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Chủ đề: Ma hoa chuoi

  1. #1
    Tham gia
    07-06-2005
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    Vui lắm ! Ma hoa chuoi

    Cac anh, cac chi co ai biet chi gium em cho nay:
    - Nguoi dung tao mot UserName moi va Password moi. UserName thi khoi noi, nhung lam sao khi luu Password xuong CSDL SQL la o dang ki tu da duoc ma hoa (khong doc duoc).

    - Con nua: De phuc vu cho viec lay lai Password cua nguoi dung. Em khong lay lai Password cu ma co tinh tao ra Password moi, Password moi nay chang han co 10 ki tu. Cac anh chi chi cho em cach dung ham Random de tao ra chuoi 10 ki tu ngau nhien ke ca ki tu so.

    - Con nua: Em cam con cac anh, cac chi. Chuc anh chi manh khoe va vui nha. Chao nha.
    Quote Quote

  2. #2
    Tham gia
    16-07-2002
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    Vui lòng gửi bài viết bằng tiếng Việt có dấu. Vì bạn mới tham gia DDTH.com nên tôi nhắc nhở bạn lần đầu tiên. Bạn có thể bấm phím F8 để chọn kiểu gõ (Telex, VNI) và không cần phải có bộ gõ Unikey hay Vietkey nếu bạn đang sử dụng Internet Explorer hoặc FireFox, AvantBrowser, MyIE2.

  3. #3
    Tham gia
    07-06-2005
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    Chuyện gì vậy? À, thì ra là chuyện bỏ dấu chữ hả. Ừ, xin lỗi nghen, tại mình chưa biết điều này mà.
    Nhưng sao mấy bạn không trả lời giùm câu hỏi của mình đi

  4. #4
    Tham gia
    23-02-2003
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    Thì bạn sử dụng bất kỳ một một thuật toán mã hóa nào cũng được, tốt nhất là dùng các thuật toán mã hóa 1 chiều để ko thể giải mã, như md5 chẳng hạn.

  5. #5
    Tham gia
    27-06-2007
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    Bạn có thể chuyển từng ký tự sang mã ASCII rồi cộng thêm tùy ý. cũng có thể được đó.

  6. #6
    Tham gia
    18-04-2005
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    Module mã hóa MD5

    1. Mã hóa password: jiSh@n nói, để an toàn, bạn có thể dùng thuật toán mã hóa 1 chiều MD5, đây là thuật toán HASH (băm) được dùng phổ biến hiện nay.
    Bạn add đoạn code sau vào 1 module trong dự án VB6, khi cần sử dụng chỉ cần gọi hàm MD5("<chuỗi mật khẩu cần mã hóa>").
    Code:
    '*******************************************************************************
    ' MODULE:       MD5Hash
    ' CREATED:      16-Feb-2001 Phil Fresle
    '*******************************************************************************
    Option Explicit
    Private Const BITS_TO_A_BYTE  As Long = 8
    Private Const BYTES_TO_A_WORD As Long = 4
    Private Const BITS_TO_A_WORD  As Long = BYTES_TO_A_WORD * BITS_TO_A_BYTE
    
