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一套相当实用的加解密字符串的函数。 |
<%
Function Encrypt(theNumber)
On Error Resume Next
Dim n, szEnc, t, HiN, LoN, i
n = CDbl((theNumber + 1570) ^ 2 - 7 * (theNumber + 1570) - 450)
If n < 0 Then szEnc = "R" Else szEnc = "J"
n = CStr(abs(n))
For i = 1 To Len(n) step 2
t = Mid(n, i, 2)
If Len(t) = 1 Then
szEnc = szEnc & t
Exit For
End If
HiN = (CInt(t) And 240) / 16
LoN = CInt(t) And 15
szEnc = szEnc & Chr(Asc("M") + HiN) & Chr(Asc("C") + LoN)
Next
Encrypt = szEnc
End Function
Function Decrypt(theNumber)
On Error Resume Next
Dim e, n, sign, t, HiN, LoN, NewN, i
e = theNumber
If Left(e, 1) = "R" Then sign = -1 Else sign = 1
e = Mid(e, 2)
NewN = ""
For i = 1 To Len(e) step 2
t = Mid(e, i, 2)
If Asc(t) >= Asc("0") And Asc(t) <= Asc("9") Then
NewN = NewN & t
Exit For
End If
HiN = Mid(t, 1, 1)
LoN = Mid(t, 2, 1)
HiN = (Asc(HiN) - Asc("M")) * 16
LoN = Asc(LoN) - Asc("C")
t = CStr(HiN Or LoN)
If Len(t) = 1 Then t = "0" & t
NewN = NewN & t
Next
e = CDbl(NewN) * sign
Decrypt = CLng((7 + sqr(49 - 4 * (-450 - e))) / 2 - 1570)
End Function
%>
<html><body>
Original number: 69 <br>
Encrypt(69) returns: JNMQMOJ8 <br>
Decrypt("JNMQMOJ8") returns: 69
<p>
Another example using variables instead: <br>
Encrypt(Request.Form("ID")) <br>
Encrypt(myVar) <br>
Decrypt(Request.QueryString("id")) <br>
Decrypt("JNMQMOJ8") <br>
Decrypt(myVar)
</body></html>
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