IOT 是一种特殊的索引结构，使用它能够解决特定场合的应用问题，但是在许多应用中很少使用，更多的是使用堆表。
我仅仅在生产系统中对3个表使用，我偏向使用静态以及"只读"的小表。
如果IOT表存在除主键外的第2索引，如果使用它存在物理猜"physical guess",我以前的理解一直是第2索引记录了主键信息，
通过这个信息就可以定位IOT表中对应的数据，一直没有很好的测试与理解。我最近也看了两个链接,介绍了IOT的内容：
http://richardfoote.wordpress.com
http://mwidlake.wordpress.com
今天测试看看，看了以前的理解存在很大的偏差！
1.测试环境:
SQL> select * from v$version ;
BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
PL/SQL Release 11.2.0.1.0 - Production
CORE 11.2.0.1.0 Production
TNS for Linux: Version 11.2.0.1.0 - Production
NLSRTL Version 11.2.0.1.0 - Production
SQL> create table t_iot ( a varchar2(10),b varchar2(10),vc varchar2(1200), constraint t_iot_pk PRIMARY KEY(a)) ORGANIZATION INDEX;
SQL> create index i_t_iot_b on t_iot(b);
insert into t_iot values ('1' ,'a','a');
insert into t_iot values ('22' ,'b','a');
insert into t_iot values ('333' ,'c','a');
insert into t_iot values ('4444' ,'d','a');
insert into t_iot values ('55555' ,'e','a');
insert into t_iot values ('666666' ,'f','a');
insert into t_iot values ('7777777' ,'g','a');
insert into t_iot values ('88888888' ,'h','a');
insert into t_iot values ('999999999','i','a');
commit ;
SQL> exec dbms_stats.gather_table_stats(ownname=>USER, tabname=>'T_iot');
2.转储IOT块的信息:
SQL> SELECT index_name, table_name, blevel, leaf_blocks FROM dba_indexes WHERE table_name = 'T_IOT';
INDEX_NAME TABLE_NAME BLEVEL LEAF_BLOCKS
------------------------------ ------------------------------ ---------- -----------
T_IOT_PK T_IOT 0 1
I_T_IOT_B T_IOT 0 1
--可以发现索引很小，blevel=0,leaf_blocks=1.
SQL> select header_file,header_block from dba_segments where segment_name='T_IOT_PK';
HEADER_FILE HEADER_BLOCK
----------- ------------
4 954
SQL> select object_id,data_object_id from dba_objects where object_name='T_IOT_PK';
OBJECT_ID DATA_OBJECT_ID
---------- --------------
83453 83453
SQL> ALTER SESSION SET EVENTS 'immediate trace name treedump level 83453';
----- begin tree dump
leaf: 0x10003bb 16778171 (0: nrow: 9 rrow: 9)
----- end tree dump
仅仅占用1个块。HEADER_BLOCK=954,根节点=955.