    Private m_lOnBits(0 To 30) As Long
    Private m_l2Power(0 To 30) As Long
    Private m_Initialized As Boolean
    '--
    Private Sub InitConstArrays()
       If m_Initialized Then Exit Sub
       ' Could have done this with a loop calculating each value, but simply
       ' assigning the values is quicker - BITS SET FROM RIGHT
       m_lOnBits(0) = 1            ' 00000000000000000000000000000001
       m_lOnBits(1) = 3            ' 00000000000000000000000000000011
       m_lOnBits(2) = 7            ' 00000000000000000000000000000111
       m_lOnBits(3) = 15           ' 00000000000000000000000000001111
       m_lOnBits(4) = 31           ' 00000000000000000000000000011111
       m_lOnBits(5) = 63           ' 00000000000000000000000000111111
       m_lOnBits(6) = 127          ' 00000000000000000000000001111111
       m_lOnBits(7) = 255          ' 00000000000000000000000011111111
       m_lOnBits(8) = 511          ' 00000000000000000000000111111111
       m_lOnBits(9) = 1023         ' 00000000000000000000001111111111
       m_lOnBits(10) = 2047        ' 00000000000000000000011111111111
       m_lOnBits(11) = 4095        ' 00000000000000000000111111111111
       m_lOnBits(12) = 8191        ' 00000000000000000001111111111111
       m_lOnBits(13) = 16383       ' 00000000000000000011111111111111
       m_lOnBits(14) = 32767       ' 00000000000000000111111111111111
       m_lOnBits(15) = 65535       ' 00000000000000001111111111111111
       m_lOnBits(16) = 131071      ' 00000000000000011111111111111111
       m_lOnBits(17) = 262143      ' 00000000000000111111111111111111
       m_lOnBits(18) = 524287      ' 00000000000001111111111111111111
       m_lOnBits(19) = 1048575     ' 00000000000011111111111111111111
       m_lOnBits(20) = 2097151     ' 00000000000111111111111111111111
       m_lOnBits(21) = 4194303     ' 00000000001111111111111111111111
       m_lOnBits(22) = 8388607     ' 00000000011111111111111111111111
       m_lOnBits(23) = 16777215    ' 00000000111111111111111111111111
       m_lOnBits(24) = 33554431    ' 00000001111111111111111111111111
       m_lOnBits(25) = 67108863    ' 00000011111111111111111111111111
       m_lOnBits(26) = 134217727   ' 00000111111111111111111111111111
       m_lOnBits(27) = 268435455   ' 00001111111111111111111111111111
       m_lOnBits(28) = 536870911   ' 00011111111111111111111111111111
       m_lOnBits(29) = 1073741823  ' 00111111111111111111111111111111
       m_lOnBits(30) = 2147483647  ' 01111111111111111111111111111111
       
       ' Could have done this with a loop calculating each value, but simply
       ' assigning the values is quicker - POWERS OF 2
       m_l2Power(0) = 1            ' 00000000000000000000000000000001
       m_l2Power(1) = 2            ' 00000000000000000000000000000010
       m_l2Power(2) = 4            ' 00000000000000000000000000000100
       m_l2Power(3) = 8            ' 00000000000000000000000000001000
       m_l2Power(4) = 16           ' 00000000000000000000000000010000
       m_l2Power(5) = 32           ' 00000000000000000000000000100000
       m_l2Power(6) = 64           ' 00000000000000000000000001000000
       m_l2Power(7) = 128          ' 00000000000000000000000010000000
       m_l2Power(8) = 256          ' 00000000000000000000000100000000
       m_l2Power(9) = 512          ' 00000000000000000000001000000000
       m_l2Power(10) = 1024        ' 00000000000000000000010000000000
       m_l2Power(11) = 2048        ' 00000000000000000000100000000000
       m_l2Power(12) = 4096        ' 00000000000000000001000000000000
       m_l2Power(13) = 8192        ' 00000000000000000010000000000000
       m_l2Power(14) = 16384       ' 00000000000000000100000000000000
       m_l2Power(15) = 32768       ' 00000000000000001000000000000000
       m_l2Power(16) = 65536       ' 00000000000000010000000000000000
       m_l2Power(17) = 131072      ' 00000000000000100000000000000000
       m_l2Power(18) = 262144      ' 00000000000001000000000000000000
       m_l2Power(19) = 524288      ' 00000000000010000000000000000000
       m_l2Power(20) = 1048576     ' 00000000000100000000000000000000
       m_l2Power(21) = 2097152     ' 00000000001000000000000000000000
       m_l2Power(22) = 4194304     ' 00000000010000000000000000000000
       m_l2Power(23) = 8388608     ' 00000000100000000000000000000000
       m_l2Power(24) = 16777216    ' 00000001000000000000000000000000
       m_l2Power(25) = 33554432    ' 00000010000000000000000000000000
       m_l2Power(26) = 67108864    ' 00000100000000000000000000000000
       m_l2Power(27) = 134217728   ' 00001000000000000000000000000000
       m_l2Power(28) = 268435456   ' 00010000000000000000000000000000
       m_l2Power(29) = 536870912   ' 00100000000000000000000000000000
       m_l2Power(30) = 1073741824  ' 01000000000000000000000000000000
       m_Initialized = True
    End Sub
    '*******************************************************************************
    ' LShift (FUNCTION)
    '
    ' PARAMETERS:
    ' (In) - lValue     - Long    - The value to be shifted
    ' (In) - iShiftBits - Integer - The number of bits to shift the value by
    '
    ' RETURN VALUE:
    ' Long - The shifted long integer
    '
    ' DESCRIPTION:
    ' A left shift takes all the set binary bits and moves them left, in-filling
    ' with zeros in the vacated bits on the right. This function is equivalent to
    ' the << operator in Java and C++
    '*******************************************************************************
    Private Function LShift(ByVal lValue As Long, ByVal iShiftBits As Integer) As Long
    
