ORACLE update 操作内部原理

    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 <--第一条row directory指针位置
sb2 kdbr[1]                                 @120      8068 <--第二条row directory指针位置
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