Ciao,
se puoi utilizzare il framework 4.5 dai un'occhiata a questi esempi che usano la classe HttpClient, che ha una API più semplice da usare rispetto a quella dell'HttpWebRequest.
http://stackoverflow.com/questions/16416601/c-sharp-httpclient-4-5-multipart-form-data-upload

E' comunque fattibile anche con HttpWebRequest, come vedi in questo esempio (un po' meno leggibile rispetto agli altri).
http://stackoverflow.com/questions/566462/upload-files-with-httpwebrequest-multipart-form-data


ciao,
Moreno

In sostanza dovrei convertire questa routine java in vb.net :

InputStream parametersStream, String parametersFileName,
InputStream indexStream, String indexMimeType, String indexFileName,
InputStream dataStream, String dataMimeType, String dataFileName) {
try {
MultipartEntity requestEntity = new MultipartEntity();
requestEntity.addPart("PARAMFILE", new InputStreamBody(parametersStream,
"conserve.xml"));
requestEntity.addPart(?INDEXFILE?, new InputStreamBody(indexStream, indexMimeType,
indexFileName));
requestEntity.addPart(?DATAFILE?, new InputStreamBody(dataStream, dataMimeType,
dataFileName));
// creazione request
HttpResponse response = this.restInvoker.callWS(conserveResourceUrl,
HttpMethod.POST, requestEntity);
if (response.getStatusLine().getStatusCode() == 201) {
// chiamata conserve ok
logger.debug("Documento conservato.");
} else
logger.debug("Documento non conservato.");
indexStream.close();
dataStream.close();
parametersStream.close();
}
catch (Exception ex) {//gestire eccezione}
}

Non riesco a trovare MultipartEntity , mi dice tipo non definito ???? che Imports devo aggiungere ?

Ma possibile che sia così complesso ?

Innanzitutto lavoro con il framework 2... non so se è utile ...

Ho creato questa routine (che non funziona)

Dim wrequest As Net.HttpWebRequest

Dim boundary As String = IO.Path.GetRandomFileName
Dim header As New System.Text.StringBuilder()
Dim Element As Object
Dim sNames(0 To 2) As String

sNames(0) = strConserveTmpPath & "conserve.xml"
sNames(1) = strConserveTmpPath & "index.xml"
sNames(2) = strConserveTmpPath & "testconservazione1.pdf"

For Each Element In sNames

If File.Exists(Element) Then

header.AppendLine("--" & boundary)
header.Append("Content-Disposition: form-data; name=""files[]"";")
header.AppendFormat("filename=""{0}""", IO.Path.GetFileName(Element))
header.AppendLine()
header.AppendLine("Content-Type: application/octet-stream")
header.AppendLine()

Dim headerbytes() As Byte = System.Text.Encoding.UTF8.GetBytes(header.ToString)
Dim endboundarybytes() As Byte = System.Text.Encoding.ASCII.GetBytes(vbNewLine & "--" & boundary & "--" & vbNewLine)

wrequest = Net.WebRequest.Create(sURL)
wrequest.ContentType = "multipart/form-data; boundary=" & boundary
wrequest.Method = "POST"
wrequest.ContentLength = headerbytes.Length + New IO.FileInfo(Element).Length + endboundarybytes.Length
wrequest.KeepAlive = True
wrequest.Headers("LDSessionID") = strldSessionID

Dim s As IO.Stream = wrequest.GetRequestStream
s.Write(headerbytes, 0, headerbytes.Length)
Dim filebytes() As Byte = My.Computer.FileSystem.ReadAllBytes(Element)
s.Write(filebytes, 0, filebytes.Length)
s.Write(endboundarybytes, 0, endboundarybytes.Length)
s.Close()

End If
Next

Dim wResponse As Net.HttpWebResponse = wrequest.GetResponse()

L'errore che compare

Errore del server remoto: (400) Richiesta non valida.

Come posso capire cosa non è corretto ?

