Oracle嵌套表很少用，下面来研究下其如何存储的。用一个例子，一个用户对应对个部门。
oracle嵌套表很少用，下面来研究下其如何存储的。用一个例子，一个用户对应对个部门。
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 type t_dept as object(
dept_id number,
dept_name varchar2(20)
)
/
SQL> create type t_dept_arry as varray(100) of t_dept
/
SQL> create table t_user(
user_id number,
user_name varchar2(10),
depts t_dept_arry
)
/
SQL> insert into t_user values(1,'张三',t_dept_arry(t_dept(100,'开发一组'),t_dept(200,'开发二组')));
SQL> insert into t_user values(2,'李四',t_dept_arry(t_dept(300,'设计一组'),t_dept(400,'设计二组')));
SQL> commit;
--要想遍历用户的所有部门信息，需要用table这种特殊的形式
SQL> select user_id,user_name,d.dept_id,d.dept_name from t_user u,table(u.depts) d;
USER_ID USER_NAME DEPT_ID DEPT_NAME
---------- ---------- ---------- --------------------
1 张三 100 开发一组
1 张三 200 开发二组
2 李四 300 设计一组
2 李四 400 设计二组
dump block看看：
block_row_dump:
tab 0, row 0, @0x3f47
tl: 81 fb: --H-FL-- lb: 0x1 cc: 3
col 0: [ 2] c1 02
col 1: [ 4] d5 c5 c8 fd
col 2: [69]
00 01 00 00 00 00 00 01 00 00 00 0a e4 23 00 37 09 00 00 00 00 00 00 27 00
00 00 00 00 01 88 01 27 01 01 00 02 0f 84 01 0f 02 c2 02 08 bf aa b7 a2 d2
bb d7 e9 0f 84 01 0f 02 c2 03 08 bf aa b7 a2 b6 fe d7 e9
tab 0, row 1, @0x3ef6
tl: 81 fb: --H-FL-- lb: 0x1 cc: 3
col 0: [ 2] c1 03
col 1: [ 4] c0 ee cb c4
col 2: [69]
00 01 00 00 00 00 00 01 00 00 00 0a e4 24 00 37 09 00 00 00 00 00 00 27 00
00 00 00 00 01 88 01 27 01 01 00 02 0f 84 01 0f 02 c2 04 08 c9 e8 bc c6 d2
bb d7 e9 0f 84 01 0f 02 c2 05 08 c9 e8 bc c6 b6 fe d7 e9
end_of_block_dump
insert into t_user values(3,'王五',t_dept_arry(t_dept(500,'家')));
commit;
block_row_dump:
tab 0, row 0, @0x3f5d
tl: 59 fb: --H-FL-- lb: 0x1 cc: 3
col 0: [ 2] c1 04
col 1: [ 4] cd f5 ce e5
col 2: [47]
00 01 00 00 00 00 00 01 00 00 00 0a e4 49 00 21 09 00 00 00 00 00 00 11 00
00 00 00 00 01 88 01 11 01 01 00 01 09 84 01 09 02 c2 06 02 bc d2
end_of_block_dump
--看了上面dump的结果，一头雾水，不知道是怎么存的，不过在user_objects中找到线索，有个lob字段
SQL> select object_name,object_type from user_objects s;
OBJECT_NAME OBJECT_TYPE
------------------------------ -------------------
SYS_LOB0000082685C00003$$ LOB
T_USER TABLE
T_DEPT_ARRY TYPE
T_DEPT TYPE
SQL> select column_name, segment_name, index_name
from user_lobs s
where s.table_name = 'T_USER';
COLUMN_NAM SEGMENT_NAME INDEX_NAME
---------- ------------------------------ ------------------------------
DEPTS SYS_LOB0000082685C00003$$ SYS_IL0000082685C00003$$
--再测试下跟普通的clob有什么区别
drop table test purge;
create table test
(
id number,
t_clob clob
);
insert into test values(1,'开发一组开发二组');
insert into test values(1,'设计一组设计二组');
insert into test values(1,'家');
commit;
tab 0, row 0, @0x3f5d
tl: 59 fb: --H-FL-- lb: 0x1 cc: 2
col 0: [ 2] c1 02
col 1: [52]
00 54 00 01 02 0c 80 00 00 02 00 00 00 01 00 00 00 0a e4 4a 00 20 09 00 00
00 00 00 00 10 00 00 00 00 00 01 5f 00 53 d1 4e 00 7e c4 5f 00 53 d1 4e 8c
7e c4
LOB
Locator:
Length: 84(52)
Version: 1
Byte Length: 2
LobID: 00.00.00.01.00.00.00.0a.e4.4a
Flags[ 0x02 0x0c 0x80 0x00 ]:
Type: CLOB
Storage: BasicFile
Enable Storage in Row
Characterset Format: IMPLICIT
Partitioned Table: No
Options: VaringWidthReadWrite
Inode:
Size: 32
Flag: 0x09 [ Valid DataInRow ]
Future: 0x00 (should be '0x00')
Blocks: 0
Bytes: 16
Version: 00000.0000000001
Inline data[16]
Dump of memory from 0x00002AE680C155EC to 0x00002AE680C155FC
2AE680C155E0 D153005F [_.S.]
