VB.NET - o upload de formulário multiparte não funciona.O que está errado?

Question

Tentei pesquisar e descobri que alguém escreveu um uploader de formulário multiparte.

Embora tivesse uma coisa errada:Ele transferiu todos os - supostamente - valores POST no URL.O que o aplicativo da web não gosta - de alguma forma.

Foi tudo

POST /index.php?page=post&s=add&title=test&tags=testtags HTTP/1.1

E apenas adicionei o arquivo ao conteúdo da postagem.

Tentei enviar com um navegador e saiu o seguinte:

POST /index.php?page=post&s=add HTTP/1.1
Host: gelbooru.com

Connection: keep-alive
User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US) AppleWebKit/532.0 (KHTML, like Gecko) Chrome/3.0.195.38 Safari/532.0
Referer: http://gelbooru.com/index.php?page=post&s=add
Content-Length: 54851
Cache-Control: max-age=0
Origin: http://gelbooru.com
Content-Type: multipart/form-data; boundary=----WebKitFormBoundaryFqdVQ+1blrPMX2py
Accept: application/xml,application/xhtml+xml,text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5
Accept-Encoding: gzip,deflate,sdch
Cookie: __utmz=52483902.1258558959.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none);user_id=7382; pass_hash=lolpasswordcensored; __utma=52483902.1527465380.1258558959.1261431208.1261504455.7; __utmc=52483902; __utmb=52483902.1.10.1261504455
Accept-Language: de-DE,de;q=0.8,en-US;q=0.6,en;q=0.4
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3

------WebKitFormBoundaryFqdVQ+1blrPMX2py
Content-Disposition: form-data; name="upload"; filename="001.jpg"
Content-Type: image/jpeg

IMGDATA
------WebKitFormBoundaryFqdVQ+1blrPMX2py
Content-Disposition: form-data; name="source"


------WebKitFormBoundaryFqdVQ+1blrPMX2py
Content-Disposition: form-data; name="title"


------WebKitFormBoundaryFqdVQ+1blrPMX2py
Content-Disposition: form-data; name="tags"

tagtests
------WebKitFormBoundaryFqdVQ+1blrPMX2py
Content-Disposition: form-data; name="rating"

e
------WebKitFormBoundaryFqdVQ+1blrPMX2py
Content-Disposition: form-data; name="submit"

Upload
------WebKitFormBoundaryFqdVQ+1blrPMX2py--

Substituí os bytes de dados da imagem por IMGDATA para torná-los mais curtos.

Este upload foi aceito.Então tentei imitar isso com o código VB.NET que mostrarei em breve.Isso saiu:

POST /index.php?page=post&s=add& HTTP/1.1
Cookie: user_id=7382; pass_hash=lolpass_hashcensor;
Origin: http://gelbooru.com
Referer: http://gelbooru.com/index.php?page=post&s=add
Content-Type: multipart/form-data; boundary=----------8cc5152e90b3c60
Host: gelbooru.com
Content-Length: 243397
Expect: 100-continue
Connection: Keep-Alive


HTTP/1.1 100 Continue


----------8cc5152e90b3c60
Content-Disposition: form-data; name="tags"

tagtest
----------8cc5152e90b3c60
Content-Disposition: form-data; name="source"


----------8cc5152e90b3c60
Content-Disposition: form-data; name="rating"

e
----------8cc5152e90b3c60
Content-Disposition: form-data; name="title"


----------8cc5152e90b3c60
Content-Disposition: form-data; name="submit"

Upload
----------8cc5152e90b3c60
Content-Disposition: form-data; name="upload"; filename="1017E.jpg"
Content-Type: application/octet-stream

FILE
----------8cc5152e90b3c60

Isso NÃO foi aceito.Então qual foi o problema?Também tentei colocar os dois extras (--) em outro lugar.E no final, não mudou nada.Este é o recorte do código VB.NET que usei para transmissão.O Uploader.Start inicia o processo de upload através de um thread que utiliza as variáveis ​​globais.