       ' NOTE: If you can guarantee that the Shift parameter will be in the
       ' range 1 to 30 you can safely strip of this first nested if structure for
       ' speed.
       '
       ' A shift of zero is no shift at all.
       If iShiftBits = 0 Then
          LShift = lValue
          Exit Function
       
       ' A shift of 31 will result in the right most bit becoming the left most
       ' bit and all other bits being cleared
       ElseIf iShiftBits = 31 Then
          If lValue And 1 Then
             LShift = &H80000000
          Else
             LShift = 0
          End If
          Exit Function
       
       ' A shift of less than zero or more than 31 is undefined
       ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
          Err.Raise 6
       End If
       
       ' If the left most bit that remains will end up in the negative bit
       ' position (&H80000000) we would end up with an overflow if we took the
       ' standard route. We need to strip the left most bit and add it back
       ' afterwards.
       If (lValue And m_l2Power(31 - iShiftBits)) Then
       
          ' (Value And OnBits(31 - (Shift + 1))) chops off the left most bits that
          ' we are shifting into, but also the left most bit we still want as this
          ' is going to end up in the negative bit marker position (&H80000000).
          ' After the multiplication/shift we Or the result with &H80000000 to
          ' turn the negative bit on.
          LShift = ((lValue And m_lOnBits(31 - (iShiftBits + 1))) * m_l2Power(iShiftBits)) Or &H80000000
       
       Else
       
          ' (Value And OnBits(31-Shift)) chops off the left most bits that we are
          ' shifting into so we do not get an overflow error when we do the
          ' multiplication/shift
          LShift = ((lValue And m_lOnBits(31 - iShiftBits)) * m_l2Power(iShiftBits))
       
       End If
    
    End Function
    
    '*******************************************************************************
    ' RShift (FUNCTION)
    '
    ' PARAMETERS:
    ' (In) - lValue     - Long    - The value to be shifted
    ' (In) - iShiftBits - Integer - The number of bits to shift the value by
    '
    ' RETURN VALUE:
    ' Long - The shifted long integer
    '
    ' DESCRIPTION:
    ' The right shift of an unsigned long integer involves shifting all the set bits
    ' to the right and in-filling on the left with zeros. This function is
    ' equivalent to the >>> operator in Java or the >> operator in C++ when used on
    ' an unsigned long.
    '*******************************************************************************
    Private Function RShift(ByVal lValue As Long, ByVal iShiftBits As Integer) As Long
        
       ' NOTE: If you can guarantee that the Shift parameter will be in the
       ' range 1 to 30 you can safely strip of this first nested if structure for
       ' speed.
       '
       ' A shift of zero is no shift at all
       If iShiftBits = 0 Then
          RShift = lValue
          Exit Function
       
       ' A shift of 31 will clear all bits and move the left most bit to the right
       ' most bit position
       ElseIf iShiftBits = 31 Then
          If lValue And &H80000000 Then
             RShift = 1
          Else
             RShift = 0
          End If
          Exit Function
       
       ' A shift of less than zero or more than 31 is undefined
       ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
          Err.Raise 6
       End If
       
       ' We do not care about the top most bit or the final bit, the top most bit
       ' will be taken into account in the next stage, the final bit (whether it
       ' is an odd number or not) is being shifted into, so we do not give a jot
       ' about it
       RShift = (lValue And &H7FFFFFFE) \ m_l2Power(iShiftBits)
       
