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我有一個加密方法，歡迎大家指正

發(fā)表時間：2024-05-29 來源：明輝站整理相關軟件相關文章人氣：

[摘要]<%IF Request("Action")=1 thenDim String,StringLen,i,StringNumTmp,StringRndNum,ResultString,jString = Request("Pass") ...

<%
IF Request("Action")=1 then
Dim String,StringLen,i,StringNumTmp,StringRndNum,ResultString,j
String = Request("Pass")            '密碼字符串

'開始計算字符數(shù)據(jù)
StringLen = Len(String)

For i = 1 to StringLen

  StringNumTmp = Asc(Mid(String,i,1))
  
  Randomize
  StringRndNum=Int((18-1)*Rnd+1)
  
  if Len(StringNumTmp + StringRndNum) < 3 then
  
  For j = 1 to 3 - Int(Len(StringNumTmp + StringRndNum))
    StringNumber = "0" & (StringNumTmp + StringRndNum)
  Next
  
  Else
  
    StringNumber = StringNumTmp + StringRndNum
  
  End if
  
  ResultString = ResultString & Chr(StringNumTmp - StringRndNum) & StringNumber

Next

Response.write "加密后結(jié)果：" & ResultString            '輸出結(jié)果

%>
<BR>
<%
'開始破解

For i = 1 to Len(ResultString) Step 4

  PString = Left(Mid(ResultString,i,4),1)
  
  PStringNum1 = Asc(PString)
  
  PStringNum2 = Int(Right(Mid(ResultString,i,4),3))
  
  PStringNum = (PStringNum1 + PStringNum2) / 2
  
  PString1 = PString1 & Chr(PStringNum)

Next

Response.write "破解后結(jié)果：" & PString1
End if
%>
<Form method="post" action="ttt.asp?action=1">
<input type=password name="pass" size=15><input type=submit>
</form>