[20140920]oracle cluster index (11g)(补充).txt
--上个星期简单研究了一下cluster表.
--应用中除了堆表,很少使用cluser表,也就仅仅在生产系统使用IOT索引组织表.
--实际上系统表中许多都是cluster表.比如SYS.TAB$,SYS.COL$等都建立在cluster中.
--没事,简单研究一下其存储结构.
1.建立测试环境:
链接
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.以下是数据块转储:
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
--说明:
ntab=3 =>表示有3个表. nrow=5 有5条记录.
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 实际上cluster主键.
SCOTT@test01p> select dump(10,16) from dual ;
DUMP(10,16)
-----------------
Typ=2 Len=2: c1,b
--tab 1 实际上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 实际上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
--可以发现信息是一致的.
--有点奇怪的是oracle如何知道tab1 对应的就是dept1.tab 2 对应的就是emp1呢?
--自己遇到一个小问题oracle如何知道tab1 对应的就是dept1.tab 2 对应的就是emp1呢? 我建立按照顺序
--先建立dept1,然后emp1,自己当然很清楚.但是oracle内部如何知道这些呢?
3.昨天查看tab视图时,无意中发现.
SCOTT@test01p> @desc tab
Name Null? Type
----------------------- -------- ----------------
TNAME NOT NULL VARCHAR2(128)
TABTYPE VARCHAR2(7)
CLUSTERID NUMBER
--可以发现包含clusterid字段.
SCOTT@test01p> select * from tab where clusterid is not null order by 3;
TNAME TABTYPE CLUSTERID
------------------------------ ------- ----------
DEPT1 TABLE 1
EMP1 TABLE 2
--clusterid的数字正好对上!
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# 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
--可以从这里看出一些线索,CLUCOLS=1表示cluster表,TAB#表示顺序.
