vb.net - تحميل نموذج متعدد الأجزاء غير يعمل. ماذا دهاك؟
سؤال
حاولت البحث ووجدت أن شخصا ما كتب تحميل نموذج متعدد الأجزاء.
على الرغم من أنه كان لديه خطأ واحد: نقل جميع القيم التي من المفترض - من المفترض في عنوان URL .. والتي لا يحب تطبيق الويب - بطريقة أو بأخرى.
ذهب كل شيء
POST /index.php?page=post&s=add&title=test&tags=testtags HTTP/1.1
وأضاف الملف فقط في محتوى البريد.
حاولت تقديم متصفح، وخرج هذا:
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--
استبدلت بايت بيانات الصورة باستخدام IMGDATA لجعلها أقصر.
تم قبول هذا التحميل. لذلك حاولت تقليد هذا مع رمز VB.NET سأعرض قريبا. جاء هذا:
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
لم يتم قبول هذا رغم ذلك. اذا ماذا كانت المشكلة؟ لقد حاولت أيضا وضع اثنين إضافي (-) في مكان آخر .. وفي النهاية، لم يغير أي شيء. هذا هو انقطاع رمز VB.NET الذي استخدمته للنقل. يبدأ التحميل "Stroader.Start عملية التحميل من خلال مؤشر ترابط يستخدم المتغيرات العالمية.
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
آمل أن يساعدني أي شخص في هذا. لا أستطيع معرفة ما هو الخطأ.
يجب أن ترجع رسالة "تحميل نجاح" مع إعادة توجيه، ولكن فقط إرجاع صفحة التحميل نفسها. راجعت مع شبكة الشم.
المحلول
لماذا لا نلقي نظرة على System.Net.WebClient
الطبقة بدلا من ذلك؟ http://msdn.microsoft.com/en-us/library/system.net.webclient.aspx. بهذه الطريقة لا داعي للقلق بشأن تنسيق بيانات البريد، أنت فقط اتصل WebClient.UploadFile()
: http://msdn.microsoft.com/en-us/library/36s52zhs.aspx.