虽然现在MD5 加密算法严格来讲已经不算安全，但一般小功能或公司内部使用应该足够了
Attribute VB_Name = "modMd5"
' MODULE: CMD5
'*******************************************************************************
Option Explicit
Public Const BITS_TO_A_BYTE As Long = 8
Public Const BYTES_TO_A_WORD As Long = 4
Public Const BITS_TO_A_WORD As Long = BYTES_TO_A_WORD * BITS_TO_A_BYTE
Public m_lOnBits(0 To 30) As Long
Public m_l2Power(0 To 30) As Long
Public Function MD5_Encrypt(ByVal sText As String) As String
MD5_Init
MD5_Encrypt = MD5(sText)
End Function
'*******************************************************************************
' Class_Initialize (SUB)
'
' DESCRIPTION:
' We will usually get quicker results by preparing arrays of bit patterns and
' powers of 2 ahead of time instead of calculating them every time,unless of
' course the methods are only ever getting called once per instantiation of the
' class.
'*******************************************************************************
Public Sub MD5_Init()
' 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
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 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 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 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,AddUnsigned(AddUnsigned(G(b,b)
End Sub
'*******************************************************************************
' HH (SUB)
'
' DESCRIPTION:
' MD5's HH procedure
'*******************************************************************************
Private Sub HH(a As Long,AddUnsigned(AddUnsigned(H(b,b)
End Sub
'*******************************************************************************
' II (SUB)
'
' DESCRIPTION:
' MD5's II procedure
'*******************************************************************************
Private Sub II(a As Long,AddUnsigned(AddUnsigned(i(b,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. Note we are using only the first byte of each character with
' the AscB function,this may well mess up in unicode/dbcs situations where you
' are comparing what was generated on two different PCs with different
' character sets.
'*******************************************************************************
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
Dim lChar 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
lChar = AscB(Mid(sMessage,lByteCount + 1,1))
lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(lChar,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
' 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,d,x(k + 0),S11,&HD76AA478
FF d,a,x(k + 1),S12,&HE8C7B756
FF c,x(k + 2),S13,&H242070DB
FF b,x(k + 3),S14,&HC1BDCEEE
FF a,x(k + 4),&HF57C0FAF
FF d,x(k + 5),&H4787C62A
FF c,x(k + 6),&HA8304613
FF b,x(k + 7),&HFD469501
FF a,x(k + 8),&H698098D8
FF d,x(k + 9),&H8B44F7AF
FF c,x(k + 10),&HFFFF5BB1
FF b,x(k + 11),&H895CD7BE
FF a,x(k + 12),&H6B901122
FF d,x(k + 13),&HFD987193
FF c,x(k + 14),&HA679438E
FF b,x(k + 15),&H49B40821
GG a,S21,&HF61E2562
GG d,S22,&HC040B340
GG c,S23,&H265E5A51
GG b,S24,&HE9B6C7AA
GG a,&HD62F105D
GG d,&H2441453
GG c,&HD8A1E681
GG b,&HE7D3FBC8
GG a,&H21E1CDE6
GG d,&HC33707D6
GG c,&HF4D50D87
GG b,&H455A14ED
GG a,&HA9E3E905
GG d,&HFCEFA3F8
GG c,&H676F02D9
GG b,&H8D2A4C8A
HH a,S31,&HFFFA3942
HH d,S32,&H8771F681
HH c,S33,&H6D9D6122
HH b,S34,&HFDE5380C
HH a,&HA4BEEA44
HH d,&H4BDECFA9
HH c,&HF6BB4B60
HH b,&HBEBFBC70
HH a,&H289B7EC6
HH d,&HEAA127FA
HH c,&HD4EF3085
HH b,&H4881D05
HH a,&HD9D4D039
HH d,&HE6DB99E5
HH c,&H1FA27CF8
HH b,&HC4AC5665
II a,S41,&HF4292244
II d,S42,&H432AFF97
II c,S43,&HAB9423A7
II b,S44,&HFC93A039
II a,&H655B59C3
II d,&H8F0CCC92
II c,&HFFEFF47D
II b,&H85845DD1
II a,&H6FA87E4F
II d,&HFE2CE6E0
II c,&HA3014314
II b,&H4E0811A1
II a,&HF7537E82
II d,&HBD3AF235
II c,&H2AD7D2BB
II b,&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
'=LCase(WordToHex(a) & WordToHex(b) & WordToHex(c) & WordToHex(d))
MD5 = LCase(WordToHex(b) & WordToHex(c))
End Function
Public Function Encrypt(ByVal sText As String,Optional ByVal bNewEncrypt As Boolean = False) As String
If bNewEncrypt = False Then
Dim m_objEncrypt As Object
Set m_objEncrypt = CreateObject("StrongEncrypt.Encrypt")
Encrypt = m_objEncrypt.Encrypt(sText)
Set m_objEncrypt = Nothing
Else
Encrypt = MD5_Encrypt(sText)
End If
End Function