找到了，解密SQL2000的存储过程

创建下面这个存储过程
然后调用
exec sp_decrypt @objectName
(@objectName就是加密过的存储过程名)
暴露无遗！
此程序不破坏原有存储过程！
呵呵呵，真好玩！
create    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