oracle update 数据库恢复,ORACLE update 操作内部原理
對于oracle的update操作,在數據塊中具體是如何出來,是直接更新原來值,還是通過插入新值修改指針的方法實現.下面通過證明:
模擬表插入數據
SQL> create table t_xifenfei(id number,name varchar2(10));
Table created.
SQL> insert into t_xifenfei values(1,'XFF');
1 row created.
SQL> insert into t_xifenfei values(2,'CHF');
1 row created.
SQL> commit;
Commit complete.
SQL> alter system checkpoint;
System altered.
SQL> select id,rowid,
2 dbms_rowid.rowid_relative_fno(rowid)rel_fno,
3 dbms_rowid.rowid_block_number(rowid)blockno,
4 dbms_rowid.rowid_row_number(rowid) rowno
5 from t_xifenfei;
ID ROWID REL_FNO BLOCKNO ROWNO
---------- ------------------ ---------- ---------- ----------
1 AAASc+AAEAAAACvAAA 4 175 0
2 AAASc+AAEAAAACvAAB 4 175 1
SQL> alter system dump datafile 4 block 175;
System altered.
SQL> select value from v$diag_info where name='Default Trace File';
VALUE
--------------------------------------------------------------------------------
/u01/oracle/diag/rdbms/ora11g/ora11g/trace/ora11g_ora_24625.trc
數據存儲對應16進制值
SQL> select dump(1,'16') from dual;
DUMP(1,'16')
-----------------
Typ=2 Len=2: c1,2
SQL> select dump(2,'16') from dual;
DUMP(2,'16')
-----------------
Typ=2 Len=2: c1,3
SQL> select dump('XFF','16') FROM DUAL;
DUMP('XFF','16')
----------------------
Typ=96 Len=3: 58,46,46
SQL> SELECT DUMP('CHF','16') FROM DUAL;
DUMP('CHF','16')
----------------------
Typ=96 Len=3: 43,48,46
得出第一條記錄對應值為:02c10203584646;第二條記錄對應值為:02c10303434846
dump 數據塊得到記錄
bdba: 0x010000af
data_block_dump,data header at 0xb683c064
===============
tsiz: 0x1f98
hsiz: 0x16
pbl: 0xb683c064
76543210
flag=--------
ntab=1
nrow=2
frre=-1
fsbo=0x16
fseo=0x1f84
avsp=0x1f6e
tosp=0x1f6e
0xe:pti[0] nrow=2 offs=0
0x12:pri[0] offs=0x1f8e ---->8078
0x14:pri[1] offs=0x1f84 ---->8068
block_row_dump:
tab 0, row 0, @0x1f8e
tl: 10 fb: --H-FL-- lb: 0x1 cc: 2
col 0: [ 2] c1 02
col 1: [ 3] 58 46 46
tab 0, row 1, @0x1f84
tl: 10 fb: --H-FL-- lb: 0x1 cc: 2
col 0: [ 2] c1 03
col 1: [ 3] 43 48 46
end_of_block_dump
End dump data blocks tsn: 4 file#: 4 minblk 175 maxblk 175
bbed查看相關記錄
BBED> p kdbr
sb2 kdbr[0] @118 8078
sb2 kdbr[1] @120 8068
BBED> p *kdbr[0]
rowdata[10]
-----------
ub1 rowdata[10] @8178 0x2c
BBED> x /rnc
rowdata[10] @8178
-----------
flag@8178: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@8179: 0x01
cols@8180: 2
col 0[2] @8181: 1
col 1[3] @8184: XFF
BBED> p *kdbr[1]
rowdata[0]
----------
ub1 rowdata[0] @8168 0x2c
BBED> x /rnc
rowdata[0] @8168
----------
flag@8168: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@8169: 0x01
cols@8170: 2
col 0[2] @8171: 2
col 1[3] @8174: CHF
BBED> d
File: /u01/oracle/oradata/ora11g/users01.dbf (4)
Block: 175 Offsets: 8168 to 8191 Dba:0x010000af
------------------------------------------------------------------------
2c010202 c1030343 48462c01 0202c102 03584646 010650e5
<32 bytes per line>
這里可以得到結論如下:
1.數據是從塊的底部開始往上存儲
2.在每一條記錄的頭部分別有flag/lock/cols對應這里的2c0102
3.這里的偏移量和dump出來的數據可以看出來兩條記錄是連續在一起(偏移量分別為:8168和8178)
更新一條記錄
SQL> update t_xifenfei set name='XIFENFEI' where id=1;
1 row updated.