Sub StartUploadJob(ByVal filepath As String, ByVal tags As String, ByVal title As String, ByVal source As String, ByVal rate As rating)
    ContinueProgressLoop = True
    processloop = New Threading.Thread(AddressOf ProgressLoopThread)
    Uploader = New Threading.Thread(AddressOf UploadFileEx)
    uploadfile = filepath

    querystring("page") = "post"
    querystring("s") = "add"
    'http://gelbooru.com/index.php?page=dapi&s=post&q=index
    'querystring("page") = "dapi"
    'querystring("s") = "post"
    'querystring("q") = "index"
    'url = "http://gelbooru.com/index.php"
    url = "http://sinni800.ath.cx/UploadTest.aspx"
    'url = "http://gelbooru.com/ok.php"
    fileFormName = "upload"
    contenttype = "application/octet-stream"
    uploadstrings("tags") = tags
    uploadstrings("source") = source
    uploadstrings("rating") = rate.ToString
    uploadstrings("title") = title
    'querystring("user_id") = My.Settings("user_id")
    'querystring("pass_hash") = My.Settings("pass_hash")
    uploadstrings("submit") = "Upload"
    working = True
    cookies = New CookieContainer()
    'cookies.Add(My.Settings.cookies)
    Uploader.Start()
    processloop.Start()

End Sub

Dim uploadstrings As New NameValueCollection
Dim uploadfile As String : Dim url As String : Dim fileFormName As String : Dim contenttype As String : Dim querystring As New NameValueCollection : Dim cookies As CookieContainer
Function UploadFileEx() '(ByVal uploadfile As String, ByVal url As String, ByVal fileFormName As String, ByVal contenttype As String, ByVal querystring As NameValueCollection, ByVal cookies As CookieContainer) As String
    If (fileFormName Is Nothing) OrElse (fileFormName.Length = 0) Then
        fileFormName = "file"
    End If

    If (contenttype Is Nothing) OrElse (contenttype.Length = 0) Then
        contenttype = "application/octet-stream"
    End If


    Dim postdata As String
    postdata = "?"
    If querystring IsNot Nothing Then
        For Each key As String In querystring.Keys
            postdata += (key & "=") + querystring.[Get](key) & "&"
        Next
    End If
    Dim uri As New Uri(url + "?page=post&s=add")


    Dim boundary As String = "----------" & DateTime.Now.Ticks.ToString("x")
    Dim webrequest__1 As HttpWebRequest = DirectCast(WebRequest.Create(uri), HttpWebRequest)
    'webrequest__1.CookieContainer = cookies
    webrequest__1.Headers(HttpRequestHeader.Cookie) = My.Settings.cookie_string
    webrequest__1.Headers.Add("Origin", "http://gelbooru.com")
    webrequest__1.Referer = "http://gelbooru.com/index.php?page=post&s=add"
    webrequest__1.ContentType = "multipart/form-data; boundary=" & boundary
    webrequest__1.Method = "POST"


    ' Build up the post message header
    Dim sb As New StringBuilder()

    For Each key As String In uploadstrings.Keys
        '    content-disposition: form-data; name="field1"
        'content-type: text/plain

        sb.Append(boundary & vbCrLf)
        sb.Append("Content-Disposition: form-data; name=""" + key + """" & vbCrLf)
        ' sb.Append("content-type: text/plain" & vbCrLf)
        sb.Append(vbCrLf)
        sb.Append(uploadstrings(key))
        sb.Append(vbCrLf)

    Next

    Dim close As Byte() = Encoding.UTF8.GetBytes("--")

    sb.Append("--")
    sb.Append(boundary)
    sb.Append(vbCr & vbLf)
    sb.Append("Content-Disposition: form-data; name=""")
    sb.Append(fileFormName)
    sb.Append("""; filename=""")
    sb.Append(Path.GetFileName(uploadfile))
    sb.Append("""")
    sb.Append(vbCr & vbLf)
    sb.Append("Content-Type: ")
    sb.Append(contenttype)
    sb.Append(vbCr & vbLf)
    sb.Append(vbCr & vbLf)