2AE680C155F0 C47E004E D153005F C47E8C4E [N.~._.S.N.~.]
tab 0, row 1, @0x3f22
tl: 59 fb: --H-FL-- lb: 0x1 cc: 2
col 0: [ 2] c1 02
col 1: [52]
00 54 00 01 02 0c 80 00 00 02 00 00 00 01 00 00 00 0a e4 4b 00 20 09 00 00
00 00 00 00 10 00 00 00 00 00 01 8b be 8b a1 4e 00 7e c4 8b be 8b a1 4e 8c
7e c4
LOB
Locator:
Length: 84(52)
Version: 1
Byte Length: 2
LobID: 00.00.00.01.00.00.00.0a.e4.4b
Flags[ 0x02 0x0c 0x80 0x00 ]:
Type: CLOB
Storage: BasicFile
Enable Storage in Row
Characterset Format: IMPLICIT
Partitioned Table: No
Options: VaringWidthReadWrite
Inode:
Size: 32
Flag: 0x09 [ Valid DataInRow ]
Future: 0x00 (should be '0x00')
Blocks: 0
Bytes: 16
Version: 00000.0000000001
Inline data[16]
Dump of memory from 0x00002AE680C155B1 to 0x00002AE680C155C1
2AE680C155B0 8BBE8B01 7E004EA1 8BBE8BC4 7E8C4EA1 [.....N.~.....N.~]
2AE680C155C0 02012CC4 [.,..]
tab 0, row 2, @0x3ef5
tl: 45 fb: --H-FL-- lb: 0x1 cc: 2
col 0: [ 2] c1 02
col 1: [38]
00 54 00 01 02 0c 80 00 00 02 00 00 00 01 00 00 00 0a e4 4c 00 12 09 00 00
00 00 00 00 02 00 00 00 00 00 01 5b b6
LOB
Locator:
Length: 84(38)
Version: 1
Byte Length: 2
LobID: 00.00.00.01.00.00.00.0a.e4.4c
Flags[ 0x02 0x0c 0x80 0x00 ]:
Type: CLOB
Storage: BasicFile
Enable Storage in Row
Characterset Format: IMPLICIT
Partitioned Table: No
Options: VaringWidthReadWrite
Inode:
Size: 18
Flag: 0x09 [ Valid DataInRow ]
Future: 0x00 (should be '0x00')
Blocks: 0
Bytes: 2
Version: 00000.0000000001
Inline data[2]
选取两次的dump的内容，，不过还是不能确定两者是否等价：
00 01 00 00 00 00 00 01 00 00 00 0a e4 49 00 21 09 00 00 00 00 00 00 11 00
00 00 00 00 01 88 01 11 01 01 00 01 09 84 01 09 02 c2 06 02 bc d2
00 54 00 01 02 0c 80 00 00 02 00 00 00 01 00 00 00 0a e4 4c 00 12 09 00 00
00 00 00 00 02 00 00 00 00 00 01 5b b6
总结：嵌套是通过lob在存储嵌套的内容的，TOM说作为一种存储机制，他更倾向于创建父子表，可以再创建一个视图，看上去像一个真正的嵌套表一样，嵌套表适合于编程构造。
在CentOS 6.4下安装Oracle 11gR2(x64)
Oracle 11gR2 在VMWare虚拟机中安装步骤
Debian 下 安装 Oracle 11g XE R2
本文永久更新链接地址：
