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<% 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> 返回类别: 教程 上一教程: ASP进阶之文章在线治理更新(6) 下一教程: BBS的树状结构算法 您可以阅读与"再来一套加解密字符串的FUNCTION"相关的教程: · 一套加解密字符串的函数 · 一套相称实用的加解密字符串的函数 · 检测含有中文字符串的实际长度 · 控制输出字符串的长度，可以区别中英文 · 用两种不同的方式来求实现数字字符串的最大，最小和平均值 |
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