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再来一套加解密字符串的function

来源:网络 作者:佚名 点击: 时间:2017-06-22 20:40
[摘要] % 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
<%
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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