zoukankan html css js c++ java

[每日一题] 11gOCP 1z0-052 :2013-09-27 bitmap index.................................................C37

转载请注明出处：http://blog.csdn.net/guoyjoe/article/details/12106027

正确答案C

这道题目是需要我们掌握位图索引知识点。

一、首先我们来看位图索引的组成

位图索引由Key、Start ROWID、End ROWID、Bitmap组成，如下图所示:

二、其次我们来看位图索引适用范围

1.适用于低基数列

2.更新键列的成本非常高

3.使用 OR/AND/NOT IN/COUNT(*) 谓词进行查询时效率较

三、最后我们来分析这道题为什么选C

1： bitmap 索引是分段存储的，也就是说很多条记录可能是分做了N段来存储，也就是有N个begin/end,当新的记录 insert 而使用以前未曾使用过的物理地址的时候，会产生一个bitmap 段来存储，就算只有一条记录
2：当删除一条记录的时候,在bitmap索引上做了一个delete 的标记并用一新的记录来标记了,下面请看具体的演示
3: 当 dml发生的时候，会lock住某个值的存储bit的那一rowid所在的记录，参考下面的 row 中lock ,这样显然会影响并发

4、实验验证

（1）创建表

gyj@OCM> Create table gyj_bitmap_t1(id int ,name varchar2(10));

Table created.

（2）插入数据

gyj@OCM>  Begin 
  2     For i in 1 .. 10000 loop
  3  Insert into gyj_bitmap_t1 values(i,'AAAAA');
  4  Commit;
  5        end loop;
  6       End;
  7       /

PL/SQL procedure successfully completed.
gyj@OCM> Begin 
  2     For i in 10001 .. 20000 loop
  3  Insert into gyj_bitmap_t1 values(i,'BBBBB');
  4  Commit;
  5        end loop;
  6       End;
  7       /

（3）创建位图索引

gyj@OCM> create bitmap index b_idx_t1 on gyj_bitmap_t1(name);

Index created.

（4）在125号会话下修改id=1的一行数据

gyj@OCM> select sid from v$mystat where rownum=1;

       SID
----------
       125

gyj@OCM> Update gyj_bitmap_t1 set name='BBBBB' where id=1;

1 row updated.

（5）在125号会话下修改id=10001的一行数据，结果被阻塞了。。。

gyj@OCM> select sid from v$mystat where rownum=1;

       SID
----------
       149

gyj@OCM> Update gyj_bitmap_t1 set name='AAAAA' where id=10001;Update

（6）查看锁

sys@OCM> select * from v$lock where sid in(125,149);

ADDR             KADDR                   SID TY        ID1        ID2      LMODE    REQUEST      CTIME      BLOCK
---------------- ---------------- ---------- -- ---------- ---------- ---------- ---------- ---------- ----------
00000000910D8760 00000000910D87B8        125 AE        100          0          4          0       1177          0
00000000910D9118 00000000910D9170        149 TX      65541       1235          0          4        194          0
00000000910DA1D0 00000000910DA228        149 AE        100          0          4          0        480          0
00002B7A1402D830 00002B7A1402D890        125 TM      77552          0          3          0        212          0
00002B7A1402D830 00002B7A1402D890        149 TM      77552          0          3          0        194          0
000000008FB03048 000000008FB030C0        125 TX      65541       1235          6          0        212          1
000000008FB1F928 000000008FB1F9A0        149 TX     131093       1616          6          0        194          0

可以看出149号会话请求4号锁，一直没有持有，这个4号锁是什么呢，锁的标记符是：TX -65541 -1235，从这里可以分析出是在哪个回滚段的第几个事务槽。

sys@OCM> select to_char('65541','xxxxxxxxxxxx') from dual;

TO_CHAR('6554
-------------
        10005

转化成0001和0005即1号回滚段的第5号事务槽上。

我去dump一下这个1号回滚段的段头（上面有事务表信息）

sys@OCM> select * from v$rollname where usn=1;

       USN NAME
---------- ------------------------------
         1 _SYSSMU1_1240252155$

sys@OCM> alter system dump undo header "_SYSSMU1_1240252155$";

System altered.