       ' If the top most bit (&H80000000) was set we need to do things differently
       ' as in a normal VB signed long integer the top most bit is used to indicate
       ' the sign of the number, when it is set it is a negative number, so just
       ' deviding by a factor of 2 as above would not work.
       ' NOTE: (lValue And  &H80000000) is equivalent to (lValue < 0), you could
       ' get a very marginal speed improvement by changing the test to (lValue < 0)
       If (lValue And &H80000000) Then
          ' We take the value computed so far, and then add the left most negative
          ' bit after it has been shifted to the right the appropriate number of
          ' places
          RShift = (RShift Or (&H40000000 \ m_l2Power(iShiftBits - 1)))
       End If
    
    End Function
    
    '*******************************************************************************
    ' RShiftSigned (FUNCTION)
    '
    ' PARAMETERS:
    ' (In) - lValue     - Long    -
    ' (In) - iShiftBits - Integer -
    '
    ' RETURN VALUE:
    ' Long -
    '
    ' DESCRIPTION:
    ' The right shift of a signed long integer involves shifting all the set bits to
    ' the right and in-filling on the left with the sign bit (0 if positive, 1 if
    ' negative. This function is equivalent to the >> operator in Java or the >>
    ' operator in C++ when used on a signed long integer. Not used in this class,
    ' but included for completeness.
    '*******************************************************************************
    Private Function RShiftSigned(ByVal lValue As Long, ByVal iShiftBits As Integer) As Long
        
       ' NOTE: If you can guarantee that the Shift parameter will be in the
       ' range 1 to 30 you can safely strip of this first nested if structure for
       ' speed.
       '
       ' A shift of zero is no shift at all
       If iShiftBits = 0 Then
          RShiftSigned = lValue
          Exit Function
       
       ' A shift of 31 will clear all bits if the left most bit was zero, and will
       ' set all bits if the left most bit was 1 (a negative indicator)
       ElseIf iShiftBits = 31 Then
       
          ' NOTE: (lValue And  &H80000000) is equivalent to (lValue < 0), you
          ' could get a very marginal speed improvement by changing the test to
          ' (lValue < 0)
          If (lValue And &H80000000) Then
             RShiftSigned = -1
          Else
             RShiftSigned = 0
          End If
          Exit Function
       
       ' A shift of less than zero or more than 31 is undefined
       ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
          Err.Raise 6
       End If
       
       ' We get the same result by dividing by the appropriate power of 2 and
       ' rounding in the negative direction
       RShiftSigned = Int(lValue / m_l2Power(iShiftBits))
    
    End Function
    
    '*******************************************************************************
    ' RotateLeft (FUNCTION)
    '
    ' PARAMETERS:
    ' (In) - lValue     - Long    - Value to act on
    ' (In) - iShiftBits - Integer - Bits to move by
    '
    ' RETURN VALUE:
    ' Long - Result
    '
    ' DESCRIPTION:
    ' Rotates the bits in a long integer to the left, those bits falling off the
    ' left edge are put back on the right edge
    '*******************************************************************************
    Private Function RotateLeft(ByVal lValue As Long, ByVal iShiftBits As Integer) As Long
    
       RotateLeft = LShift(lValue, iShiftBits) Or RShift(lValue, (32 - iShiftBits))
    
    End Function
    
    '*******************************************************************************
    ' AddUnsigned (FUNCTION)
    '
    ' PARAMETERS:
    ' (In) - lX - Long - First value
    ' (In) - lY - Long - Second value
    '
    ' RETURN VALUE:
    ' Long - Result
    '
    ' DESCRIPTION:
    ' Adds two potentially large unsigned numbers without overflowing
    '*******************************************************************************
    Private Function AddUnsigned(ByVal lX As Long, ByVal lY As Long) As Long
        
       Dim lX4     As Long
       Dim lY4     As Long
       Dim lX8     As Long
       Dim lY8     As Long
       Dim lResult As Long
       
       lX8 = lX And &H80000000
       lY8 = lY And &H80000000
       lX4 = lX And &H40000000
       lY4 = lY And &H40000000
       
       lResult = (lX And &H3FFFFFFF) + (lY And &H3FFFFFFF)
       