Ho fatto come mi hai detto ... in effetti della routine segnalata non ho dovuto modificare nulla se non aggiungere il parametro da passare al server per l'autenticazione.
Ho dovuto anche aggiungere la funzione :

Private Shared Function InlineAssignHelper(Of T)(ByRef target As T, ByVal value As T) As T
target = value
Return value
End Function

Mi segnalava l'errore (nome "InlineAssignHelper" no dichiarato)

Public Shared Sub UploadFilesToRemoteUrl(ByVal url As String, ByVal files As String(), ByVal logpath As String, ByVal nvc As NameValueCollection, ByVal strldSessionID As String)

Dim length As Long = 0
Dim boundary As String = "----------------------------" + DateTime.Now.Ticks.ToString("x")


Dim httpWebRequest2 As HttpWebRequest = DirectCast(WebRequest.Create(url), HttpWebRequest)
httpWebRequest2.ContentType = Convert.ToString("multipart/form-data; boundary=") & boundary
httpWebRequest2.Method = "POST"
httpWebRequest2.KeepAlive = True
httpWebRequest2.Headers("LDSessionID") = strldSessionID

Dim memStream As Stream = New System.IO.MemoryStream()

Dim boundarybytes As Byte() = System.Text.Encoding.ASCII.GetBytes((Convert.ToString(vbCr & vbLf & "--") & boundary) + vbCr & vbLf)


Dim formdataTemplate As String = (Convert.ToString(vbCr & vbLf & "--") & boundary) + vbCr & vbLf & "Content-Disposition: form-data; name=""{0}"";" & vbCr & vbLf & vbCr & vbLf & "{1}"

For Each key As String In nvc.Keys
Dim formitem As String = String.Format(formdataTemplate, key, nvc(key))
Dim formitembytes As Byte() = System.Text.Encoding.UTF8.GetBytes(formitem)
memStream.Write(formitembytes, 0, formitembytes.Length)
Next

memStream.Write(boundarybytes, 0, boundarybytes.Length)

Dim headerTemplate As String = "Content-Disposition: form-data; name=""{0}""; filename=""{1}""" & vbCr & vbLf & " Content-Type: application/octet-stream" & vbCr & vbLf & vbCr & vbLf

For i As Integer = 0 To files.Length - 1

'string header = string.Format(headerTemplate, "file" + i, files[i]);
Dim header As String = String.Format(headerTemplate, "uplTheFile", files(i))

Dim headerbytes As Byte() = System.Text.Encoding.UTF8.GetBytes(header)

memStream.Write(headerbytes, 0, headerbytes.Length)


Dim fileStream As New FileStream(files(i), FileMode.Open, FileAccess.Read)
Dim buffer As Byte() = New Byte(1023) {}

Dim bytesRead As Integer = 0

While (InlineAssignHelper(bytesRead, fileStream.Read(buffer, 0, buffer.Length))) <> 0

memStream.Write(buffer, 0, bytesRead)
End While


memStream.Write(boundarybytes, 0, boundarybytes.Length)


fileStream.Close()
Next

httpWebRequest2.ContentLength = memStream.Length

Dim requestStream As Stream = httpWebRequest2.GetRequestStream()

memStream.Position = 0
Dim tempBuffer As Byte() = New Byte(memStream.Length - 1) {}
memStream.Read(tempBuffer, 0, tempBuffer.Length)
memStream.Close()
requestStream.Write(tempBuffer, 0, tempBuffer.Length)
requestStream.Close()


Dim webResponse2 As WebResponse = httpWebRequest2.GetResponse()

Dim stream2 As Stream = webResponse2.GetResponseStream()
Dim reader2 As New StreamReader(stream2)


MessageBox.Show(reader2.ReadToEnd())

webResponse2.Close()
httpWebRequest2 = Nothing
webResponse2 = Nothing
End Sub

Mi restituisce sempre l'errore "Errore del server remoto : 8400) richiesta non valida"

Ora provo ad utilizzare il programma che mi hai indicato

Grazie