对于oracle的update操作,在数据块中具体是如何出来,是直接更新原来值,还是通过插入新值修改指针的方法实现.下面通过证明:

模拟表插入数据

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

数据存储对应16进制值

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

得出第一条记录对应值为:02c10203584646;第二条记录对应值为:02c10303434846

dump 数据块得到记录

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查看相关记录

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>

这里可以得到结论如下:

1.数据是从块的底部开始往上存储

2.在每一条记录的头部分别有flag/lock/cols对应这里的2c0102

3.这里的偏移量和dump出来的数据可以看出来两条记录是连续在一起(偏移量分别为:8168和8178)

更新一条记录

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

我们可以但看到值有XFF改变为XIFENFEI,存储长度变大

dump数据块信息

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

通过对比第一次dump出来的数据块发现:row 0的值和偏移量发生了变化

bbed查看相关记录

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>

从这里可以看到

1.这里可以看到三个值(XFF,CHF,XIFENFEI)均存在,但是通过p kdbr和dump block不能看到,因为row directory中无指针指定到该值上

2.也是通过row directory指针使得我们从原先看到的第一条记录处于数据块最底部变成了现在相对而言的数据部分最上层,

3.绝大多数情况:数据库更新一条记录,不是直接修改数据值,而是重新插入一条新记录,然后修改row directory指针指定到新的offset上

4.不是直接update,而是insert+指针来实现,这样做的好处:1)如果修改记录update值的长度发生变化(变大或者变小)那么该值之前的数据都要发生变动,对数据库来说成本太高.2)如果直接更新值可能导致其他数据变动,使得其他行受到影响.

5.由于是修改row directory指针,所以该处理方法的rowid值不会发生变化