从oracle8i起，oracle推出了本地管理表空间的来代替数据字典管理的表空间，数据字典管理表空间是用uet$,fet$这两个表来管理的，它们现在依然存在于数据库中，不过不起任何作用。
SQL> select count(*) from uet$;
COUNT(*)
----------
0
SQL> select count(*) from fet$;
COUNT(*)
----------
0
下面来探究自动分配的本地管理表空间中一个bit map block 的一bit能管理多少？
SQL> select * from v$version;
BANNER
----------------------------------------------------------------
Oracle Database 10g Enterprise Edition Release 10.2.0.3.0 - Prod
PL/SQL Release 10.2.0.3.0 - Production
CORE 10.2.0.3.0 Production
TNS for 32-bit Windows: Version 10.2.0.3.0 - Production
NLSRTL Version 10.2.0.3.0 - Production
SQL> show parameter db_block_size
NAME TYPE VALUE
------------------------------------ ----------- ------------------------------
db_block_size integer 8192
SQL> create tablespace lmt datafile 'C:/ORACLE/PRODUCT/10.2.0/ORADATA/ROBINSON/DATAFILE/lmt.dbf' size 20m extent
2 management local autoallocate;
Tablespace created
SQL> create table test(name varchar2(20)) tablespace lmt;
Table created
SQL> select segment_name,tablespace_name,header_file,header_block,blocks from dba_segments where tablespace_name='LMT';
SEGMENT_NA TABLESPACE_NAME HEADER_FILE HEADER_BLOCK BLOCKS
---------- -------------------- ----------- ------------ ----------
TEST LMT 6 11 8
可以看到表test位于LMT表空间，他的文件号为6，段头为11，知道ASSM段管理方式的人都知道，段头是第三个位图块，所以这里第9个block是一级位图块，第10个block是二级位图块，因此我们知道oracle预留下了8个block来管理这个数据文件。
SQL> alter system dump datafile 6 block min 1 block max 11;
System altered
部分DUMP文件
Start dump data blocks tsn: 10 file#: 6 minblk 1 maxblk 11
Block 1 (file header) not dumped: use dump file header command -----第一个block没有dump出来
buffer tsn: 10 rdba: 0x01800002 (6/2)
scn: 0x0000.001bb0c6 seq: 0x02 flg: 0x04 tail: 0xb0c61d02
frmt: 0x02 chkval: 0xbbb2 type: 0x1d=KTFB Bitmapped File Space Header
Hex dump of block: st=0, typ_found=1
Dump of memory from 0x094AFE00 to 0x094B1E00
94AFE00 0000A21D 01800002 001BB0C6 04020000 [................]
94AFE10 0000BBB2 00000006 00000008 00000A00 [................]
94AFE20 00000001 00000000 00000000 00000007 [................]
94AFE30 00000A00 00000001 0000013E 00000000 [........>.......]
94AFE40 00000000 00000000 00000000 00000000 [................]
94AFE50 00000009 00000008 00000000 00000000 [................]
94AFE60 00000000 00000000 00000000 00000000 [................]
Repeat 504 times
94B1DF0 00000000 00000000 00000000 B0C61D02 [................]
File Space Header Block: ----1，2两个块是数据文件的块头
Header Control:
RelFno: 6, Unit: 8, Size: 2560, Flag: 1
AutoExtend: NO, Increment: 0, MaxSize: 0
Initial Area: 7, Tail: 2560, First: 1, Free: 318
Deallocation scn: 0.0
Header Opcode:
Save: No Pending Op
buffer tsn: 10 rdba: 0x01800003 (6/3)
scn: 0x0000.001bb0c6 seq: 0x01 flg: 0x04 tail: 0xb0c61e01
frmt: 0x02 chkval: 0x4e76 type: 0x1e=KTFB Bitmapped File Space Bitmap
Hex dump of block: st=0, typ_found=1
Dump of memory from 0x094AFE00 to 0x094B1E00
94AFE00 0000A21E 01800003 001BB0C6 04010000 [................]
