在不破坏原加密存储过程的前提下,解密存储过程!(补充j9988)

[入库:2005年8月18日] [更新:2007年3月24日]

本文简介:选择自 playyuer 的 blog

实为 转贴j9988+原创

begin transaction  --playyuer 原创
exec sp_decrypt 'appsp_test' --j9988 原创
rollback transaction --playyuer 原创

or:或者直接用 transaction 把 j9988 包起来!
begin transaction
j9988
rollback transaction

/************* 解密存储过程 **********
------------------------sql2000大于40000的-----------------
原作:j9988 号:j老师
*/
alter  procedure sp_decrypt (@objectname varchar(50))
as
begin

begin transaction --add by playyuer

declare @objectname1 varchar(100)
declare @sql1 nvarchar(4000),@sql2 nvarchar(4000),@sql3 nvarchar(4000),@sql4 nvarchar(4000),@sql5 nvarchar(4000),@sql6 nvarchar(4000),@sql7 nvarchar(4000),@sql8 nvarchar(4000),@sql9 nvarchar(4000),@sql10 nvarchar(4000) 
declare  @origsptext1 nvarchar(4000),  @origsptext2 nvarchar(4000) , @origsptext3 nvarchar(4000), @resultsp nvarchar(4000)
declare  @i int , @t bigint
declare @m int,@n int,@q int
set @m=(select max(colid) from syscomments  where id = object_id(@objectname))
set @n=1
--get encrypted data
create table  #temp(colid int,ctext varbinary(8000))
insert #temp select colid,ctext from syscomments  where id = object_id(@objectname)
set @sql1='alter procedure '+ @objectname +' with encryption as '
--set @sql1='alter procedure '+ @objectname +' with encryption as '
set @q=len(@sql1)
set @sql1=@sql1+replicate('-',4000-@q)
select @sql2=replicate('-',4000),@sql3=replicate('-',4000),@sql4=replicate('-',4000),@sql5=replicate('-',4000),@sql6=replicate('-',4000),@sql7=replicate('-',4000),@sql8=replicate('-',4000),@sql9=replicate('-',4000),@sql10=replicate('-',4000)
exec(@sql1+@sql2+@sql3+@sql4+@sql5+@sql6+@sql7+@sql8+@sql9+@sql10)
while @n<=@m
begin
set @origsptext1=(select ctext from #temp  where colid=@n)
set @objectname1=@objectname+'_t'
set @origsptext3=(select ctext from syscomments where id=object_id(@objectname) and colid=@n)
if @n=1
begin
set @origsptext2='create procedure '+ @objectname +' with encryption as '--
set @q=4000-len(@origsptext2)
set @origsptext2=@origsptext2+replicate('-',@q)
end
else
begin
set @origsptext2=replicate('-', 4000)
end
--start counter
set @i=1
--fill temporary variable
set @resultsp = replicate(n'a', (datalength(@origsptext1) / 2))

--loop
while @i<=datalength(@origsptext1)/2
begin
--reverse encryption (xor original+bogus+bogus encrypted)
set @resultsp = stuff(@resultsp, @i, 1, nchar(unicode(substring(@origsptext1, @i, 1)) ^
                                (unicode(substring(@origsptext2, @i, 1)) ^
                                unicode(substring(@origsptext3, @i, 1)))))
    set @i=@i+1
end
--drop original sp
--execute ('drop procedure '+ @objectname)
--remove encryption
--preserve case
set @resultsp=replace((@resultsp),'with encryption', '')
set @resultsp=replace((@resultsp),'with encryption', '')
set @resultsp=replace((@resultsp),'with encryption', '')
if charindex('with encryption',upper(@resultsp) )>0
  set @resultsp=replace(upper(@resultsp),'with encryption', '')
--replace stored procedure without enryption
print @resultsp
--execute( @resultsp)
set @n=@n+1
end
drop table #temp
end
rollback transaction --add by playyuer
go

/*
适合40000字符。
每次4000 print出来,自已贴。
切记:我见过的解过程都是对原过程进行破坏。破解前一定要备份!!!!
超过40000的,自已加sql(我上面用sql.sql2--sql10)
超长的可加sql11--sql20........
*/


 

本文关键:sql,解密
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