因为这两天在弄自己的一个问答程序 ，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。