在不破坏原加密存储过程的前提下,解密存储过程!(补充j9988)

    技术2022-05-11  113

    实为 转贴j9988+原创

    begin transaction? --playyuer 原创exec sp_decrypt 'AppSP_test' --j9988 原创rollback transaction --playyuer 原创

    or:或者直接用 transaction 把 j9988 包起来!begin transactionj9988rollback transaction

    /************* 解密存储过程 **********------------------------sql2000大于40000的-----------------原作:j9988 号:J老师*/alter? PROCEDURE sp_decrypt (@objectName varchar(50))ASbegin

    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 bigintdeclare @m int,@n int,@q intset @m=(SELECT max(colid) FROM syscomments? WHERE id = object_id(@objectName))set @n=1--get encrypted datacreate 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<=@mbeginSET @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=1beginSET @OrigSpText2='CREATE PROCEDURE '+ @objectName +' WITH ENCRYPTION AS '--set @q=4000-len(@OrigSpText2)set @OrigSpText2=@OrigSpText2+REPLICATE('-',@q)endelsebeginSET @OrigSpText2=REPLICATE('-', 4000)end--start counterSET @i=1--fill temporary variableSET @resultsp = replicate(N'A', (datalength(@OrigSpText1) / 2))

    --loopWHILE @i<=datalength(@OrigSpText1)/2BEGIN--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+1END--drop original SP--EXECUTE ('drop PROCEDURE '+ @objectName)--remove encryption--preserve caseSET @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 enryptionprint @resultsp--execute( @resultsp)set @n=@n+1enddrop table #tempendrollback transaction --add by playyuerGO

    /*适合40000字符。每次4000 print出来,自已贴。切记:我见过的解过程都是对原过程进行破坏。破解前一定要备份!!!!超过40000的,自已加SQL(我上面用SQL.SQL2--SQL10)超长的可加SQL11--sql20........*/

    ?


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