       If lX4 And lY4 Then
          lResult = lResult Xor &H80000000 Xor lX8 Xor lY8
       ElseIf lX4 Or lY4 Then
          If lResult And &H40000000 Then
             lResult = lResult Xor &HC0000000 Xor lX8 Xor lY8
          Else
             lResult = lResult Xor &H40000000 Xor lX8 Xor lY8
          End If
       Else
          lResult = lResult Xor lX8 Xor lY8
       End If
       
       AddUnsigned = lResult
    
    End Function
    
    '*******************************************************************************
    ' F (FUNCTION)
    '
    ' DESCRIPTION:
    ' MD5's F function
    '*******************************************************************************
    Private Function F(ByVal x As Long, ByVal y As Long, ByVal z As Long) As Long
    
       F = (x And y) Or ((Not x) And z)
    
    End Function
    
    '*******************************************************************************
    ' G (FUNCTION)
    '
    ' DESCRIPTION:
    ' MD5's G function
    '*******************************************************************************
    Private Function G(ByVal x As Long, ByVal y As Long, ByVal z As Long) As Long
       G = (x And z) Or (y And (Not z))
    End Function
    
    '*******************************************************************************
    ' H (FUNCTION)
    '
    ' DESCRIPTION:
    ' MD5's H function
    '*******************************************************************************
    Private Function H(ByVal x As Long, ByVal y As Long, ByVal z As Long) As Long
       H = (x Xor y Xor z)
    End Function
    
    '*******************************************************************************
    ' I (FUNCTION)
    '
    ' DESCRIPTION:
    ' MD5's I function
    '*******************************************************************************
    Private Function I(ByVal x As Long, ByVal y As Long, ByVal z As Long) As Long
       I = (y Xor (x Or (Not z)))
    End Function
    
    '*******************************************************************************
    ' FF (SUB)
    '
    ' DESCRIPTION:
    ' MD5's FF procedure
    '*******************************************************************************
    Private Sub FF(a As Long, ByVal b As Long, ByVal c As Long, ByVal d As Long, ByVal x As Long, ByVal s As Long, ByVal ac As Long)
       a = AddUnsigned(a, AddUnsigned(AddUnsigned(F(b, c, d), x), ac))
       a = RotateLeft(a, s)
       a = AddUnsigned(a, b)
    End Sub
    '*******************************************************************************
    ' GG (SUB)
    '
    ' DESCRIPTION:
    ' MD5's GG procedure
    '*******************************************************************************
    Private Sub GG(a As Long, ByVal b As Long, ByVal c As Long, ByVal d As Long, ByVal x As Long, ByVal s As Long, ByVal ac As Long)
       a = AddUnsigned(a, AddUnsigned(AddUnsigned(G(b, c, d), x), ac))
       a = RotateLeft(a, s)
       a = AddUnsigned(a, b)
    End Sub
    
    '*******************************************************************************
    ' HH (SUB)
    '
    ' DESCRIPTION:
    ' MD5's HH procedure
    '*******************************************************************************
    Private Sub HH(a As Long, ByVal b As Long, ByVal c As Long, ByVal d As Long, ByVal x As Long, ByVal s As Long, ByVal ac As Long)
    
       a = AddUnsigned(a, AddUnsigned(AddUnsigned(H(b, c, d), x), ac))
       a = RotateLeft(a, s)
       a = AddUnsigned(a, b)
    
    End Sub
    
    '*******************************************************************************
    ' II (SUB)
    '
    ' DESCRIPTION:
    ' MD5's II procedure
    '*******************************************************************************
    Private Sub II(a As Long, ByVal b As Long, ByVal c As Long, ByVal d As Long, ByVal x As Long, ByVal s As Long, ByVal ac As Long)
    
       a = AddUnsigned(a, AddUnsigned(AddUnsigned(I(b, c, d), x), ac))
       a = RotateLeft(a, s)
       a = AddUnsigned(a, b)
    
    End Sub
    
    '*******************************************************************************
    ' ConvertToWordArray (FUNCTION)
    '
    ' PARAMETERS:
    ' (In/Out) - sMessage - String - String message
    '
    ' RETURN VALUE:
    ' Long() - Converted message as long array
    '
    ' DESCRIPTION:
    ' Takes the string message and puts it in a long array with padding according to
    ' the MD5 rules.
    '*******************************************************************************
    Private Function ConvertToWordArray(sMessage As String) As Long()
    