    Dim postHeader As String = sb.ToString()
    Dim postHeaderBytes As Byte() = Encoding.UTF8.GetBytes(postHeader)

    ' Build the trailing boundary string as a byte array
    ' ensuring the boundary appears on a line by itself
    Dim boundaryBytes As Byte() = Encoding.ASCII.GetBytes(vbCr & vbLf & boundary & "--" & vbCr & vbLf)

    Dim fileStream As New FileStream(uploadfile, FileMode.Open, FileAccess.Read)
    Dim length As Long = postHeaderBytes.Length + fileStream.Length + boundaryBytes.Length
    webrequest__1.ContentLength = length

    Dim requestStream As Stream = webrequest__1.GetRequestStream()

    Dim fulllength As Integer = postHeaderBytes.Length + fileStream.Length + boundaryBytes.Length

    RaiseEvent TextMessage("")

    ' Write out our post header
    RaiseEvent TextMessage("Write out our post header")
    requestStream.Write(postHeaderBytes, 0, postHeaderBytes.Length)
    progress = New Progress_Struct(100 / fulllength * postHeaderBytes.Length, postHeaderBytes.Length, fulllength)
    ' Write out the file contents
    RaiseEvent TextMessage("Write out the file contents")
    Dim buffer As Byte() = New [Byte](CUInt(Math.Min(4096, CInt(fileStream.Length))) - 1) {}
    Dim bytesRead As Integer = 0
    While (InlineAssignHelper(bytesRead, fileStream.Read(buffer, 0, buffer.Length))) <> 0
        requestStream.Write(buffer, 0, bytesRead)
        Progress = New Progress_Struct(100 / fulllength * (Progress.bytes_read + buffer.Length), Progress.bytes_read + buffer.Length, fulllength)
    End While
    RaiseEvent TextMessage("bytesRead = " + bytesRead.ToString)
    RaiseEvent TextMessage("postHeaderBytes.length = " + postHeaderBytes.Length.ToString)
    RaiseEvent TextMessage("Main File bytes written")
    ' Write out the trailing boundary
    requestStream.Write(boundaryBytes, 0, boundaryBytes.Length)
    Progress = New Progress_Struct(100 / fulllength * (Progress.bytes_read + boundaryBytes.Length), Progress.bytes_read + boundaryBytes.Length, fulllength)

    Dim responce As WebResponse
    Try
        RaiseEvent TextMessage("Trying normal response...")
        responce = webrequest__1.GetResponse()
    Catch ex As Net.WebException
        RaiseEvent TextMessage("ERROR " + ex.Status.ToString + " Trying to get Error response.")
        responce = ex.Response
    End Try

    Dim s As Stream = responce.GetResponseStream()
    Dim sr As New StreamReader(s)

    ContinueProgressLoop = False
    Dim returnstr As String = sr.ReadToEnd()
    'SendHtmlMessage(returnstr)

    RaiseEvent ProgressChange(Progress)
    RaiseEvent UploadFinished(returnstr)
    RaiseEvent TextMessage("Working FALSE")
    working = False
    Return returnstr
End Function

Espero que alguém possa me ajudar com isso.Não consigo descobrir o que há de errado.

Ele deve retornar uma mensagem "Sucesso no upload" com um redirecionamento, mas retorna apenas a própria página de upload.Verifiquei com um sniffer de rede.

Answer 1

Por que você não dá uma olhada System.Net.WebClient classe em vez disso? http://msdn.microsoft.com/en-us/library/system.net.webclient.aspx Desta forma você não precisa se preocupar em formatar os dados do POST, basta chamar WebClient.UploadFile(): http://msdn.microsoft.com/en-us/library/36s52zhs.aspx