使用UltraEdit修改oracle二进制文件

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标题:使用UltraEdit修改oracle二进制文件

作者:惜分飞©版权所有[未经本人同意,不得以任何形式转载,否则有进一步追究法律责任的权利.]

对数据库OPEN的过程做10046,因为数据库没有open,那写着语句是写在什么地方的呢?通过使用UltraEdit打开$ORACLE_HOME/bin/oracle文件,可以发现很多sql语句,而且与open过程的10046很多一致.

044acdc0h: 73 65 6C 65 63 74 20 75 2E 6E 61 6D 65 2C 20 6F ; select u.name, o
044acdd0h: 32 2E 6E 61 6D 65 2C 20 6F 32 2E 6F 62 6A 23 20 ; 2.name, o2.obj# 
044acde0h: 66 72 6F 6D 20 69 6E 64 24 20 69 2C 20 6F 62 6A ; from ind$ i, obj
044acdf0h: 24 20 6F 31 2C 20 6F 62 6A 24 20 6F 32 2C 20 75 ; $ o1, obj$ o2, u
044ace00h: 73 65 72 24 20 75 20 77 68 65 72 65 20 6F 31 2E ; ser$ u where o1.
044ace10h: 6F 77 6E 65 72 23 20 3D 20 3A 31 20 20 61 6E 64 ; owner# = :1  and
044ace20h: 20 6F 31 2E 74 79 70 65 23 20 3D 20 32 20 61 6E ;  o1.type# = 2 an
044ace30h: 64 20 69 2E 74 79 70 65 23 20 3D 20 39 20 61 6E ; d i.type# = 9 an
044ace40h: 64 20 69 2E 62 6F 23 20 3D 20 6F 31 2E 6F 62 6A ; d i.bo# = o1.obj
044ace50h: 23 20 20 61 6E 64 20 69 2E 6F 62 6A 23 20 3D 20 ; #  and i.obj# = 
044ace60h: 6F 32 2E 6F 62 6A 23 20 61 6E 64 20 6F 32 2E 6F ; o2.obj# and o2.o
044ace70h: 77 6E 65 72 23 20 3D 20 75 2E 75 73 65 72 23 20 ; wner# = u.user# 
044ace80h: 61 6E 64 20 69 2E 74 79 70 65 23 20 3D 20 39 20 ; and i.type# = 9 
044ace90h: 61 6E 64 20 6F 31 2E 6F 77 6E 65 72 23 20 21 3D ; and o1.owner# !=
044acea0h: 20 6F 32 2E 6F 77 6E 65 72 23 00 00 00 00 00 00 ;  o2.owner#......
044aceb0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
044acec0h: 73 65 6C 65 63 74 20 75 2E 6E 61 6D 65 2C 20 6F ; select u.name, o
044aced0h: 2E 6E 61 6D 65 2C 20 6F 2E 6F 62 6A 23 20 66 72 ; .name, o.obj# fr
044acee0h: 6F 6D 20 6F 62 6A 24 20 6F 2C 20 75 73 65 72 24 ; om obj$ o, user$
044acef0h: 20 75 2C 20 69 6E 64 24 20 69 20 77 68 65 72 65 ;  u, ind$ i where
044acf00h: 20 6F 2E 6F 77 6E 65 72 23 3D 3A 31 20 61 6E 64 ;  o.owner#=:1 and
044acf10h: 20 6F 2E 6F 77 6E 65 72 23 3D 75 2E 75 73 65 72 ;  o.owner#=u.user
044acf20h: 23 20 61 6E 64 20 6F 2E 6F 62 6A 23 3D 69 2E 6F ; # and o.obj#=i.o
044acf30h: 62 6A 23 20 61 6E 64 20 69 2E 74 79 70 65 23 3D ; bj# and i.type#=
044acf40h: 39 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; 9...............
044acf50h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
044acf60h: 73 65 6C 65 63 74 20 76 6E 61 6D 65 20 66 72 6F ; select vname fro
044acf70h: 6D 20 73 79 73 2E 73 6E 61 70 24 20 77 68 65 72 ; m sys.snap$ wher
044acf80h: 65 20 73 6F 77 6E 65 72 20 3D 20 3A 31 20 61 6E ; e sowner = :1 an
044acf90h: 64 20 69 6E 73 74 73 69 74 65 20 3D 20 30 20 20 ; d instsite = 0  
044acfa0h: 61 6E 64 20 28 62 69 74 61 6E 64 28 66 6C 61 67 ; and (bitand(flag
044acfb0h: 2C 20 32 36 38 34 33 35 34 35 36 29 20 3D 20 30 ; , 268435456) = 0
044acfc0h: 20 6F 72 20 62 69 74 61 6E 64 28 6F 62 6A 66 6C ;  or bitand(objfl
044acfd0h: 61 67 2C 20 33 32 29 20 3D 20 30 29 00 00 00 00 ; ag, 32) = 0)....

