Vb.net code? Thank you very much for your time and help!!!

find image of the thing I want to add here,

P.S. I cannot use PUT, I have to use POST (server restriction)

I also add HTML code that is working for my purpose, with the server from my pc (see the “file” key, again)

<html>
<body>
<form enctype="multipart/form-data" action="http://URI" method="POST">
<table border=0>

<tr>
<td align="right">File&nbsp;</td>
<td><input type="FILE" name="file"></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input type="submit"></td>
</tr>
</table>
</form>
</body>
</html>

Also I paste a script in PERL that is working, too, with the server, from my computer.

  #!perl

use strict;
use LWP; # Loads all important LWP classes

my $client_id         = 1234;
my $filename          = "new_file.xml";

### Prepare to make a request

my $browser = LWP::UserAgent->new;

my $url = "http://uri.com?&xx=$client_id";

my @post_pairs = (
    #'client_id_in' => $client_id,
    'file' => [$filename],
);

my @ns_headers = (
    'User-Agent' => 'Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.0.10) Gecko/2009042316 Firefox/3.0.10',
    'Accept' => 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
    'Accept-Language' => 'en-us,en;q=0.5',
    'Accept-Encoding' => 'gzip,deflate',
    'Accept-Charset' => 'ISO-8859-1,utf-8;q=0.7,*;q=0.7',
    'Authorization' => 'Basic base64EncodedCredentialsHere',
    'Content_Type' => 'form-data',
);

### Make a request
my $response = $browser->post($url, \@post_pairs, @ns_headers);
die "Can't get $url -- ", $response->status_line
        unless $response->is_success;

### Display the response
print STDOUT $response->content;

Answer 1

I think your problem is that you don’t really know how the request should look like, please use fiddler to take a look to the request sent by cURL and try to implement the same using vb.net. In my opinion your secons piece of code (with the buffer, not with the xml serializers) should work… but only if this structure can be understood at the server side.