puskin Posted April 20, 2008 Report Posted April 20, 2008 Salve!Am nevoie de ajutor pentru a decrypta un text codat in limbajul VB6Am modulul de encryptare-decryptare, am si textul care trebuie decriptat..(textul nu este important...vreau numai sa stiu daca functioneaza decryptorul...).Problema este urmatoarea am codat textul dar cand dau sa-l decodez surpriza...imi scoate fisierul cu txt "0 Biti".Cel mai bine ar fi sa se poata decoda direct textul, nu neaparat un fisier. Ii multumesc anticipat celui care ma ajuta.Link pt Download module :http://rapidshare.com/files/108973007/modules.zip.htmlMai jos sa vedeti cum arata incriptarea si textul encryptat:P.S cine ma ajuta sa-l decodez primeste codul sursa + programul aplicatiei...care este 100% nedetectata 100% automataPublic Function UUEncodeFile(strFilePath As String) As String Dim intFile As Integer 'file handler Dim intTempFile As Integer 'temp file Dim lFileSize As Long 'size of the file Dim strFileName As String 'name of the file Dim strFileData As String 'file data chunk Dim lEncodedLines As Long 'number of encoded lines Dim strTempLine As String 'temporary string Dim i As Long 'loop counter Dim j As Integer 'loop counter Dim strResult As String 'Get file name strFileName = Mid$(strFilePath, InStrRev(strFilePath, "\") + 1) 'This important: "begin 664" strResult = "begin 664 " + strFileName + vbLf 'Get file size lFileSize = FileLen(strFilePath) lEncodedLines = lFileSize / 45 + 1 'you need to encode every 45 bytes strFileData = Space(45) intFile = FreeFile 'open the output file Open strFilePath For Binary As intFile For i = 1 To lEncodedLines If i = lEncodedLines Then strFileData = Space(lFileSize Mod 45) End If 'get data Get intFile, , strFileData 'the first byte in a line is a char, which number describes 'how many bytes are in the line strTempLine = Chr(Len(strFileData) + 32) If i = lEncodedLines And (Len(strFileData) Mod 3) Then strFileData = strFileData + Space(3 - (Len(strFileData) Mod 3)) End If 'now some encoding For j = 1 To Len(strFileData) Step 3 strTempLine = strTempLine + Chr(Asc(Mid(strFileData, j, 1)) \ 4 + 32) strTempLine = strTempLine + Chr((Asc(Mid(strFileData, j, 1)) Mod 4) * 16 _ + Asc(Mid(strFileData, j + 1, 1)) \ 16 + 32) strTempLine = strTempLine + Chr((Asc(Mid(strFileData, j + 1, 1)) Mod 16) * 4 _ + Asc(Mid(strFileData, j + 2, 1)) \ 64 + 32) strTempLine = strTempLine + Chr(Asc(Mid(strFileData, j + 2, 1)) Mod 64 + 32) Next j strResult = strResult + strTempLine + vbLf strTempLine = "" 'get next line Next i 'close the file Close intFile 'add the "end" string strResult = strResult & "'" & vbLf + "end" + vbLf 'return the encoded string UUEncodeFile = strResultEnd FunctionAici decryptorul:Public Function UUDecodeToFile(strUUCodeData As String, strFilePath As String) On Error Resume Next Dim vDataLine As Variant 'some variables needed for decoding Dim vDataLines As Variant Dim strDataLine As String Dim intSymbols As Integer Dim intFile As Integer Dim strTemp As String If Left$(strUUCodeData, 6) = "begin " Then 'check if it is a encoded file strUUCodeData = Mid$(strUUCodeData, InStr(1, strUUCodeData, vbLf) + 1) End If If Right$(strUUCodeData, 4) = "end" + vbLf Then 'check if "end" is available strUUCodeData = Left$(strUUCodeData, Len(strUUCodeData) - 7) End If intFile = FreeFile Open strFilePath For Binary As intFile 'open output file vDataLines = Split(strUUCodeData, vbLf) For Each vDataLine In vDataLines 'get every line strDataLine = CStr(vDataLine) intSymbols = Asc(Left$(strDataLine, 1)) 'get number of chars in 'one line. This is important 'for decoding strDataLine = Mid$(strDataLine, 2, intSymbols) For i = 1 To Len(strDataLine) Step 4 'now some decoding strTemp = strTemp + Chr((Asc(Mid(strDataLine, i, 1)) - 32) * 4 + _ (Asc(Mid(strDataLine, i + 1, 1)) - 32) \ 16) strTemp = strTemp + Chr((Asc(Mid(strDataLine, i + 1, 1)) Mod 16) * 16 + _ (Asc(Mid(strDataLine, i + 2, 1)) - 32) \ 4) strTemp = strTemp + Chr((Asc(Mid(strDataLine, i + 2, 1)) Mod 4) * 64 + _ Asc(Mid(strDataLine, i + 3, 1)) - 32) Next i 'put the decoded data in the file Put intFile, , strTemp strTemp = "" Next 'close the file Close intFileEnd FunctionSi textul :begin 664 sysdll.txt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M($Y/5"!214=)4U1%4D5$/CX)#0I;,SHT-#HU,"!0M35T\/%-T87)T($UE;G4^/@D-"ELS.C0T.C4U(%!-73P\56YT:71L960@+2!.M;W1E<&%D/CX)4E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)2M4E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)2#0H-"D5N9&5D.B TM+S(P+S(P,#@@,SHT-#HU-2!030T*#0H 7 @'endCa sa fiu mai explicit:-programul genereaza fisierul pe calculatorul iar cand atinge dimeniunea de 5000 de biti (5Kb) este incryptat si trimis ca attach pe mail...numai ca el imi vine pe email in urmatorul mod (in nici un caz attach!)...e vreo posibilitate sa decodez textul?: Quote