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标签归档:编辑oracle文件
使用UltraEdit修改oracle二进制文件
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