[20140920]oracle cluster index (11g)(2)

[20140920]oracle cluster index (11g)(補充).txt

--上個星期簡單研究了一下cluster表.

--應用中除了堆表,很少使用cluser表,也就僅僅在生產系統使用IOT索引組織表.
--實際上系統表中許多都是cluster表.比如SYS.TAB$,SYS.COL$等都建立在cluster中.

--沒事,簡單研究一下其儲存結構.

1.建立測試環境:
連結
http://blog.itpub.net/267265/viewspace-1266411/

SCOTT@test> @ver
BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production

--create cluster cluster_dept (deptno NUMBER(2)) index ;
create cluster cluster_dept (deptno NUMBER(2)) ;
create index i_cluster_deptno on cluster cluster_dept;

create table dept1 cluster cluster_dept(deptno) as select * from dept;
create table emp1  cluster cluster_dept(deptno) as select * from emp;

SCOTT@test> alter system dump datafile 4 block 1983 ;
System altered.

2.以下是資料塊轉儲:
Block header dump:  0x010007bf
Object id on Block? Y
seg/obj: 0x470c8  csc: 0x02.a63f479c  itc: 2  flg: E  typ: 1 - DATA
     brn: 0  bdba: 0x10007b8 ver: 0x01 opc: 0
     inc: 0  exflg: 0

Itl           Xid                  Uba         Flag  Lck        Scn/Fsc
0x01   0x0001.021.000044c0  0x00c005b5.1c5b.0d  C---    0  scn 0x0002.a63f4788
0x02   0x0005.008.0000725e  0x00c007a0.2518.1b  --U-    3  fsc 0x0000.a63f47a3
bdba: 0x010007bf
data_block_dump,data header at 0x2a9752e064
===============
tsiz: 0x1f98
hsiz: 0x24
pbl: 0x2a9752e064
     76543210
flag=--------
ntab=3
nrow=5
frre=-1
fsbo=0x24
fseo=0x1efd
avsp=0x1ed9
tosp=0x1ed9
0xe:pti[0]  nrow=1  offs=0
0x12:pti[1] nrow=1  offs=1
0x16:pti[2] nrow=3  offs=2
0x1a:pri[0] offs=0x1f82
0x1c:pri[1] offs=0x1f6a
0x1e:pri[2] offs=0x1f44
0x20:pri[3] offs=0x1f21
0x22:pri[4] offs=0x1efd
block_row_dump:
tab 0, row 0, @0x1f82
tl: 22 fb: K-H-FL-- lb: 0x0  cc: 1
curc: 4 comc: 4 pk: 0x010007bf.0 nk: 0x010007bf.0
col  0: [ 2]  c1 0b
tab 1, row 0, @0x1f6a
tl: 24 fb: -CH-FL-- lb: 0x0  cc: 2 cki: 0
col  0: [10]  41 43 43 4f 55 4e 54 49 4e 47
col  1: [ 8]  4e 45 57 20 59 4f 52 4b
tab 2, row 0, @0x1f44
tl: 38 fb: -CH-FL-- lb: 0x2  cc: 6 cki: 0
col  0: [ 3]  c2 4e 53
col  1: [ 5]  43 4c 41 52 4b
col  2: [ 7]  4d 41 4e 41 47 45 52
col  3: [ 3]  c2 4f 28
col  4: [ 7]  77 b5 06 09 01 01 01
col  5: [ 3]  c2 19 33
tab 2, row 1, @0x1f21
tl: 35 fb: -CH-FL-- lb: 0x2  cc: 6 cki: 0
col  0: [ 3]  c2 4f 28
col  1: [ 4]  4b 49 4e 47
col  2: [ 9]  50 52 45 53 49 44 45 4e 54
col  3: *NULL*
col  4: [ 7]  77 b5 0b 11 01 01 01
col  5: [ 2]  c2 33
tab 2, row 2, @0x1efd
tl: 36 fb: -CH-FL-- lb: 0x2  cc: 6 cki: 0
col  0: [ 3]  c2 50 23
col  1: [ 6]  4d 49 4c 4c 45 52
col  2: [ 5]  43 4c 45 52 4b
col  3: [ 3]  c2 4e 53
col  4: [ 7]  77 b6 01 17 01 01 01
col  5: [ 2]  c2 0e
end_of_block_dump
End dump data blocks tsn: 4 file#: 4 minblk 1983 maxblk 1983

--說明:
ntab=3 =>表示有3個表. nrow=5 有5條記錄.

tab 0, row 0, @0x1f82
tl: 22 fb: K-H-FL-- lb: 0x0  cc: 1
curc: 4 comc: 4 pk: 0x010007bf.0 nk: 0x010007bf.0
col  0: [ 2]  c1 0b
--tab 0 實際上cluster主鍵.

