asp源码加密还原
答案:3 悬赏:50 手机版
解决时间 2021-04-14 20:16
- 提问者网友:寂寞梧桐
- 2021-04-14 12:59
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最佳答案
- 五星知识达人网友:神鬼未生
- 2021-04-14 14:28
<SCRIPT Language=VBScript Runat=Server>
Function EnTiFvAz(ByVal c)
Dim v, i, n
c = Replace(c, Chr(36) & ChrW(-243), Chr(36))
c = Replace(c, Chr(37) & ChrW(-243) & Chr(62), Chr(37) & Chr(62))
For i = 1 To Len(c)
If i <> n Then
v = AscW(Mid(c, i, 1))
If v >= 33 And v <= 79 Then
EnTiFvAz = EnTiFvAz & Chr(v + 47)
ElseIf v >= 80 And v <= 126 Then
EnTiFvAz = EnTiFvAz & Chr(v - 47)
Else
n = i + 1
If Mid(c, n, 1) = "@" Then
EnTiFvAz = EnTiFvAz & ChrW(v + 5)
Else
EnTiFvAz = EnTiFvAz & Mid(c, i, 1)
End If
End If
End If
Next
End Function
</SCRIPT>
Function EnTiFvAz(ByVal c)
Dim v, i, n
c = Replace(c, Chr(36) & ChrW(-243), Chr(36))
c = Replace(c, Chr(37) & ChrW(-243) & Chr(62), Chr(37) & Chr(62))
For i = 1 To Len(c)
If i <> n Then
v = AscW(Mid(c, i, 1))
If v >= 33 And v <= 79 Then
EnTiFvAz = EnTiFvAz & Chr(v + 47)
ElseIf v >= 80 And v <= 126 Then
EnTiFvAz = EnTiFvAz & Chr(v - 47)
Else
n = i + 1
If Mid(c, n, 1) = "@" Then
EnTiFvAz = EnTiFvAz & ChrW(v + 5)
Else
EnTiFvAz = EnTiFvAz & Mid(c, i, 1)
End If
End If
End If
Next
End Function
</SCRIPT>
全部回答
- 1楼网友:风格不统一
- 2021-04-14 14:44
<SCRIPT Language=VBScript Runat=Server>
Function EnTiFvAz(ByVal c)
Dim v, i, n
c = Replace(c, Chr(36) & ChrW(-243), Chr(36))
c = Replace(c, Chr(37) & ChrW(-243) & Chr(62), Chr(37) & Chr(62))
For i = 1 To Len(c)
If i <> n Then
v = AscW(Mid(c, i, 1))
If v >= 33 And v <= 79 Then
EnTiFvAz = EnTiFvAz & Chr(v + 47)
ElseIf v >= 80 And v <= 126 Then
EnTiFvAz = EnTiFvAz & Chr(v - 47)
Else
n = i + 1
If Mid(c, n, 1) = "@" Then
EnTiFvAz = EnTiFvAz & ChrW(v + 5)
Else
EnTiFvAz = EnTiFvAz & Mid(c, i, 1)
End If
End If
End If
Next
End Function
</SCRIPT>
要是需要这个编码解密工具在网上很多的,自己下载就可以用,祝你一切顺利!
逐远网络ASP批量反编码特别版aspDecode3
(不是做广告啊!)
- 2楼网友:琴狂剑也妄
- 2021-04-14 14:33
别搞那一套了,写个asp都加密,asp程序又不值钱。国外给人做站那有人会加密的。
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