SQL> alter system dump datafile 4 block 955 ;
Block header dump: 0x010003bb
Object id on Block? Y
seg/obj: 0x145fd csc: 0x00.63a965 itc: 2 flg: E typ: 2 - INDEX
brn: 0 bdba: 0x10003b8 ver: 0x01 opc: 0
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0000.000.00000000 0x00000000.0000.00 ---- 0 fsc 0x0000.00000000
0x02 0x0002.020.00000c9f 0x00c00b1b.066e.02 --U- 9 fsc 0x0000.0063a9ce
Leaf block dump
===============
header address 182924431972=0x2a97255264
kdxcolev 0
KDXCOLEV Flags = - - -
kdxcolok 0
kdxcoopc 0x90: pcode=0: iot flags=I-- is converted=Y
kdxconco 1
kdxcosdc 0
kdxconro 9
kdxcofbo 54=0x36
kdxcofeo 7897=0x1ed9
kdxcoavs 7843
kdxlespl 0
kdxlende 0
kdxlenxt 0=0x0
kdxleprv 0=0x0
kdxledsz 0
kdxlebksz 8032
row#0[8021] flag: K-----, lock: 2, len=11
col 0; len 1; (1): 31
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 61
col 1: [ 1] 61
row#1[8009] flag: K-----, lock: 2, len=12
col 0; len 2; (2): 32 32
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 62
col 1: [ 1] 61
row#2[7996] flag: K-----, lock: 2, len=13
col 0; len 3; (3): 33 33 33
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 63
col 1: [ 1] 61
row#3[7982] flag: K-----, lock: 2, len=14
col 0; len 4; (4): 34 34 34 34
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 64
col 1: [ 1] 61
row#4[7967] flag: K-----, lock: 2, len=15
col 0; len 5; (5): 35 35 35 35 35
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 65
col 1: [ 1] 61
row#5[7951] flag: K-----, lock: 2, len=16
col 0; len 6; (6): 36 36 36 36 36 36
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 66
col 1: [ 1] 61
row#6[7934] flag: K-----, lock: 2, len=17
col 0; len 7; (7): 37 37 37 37 37 37 37
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 67
col 1: [ 1] 61
row#7[7916] flag: K-----, lock: 2, len=18
col 0; len 8; (8): 38 38 38 38 38 38 38 38
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 68
col 1: [ 1] 61
row#8[7897] flag: K-----, lock: 2, len=19
col 0; len 9; (9): 39 39 39 39 39 39 39 39 39
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 69
col 1: [ 1] 61
----- end of leaf block dump -----
End dump data blocks tsn: 4 file#: 4 minblk 955 maxblk 955
--可以发现没有记录rowid信息，
tl: 7 fb: --H-FL-- lb: 0x0 cc: 2
col 0: [ 1] 69
col 1: [ 1] 61
--记录了主键外的信息。
3.转储第2索引的信息:
SQL> select header_file,header_block from dba_segments where segment_name='I_T_IOT_B';
HEADER_FILE HEADER_BLOCK
----------- ------------
4 962
SQL> select object_id,data_object_id from dba_objects where object_name='I_T_IOT_B';
OBJECT_ID DATA_OBJECT_ID
---------- --------------
83454 83454
SQL> ALTER SESSION SET EVENTS 'immediate trace name treedump level 83454';
----- begin tree dump
leaf: 0x10003c3 16778179 (0: nrow: 9 rrow: 9)
----- end tree dump
仅仅占用1个块。HEADER_BLOCK=962,根节点=963.
SQL> alter system dump datafile 4 block 963 ;
Block header dump: 0x010003c3
Object id on Block? Y
seg/obj: 0x145fe csc: 0x00.63a9bd itc: 2 flg: E typ: 2 - INDEX
brn: 0 bdba: 0x10003c0 ver: 0x01 opc: 0
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0000.000.00000000 0x00000000.0000.00 ---- 0 fsc 0x0000.00000000
0x02 0x0002.020.00000c9f 0x00c00b1b.066e.03 --U- 9 fsc 0x0000.0063a9ce
Leaf block dump
===============
header address 182924431972=0x2a97255264
kdxcolev 0
KDXCOLEV Flags = - - -
kdxcolok 0
kdxcoopc 0x90: pcode=0: iot flags=I-- is converted=Y
kdxconco 2
kdxcosdc 0
kdxconro 9
kdxcofbo 54=0x36
kdxcofeo 7870=0x1ebe
kdxcoavs 7816
kdxlespl 0
kdxlende 0
kdxlenxt 0=0x0
kdxleprv 0=0x0
kdxledsz 0
kdxlebksz 8032
row#0[8018] flag: K-----, lock: 2, len=14