SCOTT@test01p> select dump(10,16) from dual ;
DUMP(10,16)
-----------------
Typ=2 Len=2: c1,b

--tab 1 實際上dept1表.
tab 1, row 0, @0x1f6a
tl: 24 fb: -CH-FL-- lb: 0x0  cc: 2 cki: 0
col  0: [10]  41 43 43 4f 55 4e 54 49 4e 47
col  1: [ 8]  4e 45 57 20 59 4f 52 4b

SCOTT@test01p> select dump(dname,16) c50 ,dump(loc,16) c40 from dept1 where  deptno=10 ;
C50                                                C40
-------------------------------------------------- ----------------------------------------
Typ=1 Len=10: 41,43,43,4f,55,4e,54,49,4e,47        Typ=1 Len=8: 4e,45,57,20,59,4f,52,4b

--tab 2 實際上emp1表.
tab 2, row 0, @0x1f44
tl: 38 fb: -CH-FL-- lb: 0x2  cc: 6 cki: 0
col  0: [ 3]  c2 4e 53
col  1: [ 5]  43 4c 41 52 4b
col  2: [ 7]  4d 41 4e 41 47 45 52
col  3: [ 3]  c2 4f 28
col  4: [ 7]  77 b5 06 09 01 01 01
col  5: [ 3]  c2 19 33

SCOTT@test01p> select dump(empno,16) c30 ,dump(ename,16) c30 from emp1 where empno=7782;
C30                            C30
------------------------------ ------------------------------
Typ=2 Len=3: c2,4e,53          Typ=1 Len=5: 43,4c,41,52,4b

--可以發現資訊是一致的.
--有點奇怪的是oracle如何知道tab1 對應的就是dept1.tab 2 對應的就是emp1呢?

--自己遇到一個小問題oracle如何知道tab1 對應的就是dept1.tab 2 對應的就是emp1呢? 我建立按照順序
--先建立dept1,然後emp1,自己當然很清楚.但是oracle內部如何知道這些呢?

3.昨天檢視tab檢視時,無意中發現.
SCOTT@test01p> @desc tab
Name                    Null?    Type
----------------------- -------- ----------------
TNAME                   NOT NULL VARCHAR2(128)
TABTYPE                          VARCHAR2(7)
CLUSTERID                        NUMBER

--可以發現包含clusterid欄位.

SCOTT@test01p> select * from tab where clusterid is not null order by 3;
TNAME                          TABTYPE  CLUSTERID
------------------------------ ------- ----------
DEPT1                          TABLE            1
EMP1                           TABLE            2
--clusterid的數字正好對上!

SCOTT@test01p> select text_vc from dba_views where owner='SYS' and view_name='TAB';
TEXT_VC
------------------------------------------------------------------------------------
select o.name,
      decode(o.type#, 2, 'TABLE', 3, 'CLUSTER',
             4, 'VIEW', 5, 'SYNONYM'), t.tab#
  from  sys.tab$ t, sys."_CURRENT_EDITION_OBJ" o
  where o.owner# = userenv('SCHEMAID')
  and o.type# >=2
  and o.type# <=5
  and o.linkname is null
  and o.obj# = t.obj# (+)

SELECT object_id, data_object_id, object_name
  FROM dba_objects
WHERE owner = USER AND object_name IN ('EMP1', 'DEPT1', 'DEPT')

OBJECT_ID DATA_OBJECT_ID OBJECT_NAME
---------- -------------- -------------
     96173          96170 EMP1
     96172          96170 DEPT1
     92285          92285 DEPT

SELECT a.obj#,
         a.dataobj#,
         b.object_name,
         a.bobj#,
         a.tab#,
         a.cols,
         a.clucols
    FROM sys.tab$ a, dba_objects b
   WHERE     a.obj# = b.object_id
         AND a.dataobj# = b.data_object_id
         AND b.owner = USER
         AND b.object_name IN ('EMP1', 'DEPT1', 'DEPT')
ORDER BY 1

      OBJ#   DATAOBJ# OBJECT_NAME                         BOBJ#       TAB#       COLS    CLUCOLS
---------- ---------- ------------------------------ ---------- ---------- ---------- ----------
     92285      92285 DEPT                                                          3
     96172      96170 DEPT1                               96170          1          3          1
     96173      96170 EMP1                                96170          2          8          1

--可以從這裡看出一些線索,CLUCOLS=1表示cluster表,TAB#表示順序.