如何破解函数加密
如何破解函数加密。听说有个系统自代的方法。

------解决方案--------------------
破解SQL SERVER 加密存储过程,函数,触发器,视图
create PROCEDURE sp_decrypt(@objectName varchar(50))
AS
begin
set nocount on
--CSDN：j9988 copyright:2004.01.05
--V3.1
--破解字节不受限制，适用于SQLSERVER2000存储过程，函数，视图，触发器
--发现有错，请E_MAIL：CSDNj9988@tom.com
begin tran
declare @objectname1 varchar(100),@orgvarbin varbinary(8000)
declare @sql1 nvarchar(4000),@sql2 varchar(8000),@sql3 nvarchar(4000),@sql4 nvarchar(4000)
DECLARE @OrigSpText1 nvarchar(4000), @OrigSpText2 nvarchar(4000) , @OrigSpText3 nvarchar(4000), @resultsp nvarchar(4000)
declare @i int,@status int,@type varchar(10),@parentid int
declare @colid int,@n int,@q int,@j int,@k int,@encrypted int,@number int
select @type=xtype,@parentid=parent_obj from sysobjects where id=object_id(@ObjectName)

create table #temp(number int,colid int,ctext varbinary(8000),encrypted int,status int)
insert #temp SELECT number,colid,ctext,encrypted,status FROM syscomments WHERE id = object_id(@objectName)
select @number=max(number) from #temp
set @k=0

while @k <=@number
begin
if exists(select 1 from syscomments where id=object_id(@objectname) and number=@k)
begin
if @type=\ 'P\ '
set @sql1=(case when @number> 1 then \ 'ALTER PROCEDURE \ '+ @objectName +\ ';\ '+rtrim(@k)+\ ' WITH ENCRYPTION AS \ '
else \ 'ALTER PROCEDURE \ '+ @objectName+\ ' WITH ENCRYPTION AS \ '
end)

if @type=\ 'TR\ '
set @sql1=\ 'ALTER TRIGGER \ '+@objectname+\ ' ON \ '+OBJECT_NAME(@parentid)+\ ' WITH ENCRYPTION FOR INSERT AS PRINT 1 \ '

if @type=\ 'FN\ ' or @type=\ 'TF\ ' or @type=\ 'IF\ '
set @sql1=(case @type when \ 'TF\ ' then
\ 'ALTER FUNCTION \ '+ @objectName+\ '(@a char(1)) returns @b table(a varchar(10)) with encryption as begin insert @b select @a return end \ '
when \ 'FN\ ' then
\ 'ALTER FUNCTION \ '+ @objectName+\ '(@a char(1)) returns char(1) with encryption as begin return @a end\ '
when \ 'IF\ ' then
\ 'ALTER FUNCTION \ '+ @objectName+\ '(@a char(1)) returns table with encryption as return select @a as a\ '
end)

if @type=\ 'V\ '
set @sql1=\ 'ALTER VIEW \ '+@objectname+\ ' WITH ENCRYPTION AS SELECT 1 as f\ '

set @q=len(@sql1)
set @sql1=@sql1+REPLICATE(\ '-\ ',4000-@q)
select @sql2=REPLICATE(\ '-\ ',8000)
set @sql3=\ 'exec(@sql1\ '
select @colid=max(colid) from #temp where number=@k
set @n=1
while @n <=CEILING(1.0*(@colid-1)/2) and len(@sQL3) <=3996
begin
set @sql3=@sql3+\ '+@\ '
set @n=@n+1
end
set @sql3=@sql3+\ ')\ '
exec sp_executesql @sql3,N\ '@Sql1 nvarchar(4000),@ varchar(8000)\ ',@sql1=@sql1,@=@sql2

end
set @k=@k+1
end

set @k=0
while @k <=@number
begin

if exists(select 1 from syscomments where id=object_id(@objectname) and number=@k)
begin
select @colid=max(colid) from #temp where number=@k
set @n=1

while @n <=@colid
begin
select @OrigSpText1=ctext,@encrypted=encrypted,@status=status FROM #temp WHERE colid=@n and number=@k

SET @OrigSpText3=(SELECT ctext FROM syscomments WHERE id=object_id(@objectName) and colid=@n and number=@k)
if @n=1
begin
if @type=\ 'P\ '
SET @OrigSpText2=(case when @number> 1 then \ 'CREATE PROCEDURE \ '+ @objectName +\ ';\ '+rtrim(@k)+\ '