col 0; len 1; (1): 61
col 1; len 1; (1): 31
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#1[8003] flag: K-----, lock: 2, len=15
col 0; len 1; (1): 62
col 1; len 2; (2): 32 32
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#2[7987] flag: K-----, lock: 2, len=16
col 0; len 1; (1): 63
col 1; len 3; (3): 33 33 33
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#3[7970] flag: K-----, lock: 2, len=17
col 0; len 1; (1): 64
col 1; len 4; (4): 34 34 34 34
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#4[7952] flag: K-----, lock: 2, len=18
col 0; len 1; (1): 65
col 1; len 5; (5): 35 35 35 35 35
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#5[7933] flag: K-----, lock: 2, len=19
col 0; len 1; (1): 66
col 1; len 6; (6): 36 36 36 36 36 36
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#6[7913] flag: K-----, lock: 2, len=20
col 0; len 1; (1): 67
col 1; len 7; (7): 37 37 37 37 37 37 37
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#7[7892] flag: K-----, lock: 2, len=21
col 0; len 1; (1): 68
col 1; len 8; (8): 38 38 38 38 38 38 38 38
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#8[7870] flag: K-----, lock: 2, len=22
col 0; len 1; (1): 69
col 1; len 9; (9): 39 39 39 39 39 39 39 39 39
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
----- end of leaf block dump -----
End dump data blocks tsn: 4 file#: 4 minblk 963 maxblk 963
-- 可以发现第2索引的记录信息如下：
col0 ＝》对应的是字段b
col1 ＝》对应的是字段a，也就是主键。
而后面还存在如下信息：
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
-- 可以发现记录的信息都是一样的,col0: [ 4] 01 00 03 bb,应该跟上面的转储对上:
SQL> ALTER SESSION SET EVENTS 'immediate trace name treedump level 83453';
----- begin tree dump
leaf: 0x10003bb 16778171 (0: nrow: 9 rrow: 9)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
----- end tree dump
也就是T_IOT_PK的根节点。
可以得到一些推测：
1.如果通过第2索引查询主键，是不需要查询IOT表的，因为第2索引里面已经存在。
2.如果通过第2索引查找数据信息，IOT先通过索引节点中记录的数据块信息查询(这个就应该就是物理猜)，如果不行可以通过主键也就是对应例子的col1键值，来探查IOT索引组织表。
4.做一些测试：
SQL> set autotrace traceonly ;
SQL> select a from t_iot where b='a';
Execution Plan
----------------------------------------------------------
Plan hash value: 3946844304
------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 8 | 1 (0)| 00:00:01 |
|* 1 | INDEX RANGE SCAN| I_T_IOT_B | 1 | 8 | 1 (0)| 00:00:01 |
------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - access("B"='a')
--可以发现并没有查询IOT表，而是直接访问索引。
SQL> select vc from t_iot where b='a';
Execution Plan
----------------------------------------------------------
Plan hash value: 1095339046
-------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 4 | 1 (0)| 00:00:01 |
|* 1 | INDEX UNIQUE SCAN| T_IOT_PK | 1 | 4 | 1 (0)| 00:00:01 |
|* 2 | INDEX RANGE SCAN| I_T_IOT_B | 1 | | 1 (0)| 00:00:01 |
-------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - access("B"='a')
2 - access("B"='a')
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
3 consistent gets
0 physical reads
0 redo size
519 bytes sent via SQL*Net to client
519 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
--可以发现先查询i_t_iot_b索引，然后在查询T_IOT索引组织表。
5.再来了解一下逻辑rowid。
SQL> column vc format a30
SQL> select rowid,a.* from t_iot a;
ROWID A B VC
------------------------- ---------- ---------- ------------------------------
*BAEAA7sBMf4 1 a a
*BAEAA7sCMjL+ 22 b a
*BAEAA7sDMzMz/g 333 c a
*BAEAA7sENDQ0NP4 4444 d a
*BAEAA7sFNTU1NTX+ 55555 e a
*BAEAA7sGNjY2NjY2/g 666666 f a
*BAEAA7sHNzc3Nzc3N/4 7777777 g a
*BAEAA7sIODg4ODg4ODj+ 88888888 h a
*BAEAA7sJOTk5OTk5OTk5/g 999999999 i a
9 rows selected.