       Dim lMessageLength  As Long
       Dim lNumberOfWords  As Long
       Dim lWordArray()    As Long
       Dim lBytePosition   As Long
       Dim lByteCount      As Long
       Dim lWordCount      As Long
           
       Const MODULUS_BITS      As Long = 512
       Const CONGRUENT_BITS    As Long = 448
       
       lMessageLength = Len(sMessage)
           
       ' Get padded number of words. Message needs to be congruent to 448 bits,
       ' modulo 512 bits. If it is exactly congruent to 448 bits, modulo 512 bits
       ' it must still have another 512 bits added. 512 bits = 64 bytes
       ' (or 16 * 4 byte words), 448 bits = 56 bytes. This means lMessageSize must
       ' be a multiple of 16 (i.e. 16 * 4 (bytes) * 8 (bits))
       lNumberOfWords = (((lMessageLength + _
          ((MODULUS_BITS - CONGRUENT_BITS) \ BITS_TO_A_BYTE)) \ _
          (MODULUS_BITS \ BITS_TO_A_BYTE)) + 1) * _
          (MODULUS_BITS \ BITS_TO_A_WORD)
       ReDim lWordArray(lNumberOfWords - 1)
           
       ' Combine each block of 4 bytes (ascii code of character) into one long
       ' value and store in the message. The high-order (most significant) bit of
       ' each byte is listed first. However, the low-order (least significant) byte
       ' is given first in each word.
       lBytePosition = 0
       lByteCount = 0
       Do Until lByteCount >= lMessageLength
          ' Each word is 4 bytes
          lWordCount = lByteCount \ BYTES_TO_A_WORD
          
          ' The bytes are put in the word from the right most edge
          lBytePosition = (lByteCount Mod BYTES_TO_A_WORD) * BITS_TO_A_BYTE
          lWordArray(lWordCount) = lWordArray(lWordCount) Or _
             LShift(AscB(Mid(sMessage, lByteCount + 1, 1)), lBytePosition)
          lByteCount = lByteCount + 1
       Loop
       
       ' Terminate according to MD5 rules with a 1 bit, zeros and the length in
       ' bits stored in the last two words
       lWordCount = lByteCount \ BYTES_TO_A_WORD
       lBytePosition = (lByteCount Mod BYTES_TO_A_WORD) * BITS_TO_A_BYTE
       
       ' Add a terminating 1 bit, all the rest of the bits to the end of the
       ' word array will default to zero
       lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(&H80, lBytePosition)
       
       ' We put the length of the message in bits into the last two words, to get
       ' the length in bits we need to multiply by 8 (or left shift 3). This left
       ' shifted value is put in the first word. Any bits shifted off the left edge
       ' need to be put in the second word, we can work out which bits by shifting
       ' right the length by 29 bits.
       lWordArray(lNumberOfWords - 2) = LShift(lMessageLength, 3)
       lWordArray(lNumberOfWords - 1) = RShift(lMessageLength, 29)
           
       ConvertToWordArray = lWordArray
    
    End Function
    
    '*******************************************************************************
    ' WordToHex (FUNCTION)
    '
    ' PARAMETERS:
    ' (In) - lValue - Long - Long value to convert
    '
    ' RETURN VALUE:
    ' String - Hex value to return
    '
    ' DESCRIPTION:
    ' Takes a long integer and due to the bytes reverse order it extracts the
    ' individual bytes and converts them to hex appending them for an overall hex
    ' value
    '*******************************************************************************
    Private Function WordToHex(ByVal lValue As Long) As String
    
       Dim lByte As Long
       Dim lCount As Long
           
       For lCount = 0 To 3
          lByte = RShift(lValue, lCount * BITS_TO_A_BYTE) And m_lOnBits(BITS_TO_A_BYTE - 1)
          WordToHex = WordToHex & Right("0" & Hex(lByte), 2)
       Next
    
