前天看了一个案例因为丢了表上的数据，从索引block中找回了值
转储了oracle block的值，如何得到它真正表中的值，也算 是dump(val,16)的逆运算
sys@ORCL>conn anbob/anbob
Connected.
anbob@ORCL>create table testdump(id number,name varchar2(10),ctime date);
Table created.
anbob@ORCL>insert into testdump values(918,’anbob’,sysdate);
1 row created.
anbob@ORCL>select * from testdump;
ID NAME CTIME
———- ———- ——————-
918 anbob 2012-05-04 09:39:23
anbob@ORCL>select dbms_rowid.ROWID_RELATIVE_FNO(rowid) fno,dbms_rowid.rowid_block_number(rowid) blkno
2 from testdump;
FNO BLKNO
———- ———-
15 2676436
sys@ORCL>alter system dump datafile 15 block 2676436;
System altered.
trace file 的部分内容
Start dump data blocks tsn: 13 file#: 15 minblk 2676436 maxblk 2676436
buffer tsn: 13 rdba: 0x03e8d6d4 (15/2676436)
Block header dump: 0x03e8d6d4
Object id on Block? Y
seg/obj: 0xf58a9 csc: 0x03.74dc2b6 itc: 2 flg: E typ: 1 – DATA
brn: 0 bdba: 0x3e8d6d1 ver: 0x01 opc: 0
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0006.013.0000eb55 0x00800060.58d9.08 –U- 1 fsc 0x0000.074dc31d
0x02 0x0000.000.00000000 0x00000000.0000.00 —- 0 fsc 0x0000.00000000
data_block_dump,data header at 0xbe72464
===============
tsiz: 0x1f98
hsiz: 0x14
pbl: 0x0be72464
bdba: 0x03e8d6d4
76543210
flag=——–
ntab=1
nrow=1
frre=-1
fsbo=0x14
fseo=0x1f83
avsp=0x1f6f
tosp=0x1f6f
0xe:pti[0] nrow=1 offs=0
0x12:pri[0] offs=0x1f83
block_row_dump:
tab 0, row 0, @0x1f83
tl: 21 fb: –H-FL– lb: 0x1 cc: 3
col 0: [ 3] c2 0a 13
col 1: [ 5] 61 6e 62 6f 62
col 2: [ 7] 78 70 05 04 0a 28 18
end_of_block_dump
End dump data blocks tsn: 13 file#: 15 minblk 2676436 maxblk 2676436
–分析ump
根据seg/obj: 0xf58a9我们先确认是这个对象
sys@ORCL>select object_id from dba_objects where owner=’ANBOB’ and object_name=’TESTDUMP';
OBJECT_ID
———-
1005737
sys@ORCL>select to_number(‘f58a9′,’xxxxxxx’) from dual;
TO_NUMBER(‘F58A9′,’XXXXXXX’)
—————————-
1005737
sys@ORCL>select objd,ts#,status from v$bh where file#=15 and block#=2676436;
OBJD TS# STATUS
———- ———- ——-
1005737 13 xcur
根据bdba: 0x03e8d6d4 也可以确认到那个块
anbob@ORCL>variable dba varchar2(30)
anbob@ORCL>exec :dba := dbms_utility.make_data_block_address(15,2676436);
PL/SQL procedure successfully completed.
anbob@ORCL>print dba
DBA
——————————–
65590996
anbob@ORCL>select to_char(65590996,’xxxxxxxxxxx’) from dual;
TO_CHAR(6559
————
3e8d6d4
现在我们向回还原表中的值
col 0: [ 3] c2 0a 13
col 1: [ 5] 61 6e 62 6f 62
col 2: [ 7] 78 70 05 04 0a 28 18
第一列为number，看到老白有这种推算方法
0a 转换成10进制-1，13同理，然后再合并，我们试一下
anbob@ORCL>select to_number(‘0a’,’xxxxxxx’)-1||to_number(’13’,’xxxxx’)-1 from dual;
TO_NUMBER(‘0A’,’XXXXXXX’)-1||TO_NUMBER(’13’,’XXXXX’)-1
——————————————————
918
我们再试下接着推算负数
anbob@ORCL>select dump(-98,16) from dual;
DUMP(-98,16)
——————–
Typ=2 Len=3: 3e,3,66
anbob@ORCL>select dump(-98571,16) from dual;
DUMP(-98571,16)
—————————
Typ=2 Len=5: 3c,5c,10,1e,66
anbob@ORCL>select 101-to_number(‘3′,’xxxxxxx’) from dual;
101-TO_NUMBER(‘3′,’XXXXXXX’)
—————————-
98
anbob@ORCL>select 101-to_number(‘5c’,’xxxxxxx’),101-to_number(’10’,’xxxxxxx’),101-to_number(‘1e’,’xxxxxxx’) from dual;
101-TO_NUMBER(‘5C’,’XXXXXXX’) 101-TO_NUMBER(’10’,’XXXXXXX’) 101-TO_NUMBER(‘1E’,’XXXXXXX’)
—————————– —————————– —————————–
9 85 71
往回走是去最后的66，和第一组3e中间用101-（16输为10进制）再组合再一起
有点意思，放开oracle的内部原理不说，其实我们可以用一个ORACLE的函数很方便就可以算出
anbob@ORCL>SELECT utl_raw.cast_to_number(replace(‘3c,5c,10,1e,66′,’,’)) value FROM dual;
VALUE
———-
-98571
anbob@ORCL>SELECT utl_raw.cast_to_number(replace(‘c2 0a 13′,’ ‘)) value FROM dual;
VALUE
———-
918
第二列为字符col 1: [ 5] 61 6e 62 6f 62
对于字符就是把每位ascII码转换为16进制
anbob@ORCL>select chr(to_number(’61’,’xxxxxx’)) from dual;
CH
—
a
anbob@ORCL>select chr(to_number(‘6e’,’xxxxxx’)) from dual;
CH
—
n
anbob@ORCL>select chr(to_number(’62’,’xxxxxx’)) from dual;
CH
—
b
anbob@ORCL>select utl_raw.CAST_TO_VARCHAR2(‘616e626f62′) from dual;
anbob@ORCL>select utl_raw.CAST_TO_VARCHAR2(‘616e626f62′) from dual;
UTL_RAW.CAST_TO_VARCHAR2(‘616E626F62′)
—————————————————-
anbob
第三列是date类型col 2: [ 7] 78 70 05 04 0a 28 18 实值为2012-05-04 09:39:23
anbob@ORCL>select to_number(val16,’xxxxx’) val10 from (select regexp_substr(’78 70 05 04 0a 28 18′,'[^ ]+’,1,rownum) val16 from dual connect by rownum<=7);
VAL10
———-
120
112
5
4
10
40
24
7 rows selected.
年份-100，月日原值，时分秒-1再组合
note:
dump block 中只会记录col 值和长度，而不会记录col datatype,列类型在数据字典中。