--可以发现一些特点，前面都是一样的都是*BAEAA7s，而且主键键值的长度越长与rowid的长度也越长。再dump(rowid)
SQL> column x format a70
SQL> select dump(rowid) x,rowid,a,b from t_iot ;
X ROWID A B
---------------------------------------------------------------------- ------------------------- ---------- ----------
Typ=208 Len=9: 2,4,1,0,3,187,1,49,254 *BAEAA7sBMf4 1 a
Typ=208 Len=10: 2,4,1,0,3,187,2,50,50,254 *BAEAA7sCMjL+ 22 b
Typ=208 Len=11: 2,4,1,0,3,187,3,51,51,51,254 *BAEAA7sDMzMz/g 333 c
Typ=208 Len=12: 2,4,1,0,3,187,4,52,52,52,52,254 *BAEAA7sENDQ0NP4 4444 d
Typ=208 Len=13: 2,4,1,0,3,187,5,53,53,53,53,53,254 *BAEAA7sFNTU1NTX+ 55555 e
Typ=208 Len=14: 2,4,1,0,3,187,6,54,54,54,54,54,54,254 *BAEAA7sGNjY2NjY2/g 666666 f
Typ=208 Len=15: 2,4,1,0,3,187,7,55,55,55,55,55,55,55,254 *BAEAA7sHNzc3Nzc3N/4 7777777 g
Typ=208 Len=16: 2,4,1,0,3,187,8,56,56,56,56,56,56,56,56,254 *BAEAA7sIODg4ODg4ODj+ 88888888 h
Typ=208 Len=17: 2,4,1,0,3,187,9,57,57,57,57,57,57,57,57,57,254 *BAEAA7sJOTk5OTk5OTk5/g 999999999 i
9 rows selected.
SQL> select dump(rowid,16) x,rowid,a,b from t_iot ;
X ROWID A B
---------------------------------------------------------------------- ------------------------- ---------- ----------
Typ=208 Len=9: 2,4,1,0,3,bb,1,31,fe *BAEAA7sBMf4 1 a
Typ=208 Len=10: 2,4,1,0,3,bb,2,32,32,fe *BAEAA7sCMjL+ 22 b
Typ=208 Len=11: 2,4,1,0,3,bb,3,33,33,33,fe *BAEAA7sDMzMz/g 333 c
Typ=208 Len=12: 2,4,1,0,3,bb,4,34,34,34,34,fe *BAEAA7sENDQ0NP4 4444 d
Typ=208 Len=13: 2,4,1,0,3,bb,5,35,35,35,35,35,fe *BAEAA7sFNTU1NTX+ 55555 e
Typ=208 Len=14: 2,4,1,0,3,bb,6,36,36,36,36,36,36,fe *BAEAA7sGNjY2NjY2/g 666666 f
Typ=208 Len=15: 2,4,1,0,3,bb,7,37,37,37,37,37,37,37,fe *BAEAA7sHNzc3Nzc3N/4 7777777 g
Typ=208 Len=16: 2,4,1,0,3,bb,8,38,38,38,38,38,38,38,38,fe *BAEAA7sIODg4ODg4ODj+ 88888888 h
Typ=208 Len=17: 2,4,1,0,3,bb,9,39,39,39,39,39,39,39,39,39,fe *BAEAA7sJOTk5OTk5OTk5/g 999999999 i
9 rows selected.