    End Function
    
    '*******************************************************************************
    ' MD5 (FUNCTION)
    '
    ' PARAMETERS:
    ' (In/Out) - sMessage - String - String to be digested
    '
    ' RETURN VALUE:
    ' String - The MD5 digest
    '
    ' DESCRIPTION:
    ' This function takes a string message and generates an MD5 digest for it.
    ' sMessage can be up to the VB string length limit of 2^31 (approx. 2 billion)
    ' characters.
    '
    ' NOTE: Due to the way in which the string is processed the routine assumes a
    ' single byte character set. VB passes unicode (2-byte) character strings, the
    ' ConvertToWordArray function uses on the first byte for each character. This
    ' has been done this way for ease of use, to make the routine truely portable
    ' you could accept a byte array instead, it would then be up to the calling
    ' routine to make sure that the byte array is generated from their string in
    ' a manner consistent with the string type.
    '*******************************************************************************
    Public Function MD5(sMessage As String) As String
       Dim x() As Long
       Dim k   As Long
       Dim AA  As Long
       Dim BB  As Long
       Dim CC  As Long
       Dim DD  As Long
       Dim a   As Long
       Dim b   As Long
       Dim c   As Long
       Dim d   As Long
           
       Const S11 As Long = 7
       Const S12 As Long = 12
       Const S13 As Long = 17
       Const S14 As Long = 22
       Const S21 As Long = 5
       Const S22 As Long = 9
       Const S23 As Long = 14
       Const S24 As Long = 20
       Const S31 As Long = 4
       Const S32 As Long = 11
       Const S33 As Long = 16
       Const S34 As Long = 23
       Const S41 As Long = 6
       Const S42 As Long = 10
       Const S43 As Long = 15
       Const S44 As Long = 21
       
       ' fill constant arrays
       InitConstArrays
           
       ' Steps 1 and 2.  Append padding bits and length and convert to words
       x = ConvertToWordArray(sMessage)
       
       ' Step 3.  Initialise
       a = &H67452301
       b = &HEFCDAB89
       c = &H98BADCFE
       d = &H10325476
       
       ' Step 4.  Process the message in 16-word blocks
       For k = 0 To UBound(x) Step 16
          AA = a
          BB = b
          CC = c
          DD = d
           
          ' The hex number on the end of each of the following procedure calls is
          ' an element from the 64 element table constructed with
          ' T(i) = Int(4294967296 * Abs(Sin(i))) where i is 1 to 64.
          '
          ' However, for speed we don't want to calculate the value every time.
          FF a, b, c, d, x(k + 0), S11, &HD76AA478
          FF d, a, b, c, x(k + 1), S12, &HE8C7B756
          FF c, d, a, b, x(k + 2), S13, &H242070DB
          FF b, c, d, a, x(k + 3), S14, &HC1BDCEEE
          FF a, b, c, d, x(k + 4), S11, &HF57C0FAF
          FF d, a, b, c, x(k + 5), S12, &H4787C62A
          FF c, d, a, b, x(k + 6), S13, &HA8304613
          FF b, c, d, a, x(k + 7), S14, &HFD469501
          FF a, b, c, d, x(k + 8), S11, &H698098D8
          FF d, a, b, c, x(k + 9), S12, &H8B44F7AF
          FF c, d, a, b, x(k + 10), S13, &HFFFF5BB1
          FF b, c, d, a, x(k + 11), S14, &H895CD7BE
          FF a, b, c, d, x(k + 12), S11, &H6B901122
          FF d, a, b, c, x(k + 13), S12, &HFD987193
          FF c, d, a, b, x(k + 14), S13, &HA679438E
          FF b, c, d, a, x(k + 15), S14, &H49B40821
           
