HTTP Form Post

[LD]

Hi,

I am trying to upload a zip file and other form data elements to a server
and get a response. The server keeps responding with Malformed Multipart
Post. Can
anyone see what might be wrong here? Any feedback much appreciated.

Dim vbCrLf As String = Convert.ToStrin g(Chr(13)) + Convert.ToStrin g(Chr(10))
Dim vbCr As String = Convert.ToStrin g(Chr(13))
Dim vbLf As String = Convert.ToStrin g(Chr(10))
Dim boundary As String = "-----------------------------7d31151970286"

Dim myWebClient As New System.Net.WebC lient()
Dim URL As String
Dim body As String
Dim body2 As String

'============RE AD FILE CONTENTS INTO BYTE ARRAY========== ==
Dim Tem() As Byte
Dim st As FileStream = File.OpenRead(" data.zip")
ReDim Tem(st.Length)
Do While st.Position < st.Length
st.Read(Tem, 0, Tem.Length - 1)
Loop
st.Close()

'============== ====SET HEADERS======== ==============
myWebClient.Hea ders.Add("Accep t", "image/gif, image/x-xbitmap, image/jpeg,
image/pjpeg, application/vnd.ms-excel, application/msword,
application/x-shockwave-flash, */*")
myWebClient.Hea ders.Add("Accep t-Language", "en-us")
myWebClient.Hea ders.Add("Conte nt-Type", "multipart/form-data; boundary=" +
boundary)
myWebClient.Hea ders.Add("Accep t-Encoding", "gzip, deflate")
myWebClient.Hea ders.Add("User-Agent", "Mozilla/4.0 (compatible; MSIE 6.0;
Windows NT 5.1; Q312461; .NET CLR 1.0.3705)")
myWebClient.Hea ders.Add("Cache-Control", "no-cache")

'============== ====CREATE BODY=========== ===========
body = boundary & vbCrLf & _
"Content-Disposition: " & "form-data; name=""username """ & vbCrLf & vbCrLf &
_
"username" & vbCrLf

body = body & boundary & vbCrLf & _
"Content-Disposition: " & "form-data; name=""password """ & vbCrLf & vbCrLf &
_
"password" & vbCrLf

body = body & boundary & vbCrLf & _
"Content-Disposition: " & "form-data; name=""client"" " & vbCrLf & vbCrLf & _
"client" & vbCrLf

body = body & boundary & vbCrLf & _
"Content-Disposition: " & "form-data; name=""data""" & ";
filename=""data .zip""" & vbCrLf & _
"Content-Type: application/x-zip-compressed" & vbCrLf & vbCrLf

body2 = vbCrLf & boundary & "--"

Dim byte1 As Byte()
Dim byte2 As Byte()
byte1 = System.Text.Enc oding.Default.G etBytes(body)
byte2 = System.Text.Enc oding.Default.G etBytes(body2)

'===========JOI N BODY AND FILE CONENTS======== ====
Dim pobjCombinedArr ays = New ArrayList(byte1 .Length + Tem.Length +
byte2.Length)

'Append the arrays here.
pobjCombinedArr ays.AddRange(by te1)
pobjCombinedArr ays.AddRange(Te m)
pobjCombinedArr ays.AddRange(by te2)

Dim pbytCombinedArr ays(pobjCombine dArrays.Count) As Byte
pobjCombinedArr ays.CopyTo(pbyt CombinedArrays)

'===========POS T DATA AND GET RESPONSE======= =====
Dim bResponse As Byte() = myWebClient.Upl oadData(URL, "POST",
pobjCombinedArr ays)

Response.Write( System.Text.Enc oding.ASCII.Get String(bRespons e))

[Cor]

Hi,
Just a quick look and maybe I am wrong,

Dim st As FileStream = File.OpenRead(" data.zip")

Are you sure this instruction finds the right text file in the zip?

Maybe you know, but if not, there is a webclient.uploa dfile methode to.
I hope this helps a little bit.
Cor

[Herfried K. Wagner [MVP]]

"LD" <he**@home.co m> schrieb:

Dim vbCrLf As String = Convert.ToStrin g(Chr(13)) + Convert.ToStrin g(Chr(10)) Dim vbCr As String = Convert.ToStrin g(Chr(13))
Dim vbLf As String = Convert.ToStrin g(Chr(10))

The constants above are already defined in the
'Microsoft.Visu alBasic.Constan ts' class.

--
Herfried K. Wagner
MVP · VB Classic, VB.NET
http://www.mvps.org/dotnet