--结合dump（rowid）,dump(rowid,16)的转出信息.很容易推导出rowid的信息。以下的例子来说明:
Typ=208 Len=17: 2,4,1,0,3,bb,9,39,39,39,39,39,39,39,39,39,fe *BAEAA7sJOTk5OTk5OTk5/g 999999999 i
1.开头的2,4，以及结尾fe实在不好猜测，不过好像是固定不变的。
2.中间的1,0,3,bb正好对应的就是IOT的根节点，也就是对应数据IOT的数据块。
3.9,39,39,39,39,39,39,39,39,39中9表示主键长度，39,39,39,39,39,39,39,39,39就是主键的信息。
6.关于物理猜:
SQL> column name format a30
SQL> column value format 999999999
SQL> SELECT n.name, s.value FROM v$mystat s, v$statname n WHERE s.statistic# = n.statistic# AND lower(n.name) like '%leaf%';
NAME VALUE
------------------------------ ----------
leaf node splits 0
leaf node 90-10 splits 0
SQL> update t_iot set vc=lpad('a',1000,'a') ;
9 rows updated.
SQL> SELECT n.name, s.value FROM v$mystat s, v$statname n WHERE s.statistic# = n.statistic# AND lower(n.name) like '%leaf%';
NAME VALUE
------------------------------ ----------
leaf node splits 1
leaf node 90-10 splits 0
SQL> commit;
--vc长度加大到1000,原来的IOT已经无法存在这些信息，可以发现IOT表发生了1次索引分裂。
SQL> select dump(rowid,16) x,rowid,a,b from t_iot ;
X ROWID A B
---------------------------------------------------------------------- ------------------------- ---------- ----------
Typ=208 Len=9: 2,4,1,0,3,bd,1,31,fe *BAEAA70BMf4 1 a
Typ=208 Len=10: 2,4,1,0,3,bd,2,32,32,fe *BAEAA70CMjL+ 22 b
Typ=208 Len=11: 2,4,1,0,3,bd,3,33,33,33,fe *BAEAA70DMzMz/g 333 c
Typ=208 Len=12: 2,4,1,0,3,bd,4,34,34,34,34,fe *BAEAA70ENDQ0NP4 4444 d
Typ=208 Len=13: 2,4,1,0,3,bd,5,35,35,35,35,35,fe *BAEAA70FNTU1NTX+ 55555 e
Typ=208 Len=14: 2,4,1,0,3,bd,6,36,36,36,36,36,36,fe *BAEAA70GNjY2NjY2/g 666666 f
Typ=208 Len=15: 2,4,1,0,3,be,7,37,37,37,37,37,37,37,fe *BAEAA74HNzc3Nzc3N/4 7777777 g
Typ=208 Len=16: 2,4,1,0,3,be,8,38,38,38,38,38,38,38,38,fe *BAEAA74IODg4ODg4ODj+ 88888888 h
Typ=208 Len=17: 2,4,1,0,3,be,9,39,39,39,39,39,39,39,39,39,fe *BAEAA74JOTk5OTk5OTk5/g 999999999 i
9 rows selected.
--可以发现逻辑rowid记录的块信息发现了变化。从1,0,3,bb变成了1,0,3,bd和1,0,3,be.