          GG a, b, c, d, x(k + 1), S21, &HF61E2562
          GG d, a, b, c, x(k + 6), S22, &HC040B340
          GG c, d, a, b, x(k + 11), S23, &H265E5A51
          GG b, c, d, a, x(k + 0), S24, &HE9B6C7AA
          GG a, b, c, d, x(k + 5), S21, &HD62F105D
          GG d, a, b, c, x(k + 10), S22, &H2441453
          GG c, d, a, b, x(k + 15), S23, &HD8A1E681
          GG b, c, d, a, x(k + 4), S24, &HE7D3FBC8
          GG a, b, c, d, x(k + 9), S21, &H21E1CDE6
          GG d, a, b, c, x(k + 14), S22, &HC33707D6
          GG c, d, a, b, x(k + 3), S23, &HF4D50D87
          GG b, c, d, a, x(k + 8), S24, &H455A14ED
          GG a, b, c, d, x(k + 13), S21, &HA9E3E905
          GG d, a, b, c, x(k + 2), S22, &HFCEFA3F8
          GG c, d, a, b, x(k + 7), S23, &H676F02D9
          GG b, c, d, a, x(k + 12), S24, &H8D2A4C8A
                   
          HH a, b, c, d, x(k + 5), S31, &HFFFA3942
          HH d, a, b, c, x(k + 8), S32, &H8771F681
          HH c, d, a, b, x(k + 11), S33, &H6D9D6122
          HH b, c, d, a, x(k + 14), S34, &HFDE5380C
          HH a, b, c, d, x(k + 1), S31, &HA4BEEA44
          HH d, a, b, c, x(k + 4), S32, &H4BDECFA9
          HH c, d, a, b, x(k + 7), S33, &HF6BB4B60
          HH b, c, d, a, x(k + 10), S34, &HBEBFBC70
          HH a, b, c, d, x(k + 13), S31, &H289B7EC6
          HH d, a, b, c, x(k + 0), S32, &HEAA127FA
          HH c, d, a, b, x(k + 3), S33, &HD4EF3085
          HH b, c, d, a, x(k + 6), S34, &H4881D05
          HH a, b, c, d, x(k + 9), S31, &HD9D4D039
          HH d, a, b, c, x(k + 12), S32, &HE6DB99E5
          HH c, d, a, b, x(k + 15), S33, &H1FA27CF8
          HH b, c, d, a, x(k + 2), S34, &HC4AC5665
           
          II a, b, c, d, x(k + 0), S41, &HF4292244
          II d, a, b, c, x(k + 7), S42, &H432AFF97
          II c, d, a, b, x(k + 14), S43, &HAB9423A7
          II b, c, d, a, x(k + 5), S44, &HFC93A039
          II a, b, c, d, x(k + 12), S41, &H655B59C3
          II d, a, b, c, x(k + 3), S42, &H8F0CCC92
          II c, d, a, b, x(k + 10), S43, &HFFEFF47D
          II b, c, d, a, x(k + 1), S44, &H85845DD1
          II a, b, c, d, x(k + 8), S41, &H6FA87E4F
          II d, a, b, c, x(k + 15), S42, &HFE2CE6E0
          II c, d, a, b, x(k + 6), S43, &HA3014314
          II b, c, d, a, x(k + 13), S44, &H4E0811A1
          II a, b, c, d, x(k + 4), S41, &HF7537E82
          II d, a, b, c, x(k + 11), S42, &HBD3AF235
          II c, d, a, b, x(k + 2), S43, &H2AD7D2BB
          II b, c, d, a, x(k + 9), S44, &HEB86D391
           
          a = AddUnsigned(a, AA)
          b = AddUnsigned(b, BB)
          c = AddUnsigned(c, CC)
          d = AddUnsigned(d, DD)
       Next
           
       ' Step 5.  Output the 128 bit digest
       MD5 = LCase(WordToHex(a) & WordToHex(b) & WordToHex(c) & WordToHex(d))
    
    End Function

    [=========> Bổ sung bài viết <=========]

    2. Còn đây là hàm tạo mật khẩu ngẫu nhiên gồm các ký tự hoa, thường, số
    Code:
    Private Function RandomPassword(Optional MAX_LEN = 10) As String
       Dim mPass As String
       Dim I As Long
       Const PASS_CHARSET = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
       Randomize
       '--
       mPass = ""
       For I = 1 To MAX_LEN
          mPass = mPass & Mid$(PASS_CHARSET, Round(Rnd * Len(PASS_CHARSET) + 1), 1)
       Next
       RandomPassword = mPass
    End Function
    Chúc bạn thành công,
    Được sửa bởi huytranaz lúc 17:04 ngày 31-07-2009 Reason: Bổ sung bài viết

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