因为这两天在弄自己的一个问答程序www.sosoask.com ，结果发现开发人员把我的存储过程加密了，郁闷，还好找到解决方法了，现在共享下：

打开查询分析器，定位到你所在的数据库，执行以下SQL语句：

CREATE PROCEDURE sp_decrypt (@objectName varchar(50))

AS

begin

begin transaction

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

rollback transaction

end

GO

说明一下：其中的@objectName要全部换成你所要破解的存储过程名称(比如我的就换成：@www.***.com)。

这样你就在你的数据库里新建了一个叫做sp_decrypt的存储过程，

现在在查询分析器里面调用sp_decrypt来解密需要破解的存储过程，语句如下：

exec sp_decrypt ‘存储过程名称’

具体的语句我来举个例子：exec sp_decrypt ‘www.***.com’ (注意名称有“单引号”)。

这样解密完成，直接拷贝出Sql脚本到你的服务器上恢复吧。

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