既然数据库OPEN的过程很多sql是在oracle二进制文件中写的,那是否可以考虑通过修改二进制文件中的sql语句来改变执行计划,甚至可以尝试修改这些语句使得异常的数据库能够正常open.这里演示通过修改sql语句来展示改变执行计划(不使用index,而改用全表扫描)
10046 捕获语句和执行计划

PARSING IN CURSOR #2 len=116 dep=2 uid=0 oct=3 lid=0 tim=1345475346332403 hv=854877822 ad='2f2be060'
select o.owner#,o.name,o.namespace,o.remoteowner,o.linkname,o.subname,o.dataobj#,o.flags from obj$ o where o.obj#=:1
END OF STMT
PARSE #2:c=1000,e=1258,p=0,cr=0,cu=0,mis=1,r=0,dep=2,og=4,tim=1345475346332394
BINDS #2:
kkscoacd
 Bind#0
  oacdty=02 mxl=22(22) mxlc=00 mal=00 scl=00 pre=00
  oacflg=08 fl2=0001 frm=00 csi=00 siz=24 off=0
  kxsbbbfp=b69ef000  bln=22  avl=03  flg=05
  value=893
EXEC #2:c=2000,e=1382,p=0,cr=0,cu=0,mis=1,r=0,dep=2,og=4,tim=1345475346333978
WAIT #2: nam='db file sequential read' ela= 17820 file#=1 block#=220 blocks=1 obj#=-1 tim=1345475346351927
FETCH #2:c=1000,e=18054,p=1,cr=3,cu=0,mis=0,r=1,dep=2,og=4,tim=1345475346352100
STAT #2 id=1 cnt=1 pid=0 pos=1 obj=18 op='TABLE ACCESS BY INDEX ROWID OBJ$ (cr=3 pr=1 pw=0 time=18047 us)'
STAT #2 id=2 cnt=1 pid=1 pos=1 obj=36 op='INDEX UNIQUE SCAN I_OBJ1 (cr=2 pr=1 pw=0 time=18014 us)'

使用UE查看该sql在二进制文件内容

0459c650h: 3A 35 2C 3A 36 2C 3A 37 29 00 00 00 00 00 00 00 ; :5,:6,:7).......
0459c660h: 73 65 6C 65 63 74 20 6F 2E 6F 77 6E 65 72 23 2C ; select o.owner#,
0459c670h: 6F 2E 6E 61 6D 65 2C 6F 2E 6E 61 6D 65 73 70 61 ; o.name,o.namespa
0459c680h: 63 65 2C 6F 2E 72 65 6D 6F 74 65 6F 77 6E 65 72 ; ce,o.remoteowner
0459c690h: 2C 6F 2E 6C 69 6E 6B 6E 61 6D 65 2C 6F 2E 73 75 ; ,o.linkname,o.su
0459c6a0h: 62 6E 61 6D 65 2C 6F 2E 64 61 74 61 6F 62 6A 23 ; bname,o.dataobj#
0459c6b0h: 2C 6F 2E 66 6C 61 67 73 20 66 72 6F 6D 20 6F 62 ; ,o.flags from ob
0459c6c0h: 6A 24 20 6F 20 77 68 65 72 65 20 6F 2E 6F 62 6A ; j$ o where o.obj
0459c6d0h: 23 3D 3A 31 00 00 00 00 00 00 00 00 00 00 00 00 ; #=:1............
0459c6e0h: 73 65 6C 65 63 74 20 74 79 70 65 23 2C 63 6F 6E ; select type#,con