94AFE10 00004E76 00000006 00000009 00000000 [vN..............]
94AFE20 00000001 0000F7FF 00000000 00000000 [................]
94AFE30 00000000 00000000 00000001 00000000 [................]
94AFE40 00000000 00000000 00000000 00000000 [................]
Repeat 506 times
94B1DF0 00000000 00000000 00000000 B0C61E01 [................]
File Space Bitmap Block: ----从第三个block开始，就是本地管理表空间的位图块
BitMap Control:
RelFno: 6, BeginBlock: 9, Flag: 0, First: 1, Free: 63487
0100000000000000 0000000000000000 0000000000000000 0000000000000000
0000000000000000 0000000000000000 0000000000000000 0000000000000000
...................省略若干.................
SQL> select segment_name,extent_id,file_id,block_id from dba_extents where segment_name='TEST' and file_id=6;
SEGMENT_NA EXTENT_ID FILE_ID BLOCK_ID
---------- ---------- ---------- ----------
TEST 0 6 9
First=1表示分配了1个bit，这里分配了一个64K的空间，也就是说分配了1个区，因为是自动扩展，那么第一个区为64K。
下面向表TEST多插入10万行数据
SQL> begin
2 for i in 1..100000 loop
3 insert into test values('robinson') ;
4 end loop;
5 commit;
6 end;
7 /
PL/SQL procedure successfully completed
SQL> select extent_id,blocks from dba_extents where file_id=6;
EXTENT_ID BLOCKS
---------- ----------
0 8
1 8
2 8
3 8
4 8
5 8
6 8
7 8
8 8
9 8
10 8
11 8
12 8
13 8
14 8
15 8
16 128 ----注意这里一个区为128个blok,那么一共有32个区为64K的空间
17 rows selected
插入10W行数据之后，TEST表所在的数据文件分配了17个区，一共有256个block，也就是说分配了256*8=2048K，
现在我们dump datafile 6 block 3
SQL> alter system dump datafile 6 block 3;
系统已更改。
部分的DUMP文件
Start dump data blocks tsn: 10 file#: 6 minblk 3 maxblk 3
buffer tsn: 10 rdba: 0x01800003 (6/3)
scn: 0x0000.001bc284 seq: 0x01 flg: 0x00 tail: 0xc2841e01
frmt: 0x02 chkval: 0x0000 type: 0x1e=KTFB Bitmapped File Space Bitmap
Hex dump of block: st=0, typ_found=1
Dump of memory from 0x07CE7800 to 0x07CE9800
7CE7800 0000A21E 01800003 001BC284 00010000 [................]
7CE7810 00000000 00000006 00000009 00000000 [................]
7CE7820 00000020 0000F7E0 00000000 00000000 [ ...............]
7CE7830 00000000 00000000 FFFFFFFF 00000000 [................]
7CE7840 00000000 00000000 00000000 00000000 [................]
Repeat 506 times
7CE97F0 00000000 00000000 00000000 C2841E01 [................]
File Space Bitmap Block:
BitMap Control:
RelFno: 6, BeginBlock: 9, Flag: 0, First: 32, Free: 63456
FFFFFFFF00000000 0000000000000000 0000000000000000 0000000000000000
............................省略若干...................................................................