SQL> commit;
Commit complete.
SQL> alter system checkpoint;
System altered.
SQL> alter system dump datafile 4 block 175;
System altered.
SQL> select dump('XIFENFEI','16') from dual;
DUMP('XIFENFEI','16')
-------------------------------------
Typ=96 Len=8: 58,49,46,45,4e,46,45,49
我們可以但看到值有XFF改變為XIFENFEI,存儲長度變大
dump數據塊信息
bdba: 0x010000af
data_block_dump,data header at 0xb683c064
===============
tsiz: 0x1f98
hsiz: 0x16
pbl: 0xb683c064
76543210
flag=--------
ntab=1
nrow=2
frre=-1
fsbo=0x16
fseo=0x1f75
avsp=0x1f69
tosp=0x1f69
0xe:pti[0] nrow=2 offs=0
0x12:pri[0] offs=0x1f75 ---->8053
0x14:pri[1] offs=0x1f84 ---->8068
block_row_dump:
tab 0, row 0, @0x1f75
tl: 15 fb: --H-FL-- lb: 0x2 cc: 2
col 0: [ 2] c1 02
col 1: [ 8] 58 49 46 45 4e 46 45 49
tab 0, row 1, @0x1f84
tl: 10 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 2] c1 03
col 1: [ 3] 43 48 46
end_of_block_dump
End dump data blocks tsn: 4 file#: 4 minblk 175 maxblk 175
通過對比第一次dump出來的數據塊發現:row 0的值和偏移量發生了變化
bbed查看相關記錄
BBED> set file 4 block 175
FILE# 4
BLOCK# 175
BBED> map
File: /u01/oracle/oradata/ora11g/users01.dbf (4)
Block: 175 Dba:0x010000af
------------------------------------------------------------
KTB Data Block (Table/Cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 72 bytes @20
struct kdbh, 14 bytes @100
struct kdbt[1], 4 bytes @114
sb2 kdbr[2] @118
ub1 freespace[8031] @122
ub1 rowdata[35] @8153
ub4 tailchk @8188
BBED> p kdbr
sb2 kdbr[0] @118 8053
sb2 kdbr[1] @120 8068
BBED> p *kdbr[1]
rowdata[15]
-----------
ub1 rowdata[15] @8168 0x2c
BBED> x /rnc
rowdata[15] @8168
-----------
flag@8168: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@8169: 0x00
cols@8170: 2
col 0[2] @8171: 2
col 1[3] @8174: CHF
BBED> p *kdbr[0]
rowdata[0]
----------
ub1 rowdata[0] @8153 0x2c
BBED> x /r
rowdata[0] @8153
----------
flag@8153: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@8154: 0x02
cols@8155: 2
col 0[2] @8156: 0xc1 0x02
col 1[8] @8159: 0x58 0x49 0x46 0x45 0x4e 0x46 0x45 0x49
BBED> set count 64
COUNT 64
<32 bytes per line>
BBED> d /v
File: /u01/oracle/oradata/ora11g/users01.dbf (4)
Block: 175 Offsets: 8153 to 8191 Dba:0x010000af
-------------------------------------------------------
2c020202 c1020858 4946454e 4645492c l ,......XIFENFEI,
000202c1 03034348 462c0002 02c10203 l ......CHF,......
58464602 068de8 l XFF....
<16 bytes per line>
從這里可以看到
1.這里可以看到三個值(XFF,CHF,XIFENFEI)均存在,但是通過p kdbr和dump block不能看到,因為row directory中無指針指定到該值上
2.也是通過row directory指針使得我們從原先看到的第一條記錄處于數據塊最底部變成了現在相對而言的數據部分最上層,
3.絕大多數情況:數據庫更新一條記錄,不是直接修改數據值,而是重新插入一條新記錄,然后修改row directory指針指定到新的offset上
4.不是直接update,而是insert+指針來實現,這樣做的好處:1)如果修改記錄update值的長度發生變化(變大或者變小)那么該值之前的數據都要發生變動,對數據庫來說成本太高.2)如果直接更新值可能導致其他數據變動,使得其他行受到影響.
5.由于是修改row directory指針,所以該處理方法的rowid值不會發生變化
總結
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