数据库加密问题?

cnepine 2005-03-04 10:18:23
在我们的软件中使用的是SQL server 2000数据库,我在SQL Server 2000中设计的数据库的表结构,存储过程,视图等,在软件安装好以后用户就可以在SQL server 2000中一览无余的看到我的设计情况,现在我想通过加密技术,使用户不能浏览到我的设计,请问,我该如何办?
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cnepine 2005-03-08
这个问题可就难办了啊
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pbsh 2005-03-04
wo kao!so chang!!!
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zjcxc 2005-03-04
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[sp_decrypt]') and OBJECTPROPERTY(id, N'IsProcedure') = 1)
drop procedure [dbo].[sp_decrypt]
GO

/*--破解函数,过程,触发器,视图.仅限于SQLSERVER2000

--作者:J9988--*/
/*--调用示例

--解密指定存储过程
exec sp_decrypt 'AppSP_test'

--对所有的存储过程解密
declare tb cursor for
select name from sysobjects where xtype='P' and status>0 and name<>'sp_decrypt'

declare @name sysname
open tb
fetch next from tb into @name
while @@fetch_status=0
begin
print '/*-------存储过程 ['+@name+'] -----------*/'
exec sp_decrypt @name
fetch next from tb into @name
end
close tb
deallocate tb
--*/

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

CREATE PROCEDURE sp_decrypt(@objectName varchar(50))
AS
begin
set nocount on
--CSDN:j9988 copyright:2004.04.15
--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
--修正上一版视图触发器不能正确解密错误
--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'
begin
declare @parent_obj varchar(255),@tr_parent_xtype varchar(10)
select @parent_obj=parent_obj from sysobjects where id=object_id(@objectName)
select @tr_parent_xtype=xtype from sysobjects where id=@parent_obj
if @tr_parent_xtype='V'
begin
set @sql1='ALTER TRIGGER '+@objectname+' ON '+OBJECT_NAME(@parentid)+' WITH ENCRYPTION INSTERD OF INSERT AS PRINT 1 '
end
else
begin
set @sql1='ALTER TRIGGER '+@objectname+' ON '+OBJECT_NAME(@parentid)+' WITH ENCRYPTION FOR INSERT AS PRINT 1 '
end

end
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)+' WITH ENCRYPTION AS '
else 'CREATE PROCEDURE '+ @objectName +' WITH ENCRYPTION AS '
end)


if @type='FN' or @type='TF' or @type='IF'
SET @OrigSpText2=(case @type when 'TF' then
'CREATE FUNCTION '+ @objectName+'(@a char(1)) returns @b table(a varchar(10)) with encryption as begin insert @b select @a return end '
when 'FN' then
'CREATE FUNCTION '+ @objectName+'(@a char(1)) returns char(1) with encryption as begin return @a end'
when 'IF' then
'CREATE FUNCTION '+ @objectName+'(@a char(1)) returns table with encryption as return select @a as a'
end)

if @type='TR'
begin

if @tr_parent_xtype='V'
begin
set @OrigSpText2='CREATE TRIGGER '+@objectname+' ON '+OBJECT_NAME(@parentid)+' WITH ENCRYPTION INSTEAD OF INSERT AS PRINT 1 '
end
else
begin
set @OrigSpText2='CREATE TRIGGER '+@objectname+' ON '+OBJECT_NAME(@parentid)+' WITH ENCRYPTION FOR INSERT AS PRINT 1 '
end

end

if @type='V'
set @OrigSpText2='CREATE VIEW '+@objectname+' WITH ENCRYPTION AS SELECT 1 as f'

set @q=4000-len(@OrigSpText2)
set @OrigSpText2=@OrigSpText2+REPLICATE('-',@q)
end
else
begin
SET @OrigSpText2=REPLICATE('-', 4000)
end
SET @i=1

SET @resultsp = replicate(N'A', (datalength(@OrigSpText1) / 2))

WHILE @i<=datalength(@OrigSpText1)/2
BEGIN

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
set @orgvarbin=cast(@OrigSpText1 as varbinary(8000))
set @resultsp=(case when @encrypted=1
then @resultsp
else convert(nvarchar(4000),case when @status&2=2 then uncompress(@orgvarbin) else @orgvarbin end)
end)
print @resultsp

set @n=@n+1

end

end
set @k=@k+1
end

drop table #temp
rollback tran
end
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pbsql 2005-03-04
那就解密吧

你应该保留创建存储过程的代码
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cnepine 2005-03-04
存储过程等加密以后就不能在对其进行编辑了,但是我还想对其进行编辑,现在怎么办呢?
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zjcxc 2005-03-04
数据安全可以通过用户来管理
你可以安装一个msde,自己设置所有用户的密码,不告诉用户密码
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zjcxc 2005-03-04
表结构也加密,那每次数据处理都要解密,其麻烦程度和效率之低可想而知
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子陌红尘 2005-03-04
没有什么办法,即使使用数据库自己的加密途径加密存储过程和函数,也还是可以破解饿。
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Softlee81307 2005-03-04
数据库的表结构沒辦法加密,
過程和視圖用 with encryption

create proc kk
with encryption
as
.....
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pbsql 2005-03-04
凉拌

开玩笑了
存储过程,视图等可以加WITH ENCRYPTION进行加密,但有办法解密,所以还是达不到效果,只能从管理上下功夫,设置权限不让别人浏览
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hglhyy 2005-03-04
收藏之!
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delphi_dcs 2005-03-04
you bad,
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TigerSuper 2005-03-04
只能通过权限控制了
数据结构没法加密
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