First=32所以，这里一共分配了32bit,但是根据上面的查询，一共分配了17个区，17个区一共等于2048K，2048/32=64,猜测一bit能管理64K
下面插入100W行数据测试一下
SQL> begin
2 for i in 1..1000000 loop
3 insert into test values('luoluo');
4 end loop;
5 commit;
6 end;
7 /
PL/SQL procedure successfully completed
SQL> select extent_id,blocks from dba_extents where file_id=6;
EXTENT_ID BLOCKS
---------- ----------
0 8
1 8
2 8
3 8
4 8
5 8
6 8
7 8
8 8
9 8
10 8
11 8
12 8
13 8
14 8
15 8
16 128
17 128
18 128
19 128
EXTENT_ID BLOCKS
---------- ----------
20 128
21 128
22 128
23 128
24 128
25 128
26 128
27 128
28 128
29 128
30 rows selected
可以看到分配了16*8+14*128=1920个block,也就是1920*8= 15360K，下面dump一下datafile 6 block 3看看first是否等于15360/64=240
SQL> alter system dump datafile 6 block 3;
系统已更改。
部分dump文件
Start dump data blocks tsn: 10 file#: 6 minblk 3 maxblk 3
buffer tsn: 10 rdba: 0x01800003 (6/3)
scn: 0x0000.001c1595 seq: 0x01 flg: 0x04 tail: 0x15951e01
frmt: 0x02 chkval: 0xb191 type: 0x1e=KTFB Bitmapped File Space Bitmap
Hex dump of block: st=0, typ_found=1
Dump of memory from 0x04807800 to 0x04809800
4807800 0000A21E 01800003 001C1595 04010000 [................]
4807810 0000B191 00000006 00000009 00000000 [................]
4807820 000000F0 0000F710 00000000 00000000 [................]
4807830 00000000 00000000 FFFFFFFF FFFFFFFF [................]
4807840 FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF [................]
4807850 FFFFFFFF 0000FFFF 00000000 00000000 [................]
4807860 00000000 00000000 00000000 00000000 [................]
Repeat 504 times
48097F0 00000000 00000000 00000000 15951E01 [................]
File Space Bitmap Block:
BitMap Control:
RelFno: 6, BeginBlock: 9, Flag: 0, First: 240, Free: 63248
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFF0000
果然First等于240，恩还得继续
下面插入500W行再测试一下
SQL> alter database datafile 'C:/ORACLE/PRODUCT/10.2.0/ORADATA/ROBINSON/DATAFILE/LMT.DBF' autoextend on maxsize 100m;
Database altered
SQL> alter table test nologging;
Table altered
SQL>
SQL> begin
2 for i in 1..5000000 loop
3 insert into test values('luoluo');
4 end loop;
5 commit;
6 end;
7 /
PL/SQL procedure successfully completed
SQL> select sum(blocks) from dba_extents where file_id=6;
SUM(BLOCKS)
-----------
11264
可以看到一共分配了11264个block，那么dump 一个datafile 6 block 3 ，看看First是否等于11264/8=1048
Start dump data blocks tsn: 10 file#: 6 minblk 3 maxblk 3
buffer tsn: 10 rdba: 0x01800003 (6/3)
scn: 0x0000.001d0f53 seq: 0x01 flg: 0x04 tail: 0x0f531e01
frmt: 0x02 chkval: 0x4e8f type: 0x1e=KTFB Bitmapped File Space Bitmap
Hex dump of block: st=0, typ_found=1
Dump of memory from 0x04807800 to 0x04809800
4807800 0000A21E 01800003 001D0F53 04010000 [........S.......]
4807810 00004E8F 00000006 00000009 00000000 [.N..............]
4807820 00000580 0000F280 00000000 00000000 [................]
4807830 00000000 00000000 FFFFFFFF FFFFFFFF [................]
4807840 FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF [................]
Repeat 9 times
48078E0 FFFFFFFF FFFFFFFF 00000000 00000000 [................]
48078F0 00000000 00000000 00000000 00000000 [................]
Repeat 495 times
48097F0 00000000 00000000 00000000 0F531E01 [..............S.]
File Space Bitmap Block:
BitMap Control:
RelFno: 6, BeginBlock: 9, Flag: 0, First: 1408, Free: 62080
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFF 0000000000000000 0000000000000000
果然如此，再继续插入我就吃不消了，刚才足足耗费了8分钟才完成插入500W行
由此可以猜测在block size =8K, 自动分配，本地管理的表空间中，位图的一位表示64K，因为不管你怎么分配，都要分配一个64K的区。那block size 为16K/32k,自动分配，本地管理的表空间呢？其实原理也应该是同样的，一位能管理多少空间应该取决于最小的一个区间的值。