--再次转储第2索引的信息。
SQL> alter system dump datafile 4 block 963 ;
Block header dump: 0x010003c3
Object id on Block? Y
seg/obj: 0x145fe csc: 0x00.63a9bd itc: 2 flg: E typ: 2 - INDEX
brn: 0 bdba: 0x10003c0 ver: 0x01 opc: 0
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0000.000.00000000 0x00000000.0000.00 ---- 0 fsc 0x0000.00000000
0x02 0x0002.020.00000c9f 0x00c00b1b.066e.03 --U- 9 fsc 0x0000.0063a9ce
Leaf block dump
===============
header address 182925401700=0x2a97341e64
kdxcolev 0
KDXCOLEV Flags = - - -
kdxcolok 0
kdxcoopc 0x90: pcode=0: iot flags=I-- is converted=Y
kdxconco 2
kdxcosdc 0
kdxconro 9
kdxcofbo 54=0x36
kdxcofeo 7870=0x1ebe
kdxcoavs 7816
kdxlespl 0
kdxlende 0
kdxlenxt 0=0x0
kdxleprv 0=0x0
kdxledsz 0
kdxlebksz 8032
row#0[8018] flag: K-----, lock: 2, len=14
col 0; len 1; (1): 61
col 1; len 1; (1): 31
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#1[8003] flag: K-----, lock: 2, len=15
col 0; len 1; (1): 62
col 1; len 2; (2): 32 32
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#2[7987] flag: K-----, lock: 2, len=16
col 0; len 1; (1): 63
col 1; len 3; (3): 33 33 33
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#3[7970] flag: K-----, lock: 2, len=17
col 0; len 1; (1): 64
col 1; len 4; (4): 34 34 34 34
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#4[7952] flag: K-----, lock: 2, len=18
col 0; len 1; (1): 65
col 1; len 5; (5): 35 35 35 35 35
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#5[7933] flag: K-----, lock: 2, len=19
col 0; len 1; (1): 66
col 1; len 6; (6): 36 36 36 36 36 36
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#6[7913] flag: K-----, lock: 2, len=20
col 0; len 1; (1): 67
col 1; len 7; (7): 37 37 37 37 37 37 37
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#7[7892] flag: K-----, lock: 2, len=21
col 0; len 1; (1): 68
col 1; len 8; (8): 38 38 38 38 38 38 38 38
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
row#8[7870] flag: K-----, lock: 2, len=22
col 0; len 1; (1): 69
col 1; len 9; (9): 39 39 39 39 39 39 39 39 39
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
----- end of leaf block dump -----
End dump data blocks tsn: 4 file#: 4 minblk 963 maxblk 963
--可以发现如下信息没有发生变化，依旧是col 0: [ 4] 01 00 03 bb.
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bb
这样如果我们执行select substr(vc,1,20) from t_iot from where b='a';通过第2索引记录的数据块查询应该找不到vc的值(可以理解物理猜失败)，
必须在通过主键来查询，这样逻辑读就会比原来加1.
SQL> set autotrace traceonly
SQL> select substr(vc,1,20) from t_iot where b='a';
Execution Plan
----------------------------------------------------------
Plan hash value: 1095339046
-------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 4 | 1 (0)| 00:00:01 |
|* 1 | INDEX UNIQUE SCAN| T_IOT_PK | 1 | 4 | 1 (0)| 00:00:01 |
|* 2 | INDEX RANGE SCAN| I_T_IOT_B | 1 | | 1 (0)| 00:00:01 |
-------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - access("B"='a')
2 - access("B"='a')
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
4 consistent gets
0 physical reads
0 redo size
551 bytes sent via SQL*Net to client
519 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
7.现在我们重建第2索引看看。
SQL> alter index i_t_iot_b rebuild ;
Index altered.
SQL> select header_file,header_block from dba_segments where segment_name='I_T_IOT_B';
HEADER_FILE HEADER_BLOCK
----------- ------------
4 970
SQL> select object_id,data_object_id from dba_objects where object_name='I_T_IOT_B';
OBJECT_ID DATA_OBJECT_ID
---------- --------------
83454 83455
SQL> exec dbms_stats.gather_table_stats(ownname=>USER, tabname=>'T_iot');
PL/SQL procedure successfully completed.
SQL> SELECT index_name, table_name, blevel, leaf_blocks FROM dba_indexes WHERE table_name = 'T_IOT';
INDEX_NAME TABLE_NAME BLEVEL LEAF_BLOCKS
------------------------------ ------------------------------ ---------- -----------
T_IOT_PK T_IOT 1 2
I_T_IOT_B T_IOT 0 1
SQL> ALTER SESSION SET EVENTS 'immediate trace name treedump level 83454';
Session altered.
----- begin tree dump
leaf: 0x10003cb 16778187 (0: nrow: 9 rrow: 9)
----- end tree dump
I_T_IOT_B仅仅占用1个块。HEADER_BLOCK=970,根节点=971.
