[20160904]在内存修改数据.txt
--测试一下,是否可以在内存crack数据块的数据.
1.环境:
SCOTT@book> @ &r/ver1
PORT_STRING VERSION BANNER
------------------------------ -------------- --------------------------------------------------------------------------------
x86_64/Linux 2.4.xx 11.2.0.4.0 Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production
create table t1 (id number,name varchar2(20));
insert into t1 values (1,'AAAA');
commit ;
--分析表略.
2.测试:
SCOTT@test01p> select rowid,t1.* from t1;
ROWID ID NAME
------------------ ---------- --------------------
AAAZuzAAJAAAACOAAA 1 AAAA
SCOTT@book> select rowid,t1.* from t1;
ROWID ID NAME
------------------ ---------- --------------------
AAAVpqAAEAAAAePAAA 1 AAAA
SCOTT@book> @ &r/rowid AAAVpqAAEAAAAePAAA
OBJECT FILE BLOCK ROW ROWID_DBA DBA TEXT
---------- ---------- ---------- ---------- -------------------- -------------------- ----------------------------------------
88682 4 1935 0 0x100078F 4,1935 alter system dump datafile 4 block 1935
SYS@book> @ &r/bh 4 1935
HLADDR DBARFIL DBABLK CLASS CLASS_TYPE STATE TCH CR_SCN_BAS CR_SCN_WRP CR_UBA_FIL CR_UBA_BLK CR_UBA_SEQ BA OBJECT_NAME
---------------- ---------- ---------- ---------- ------------------ ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------------- --------------------
0000000084B95950 4 1935 1 data block xcur 2 0 0 0 0 0 000000006BAC6000 T1
--STATE=xcur BA=000000006BAC6000
SYS@book> alter system checkpoint ;
System altered.
--//通过bbed观察:
BBED> set dba 4,1935
DBA 0x0100078f (16779151 4,1935)
BBED> x /rnc *kdbr[0]
rowdata[0] @8177
----------
flag@8177: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@8178: 0x01
cols@8179: 2
col 0[2] @8180: 1
col 1[4] @8183: AAAA
BBED> dump /v
File: /mnt/ramdisk/book/users01.dbf (4)
Block: 1935 Offsets: 8177 to 8191 Dba:0x0100078f
-----------------------------------------------------------------------------------------------------------
2c010202 c1020441 41414101 066871 l ,......AAAA..hq
<32 bytes per line>
BBED> p kcbh.chkval_kcbh
ub2 chkval_kcbh @16 0xd253
--可以发现'AAAA'现在在该块offset=8183.
--8183 = 0x1ff7,ba=0x000000006BAC6000 ba + 0x1FF7 = 0x000000006BAC6000+0x1FF7+1 = 0x000000006BAC7ff8
3.先通过oradebug确定判断是否正确.
SYS@book> oradebug setmypid
Statement processed.
SYS@book> oradebug peek 0x000000006BAC7ff8 4
[06BAC7FF8, 06BAC7FFC) = 41414141
SYS@book> oradebug poke 0x000000006BAC7ff8 4 0x41424344
BEFORE: [06BAC7FF8, 06BAC7FFC) = 41414141
AFTER: [06BAC7FF8, 06BAC7FFC) = 41424344
--依旧修改为'ABCD'.
SCOTT@book> select rowid,t1.* from t1;
ROWID ID NAME
------------------ ---------- --------------------
AAAVpqAAEAAAAePAAA 1 DCBA
--昏！顺序是颠倒的，不过已经修改过来了。并且不需要计算检查和。如果刷新buffer cache。
SCOTT@book> alter system flush buffer_cache;
System altered.
SCOTT@book> select rowid,t1.* from t1;
ROWID ID NAME
------------------ ---------- --------------------
AAAVpqAAEAAAAePAAA 1 AAAA
--可以发现有改动回来了，因为数据块的信息没有改动。