修改二进制文件中sql语句

0459c650h: 3A 35 2C 3A 36 2C 3A 37 29 00 00 00 00 00 00 00 ; :5,:6,:7).......
0459c660h: 73 65 6C 65 63 74 20 6F 2E 6F 77 6E 65 72 23 2C ; select o.owner#,
0459c670h: 6F 2E 6E 61 6D 65 2C 6F 2E 6E 61 6D 65 73 70 61 ; o.name,o.namespa
0459c680h: 63 65 2C 6F 2E 72 65 6D 6F 74 65 6F 77 6E 65 72 ; ce,o.remoteowner
0459c690h: 2C 6F 2E 6C 69 6E 6B 6E 61 6D 65 2C 6F 2E 73 75 ; ,o.linkname,o.su
0459c6a0h: 62 6E 61 6D 65 2C 6F 2E 64 61 74 61 6F 62 6A 23 ; bname,o.dataobj#
0459c6b0h: 2C 6F 2E 66 6C 61 67 73 20 66 72 6F 6D 20 6F 62 ; ,o.flags from ob
0459c6c0h: 6A 24 20 6F 20 77 68 65 72 65 20 6F 62 6A 23 2B ; j$ o where obj#+
0459c6d0h: 30 3D 3A 31 00 00 00 00 00 00 00 00 00 00 00 00 ; 0=:1............
0459c6e0h: 73 65 6C 65 63 74 20 74 79 70 65 23 2C 63 6F 6E ; select type#,con

这里通过对obj#加上一个常量0,使得该sql在执行之时不会使用obj$.obj#上的index,从而使用全表扫描,来实现我们修改执行计划,屏蔽该index的目的
修改后的10046观察

PARSING IN CURSOR #2 len=116 dep=2 uid=0 oct=3 lid=0 tim=1345475781593851 hv=493726595 ad='2f2ba76c'
select o.owner#,o.name,o.namespace,o.remoteowner,o.linkname,o.subname,o.dataobj#,o.flags from obj$ o where obj#+0=:1
END OF STMT
PARSE #2:c=1000,e=1095,p=0,cr=0,cu=0,mis=1,r=0,dep=2,og=4,tim=1345475781593840
BINDS #2:
kkscoacd
 Bind#0
  oacdty=02 mxl=22(22) mxlc=00 mal=00 scl=00 pre=00
  oacflg=08 fl2=0001 frm=00 csi=00 siz=24 off=0
  kxsbbbfp=b6a86000  bln=22  avl=03  flg=05
  value=893
EXEC #2:c=2999,e=1603,p=0,cr=0,cu=0,mis=1,r=0,dep=2,og=4,tim=1345475781595754
WAIT #2: nam='db file sequential read' ela= 18417 file#=1 block#=121 blocks=1 obj#=-1 tim=1345475781614344
WAIT #2: nam='db file sequential read' ela= 1000 file#=1 block#=123 blocks=1 obj#=-1 tim=1345475781615726
WAIT #2: nam='db file sequential read' ela= 1462 file#=1 block#=124 blocks=1 obj#=-1 tim=1345475781617720
WAIT #2: nam='db file sequential read' ela= 492 file#=1 block#=125 blocks=1 obj#=-1 tim=1345475781618452
WAIT #2: nam='db file sequential read' ela= 1358 file#=1 block#=126 blocks=1 obj#=-1 tim=1345475781620361
WAIT #2: nam='db file sequential read' ela= 500 file#=1 block#=127 blocks=1 obj#=-1 tim=1345475781621039
WAIT #2: nam='db file sequential read' ela= 63132 file#=1 block#=128 blocks=1 obj#=-1 tim=1345475781684316
WAIT #2: nam='db file sequential read' ela= 989 file#=1 block#=5233 blocks=1 obj#=-1 tim=1345475781685559
WAIT #2: nam='db file sequential read' ela= 792 file#=1 block#=5234 blocks=1 obj#=-1 tim=1345475781686583
FETCH #2:c=7999,e=90994,p=9,cr=13,cu=0,mis=0,r=1,dep=2,og=4,tim=1345475781686865
STAT #2 id=1 cnt=1 pid=0 pos=1 obj=18 op='TABLE ACCESS FULL OBJ$ (cr=13 pr=9 pw=0 time=90977 us)'

很明显,到这里我们通过UE修改oracle二进制文件,实现了sql语句执行计划的更改.
免责申明:本方法仅供在非常特殊或者测试环境下使用,一般生产环境请勿模仿,否则后果自负

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