SQL> alter system dump datafile 4 block 971 ;
Block header dump: 0x010003cb
Object id on Block? Y
seg/obj: 0x145ff csc: 0x00.63b560 itc: 2 flg: E typ: 2 - INDEX
brn: 0 bdba: 0x10003c8 ver: 0x01 opc: 0
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0000.000.00000000 0x00000000.0000.00 ---- 0 fsc 0x0000.00000000
0x02 0xffff.000.00000000 0x00000000.0000.00 C--- 0 scn 0x0000.0063b560
Leaf block dump
===============
header address 182925401700=0x2a97341e64
kdxcolev 0
KDXCOLEV Flags = - - -
kdxcolok 0
kdxcoopc 0x90: pcode=0: iot flags=I-- is converted=Y
kdxconco 2
kdxcosdc 0
kdxconro 9
kdxcofbo 54=0x36
kdxcofeo 7870=0x1ebe
kdxcoavs 7816
kdxlespl 0
kdxlende 0
kdxlenxt 0=0x0
kdxleprv 0=0x0
kdxledsz 0
kdxlebksz 8032
row#0[8018] flag: K-----, lock: 0, len=14
col 0; len 1; (1): 61
col 1; len 1; (1): 31
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bd
row#1[8003] flag: K-----, lock: 0, len=15
col 0; len 1; (1): 62
col 1; len 2; (2): 32 32
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bd
row#2[7987] flag: K-----, lock: 0, len=16
col 0; len 1; (1): 63
col 1; len 3; (3): 33 33 33
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bd
row#3[7970] flag: K-----, lock: 0, len=17
col 0; len 1; (1): 64
col 1; len 4; (4): 34 34 34 34
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bd
row#4[7952] flag: K-----, lock: 0, len=18
col 0; len 1; (1): 65
col 1; len 5; (5): 35 35 35 35 35
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bd
row#5[7933] flag: K-----, lock: 0, len=19
col 0; len 1; (1): 66
col 1; len 6; (6): 36 36 36 36 36 36
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 bd
row#6[7913] flag: K-----, lock: 0, len=20
col 0; len 1; (1): 67
col 1; len 7; (7): 37 37 37 37 37 37 37
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 be
row#7[7892] flag: K-----, lock: 0, len=21
col 0; len 1; (1): 68
col 1; len 8; (8): 38 38 38 38 38 38 38 38
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 be
row#8[7870] flag: K-----, lock: 0, len=22
col 0; len 1; (1): 69
col 1; len 9; (9): 39 39 39 39 39 39 39 39 39
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 be
----- end of leaf block dump -----
End dump data blocks tsn: 4 file#: 4 minblk 971 maxblk 971
--可以发现索引中的数据块信息指向了正确的位置。
tl: 8 fb: --H-FL-- lb: 0x0 cc: 1
col 0: [ 4] 01 00 03 be
SQL> set autotrace traceonly
SQL> select substr(vc,1,20) from t_iot where b='a';
Execution Plan
----------------------------------------------------------
Plan hash value: 1095339046
-------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 1003 | 1 (0)| 00:00:01 |
|* 1 | INDEX UNIQUE SCAN| T_IOT_PK | 1 | 1003 | 1 (0)| 00:00:01 |
|* 2 | INDEX RANGE SCAN| I_T_IOT_B | 1 | | 1 (0)| 00:00:01 |
-------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - access("B"='a')
2 - access("B"='a')
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
3 consistent gets
0 physical reads
0 redo size
551 bytes sent via SQL*Net to client
519 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
--可以发现逻辑读又变为3个.物理猜测是正确的。
8.总结：
1.IOT的第2索引包含主键信息以及对应信息的块信息。
2.当IOT索引分裂时，第2索引中记录的块信息不会发生变化。这样在使用第2索引探查IOT表时，物理猜就会失败。
3.如果物理猜失败很多，会导致逻辑读增加，可以通过重建第2索引来解决这个问题。