(7)转储出来的事务表如下：

Undo block address=0x00c00576，转化成地址是：3号文件的1398号块

sys@OCM> alter system dump datafile 3 block 1398;

System altered.

(8)转储出undo信息：

(9) 通过undo链接这里一层层往前找到事务开始：

*-----------------------------
* Rec #0x1e  slt: 0x05  objn: 77552(0x00012ef0)  objd: 77552  tblspc: 7(0x00000007)
*       Layer:  11 (Row)   opc: 1   rci 0x00   
Undo type:  Regular undo    Begin trans    Last buffer split:  No 
Temp Object:  No 
Tablespace Undo:  No 
rdba: 0x00000000Ext idx: 0
flg2: 0
*-----------------------------
uba: 0x00c00576.0145.1d ctl max scn: 0x0000.00243640 prv tx scn: 0x0000.00243658
txn start scn: scn: 0x0000.00243606 logon user: 90
 prev brb: 12584309 prev bcl: 0
KDO undo record:
KTB Redo 
op: 0x04  ver: 0x01  
compat bit: 4 (post-11) padding: 1
op: L  itl: xid:  0x000a.018.00000495 uba: 0x00c007b8.015a.15
                      flg: C---    lkc:  0     scn: 0x0000.0023aa8c
KDO Op code: URP row dependencies Disabled
  xtype: XAxtype KDO_KDOM2 flags: 0x00000080  bdba: 0x018000c7  hdba: 0x018000c2
itli: 2  ispac: 0  maxfr: 4858
tabn: 0 slot: 0(0x0) flag: 0x2c lock: 0 ckix: 0
ncol: 2 nnew: 1 size: 0
Vector content: 
col  1: [ 5]  41 41 41 41 41
 
*-----------------------------
* Rec #0x1f  slt: 0x05  objn: 77553(0x00012ef1)  objd: 77553  tblspc: 7(0x00000007)
*       Layer:  10 (Index)   opc: 22   rci 0x1e   
Undo type:  Regular undo   Last buffer split:  No 
Temp Object:  No 
Tablespace Undo:  No 
rdba: 0x00000000
*-----------------------------
index undo for leaf key operations
KTB Redo 
op: 0x04  ver: 0x01  
compat bit: 4 (post-11) padding: 1
op: L  itl: xid:  0xffff.000.00000000 uba: 0x00000000.0000.00
                      flg: C---    lkc:  0     scn: 0x0000.002435d3
Dump kdilk : itl=2, kdxlkflg=0x1 sdc=0 indexid=0x1c0008a block=0x01c0008b
(kdxlup): update keydata in row
key :(1469): 
 05 41 41 41 41 41 06 01 80 00 c3 00 00 06 01 80 00 db 01 3f 85 a7 cf ff ff
 ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf
 ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff
 ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff
 ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff
 ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff
 ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff
 cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff
 cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff
 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff
 ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cd ff ff ff ff ff 07 ff 19 ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff
 cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff
 ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf
 ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff
 ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff
 ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff
 ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff
 cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff
 ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf
 ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cc ff ff ff ff 0f ff 77 ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff
 ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff
 ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff
 ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff
 cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff
 ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff
 ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff
 ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf
 ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cc ff ff ff ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff
 ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff
 ff 0f ff 1a ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff
 ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff
 ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cc ff ff ff ff 0f ff b1 02
 ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff
 ff cf ff ff ff ff ff ff ff ff cf ff ff ff ff ff ff ff 01
 
*-----------------------------
* Rec #0x20  slt: 0x05  objn: 77553(0x00012ef1)  objd: 77553  tblspc: 7(0x00000007)
*       Layer:  10 (Index)   opc: 37   rci 0x1f   
Undo type:  Regular undo   Last buffer split:  No 
Temp Object:  No 
Tablespace Undo:  No 
rdba: 0x00000000
*-----------------------------
index undo for leaf key operations
index change (kdxIndexlogicalNonkeyUpdate): count=6
KTB Redo 
op: 0x02  ver: 0x01  
compat bit: 4 (post-11) padding: 1
op: C  uba: 0x00c00576.0145.1f
Dump kdilk : len=29 != sizeof(kdilk)=20 :(29): 
 26 02 05 00 8a 00 c0 01 8b 00 c0 01 00 00 00 00 00 00 00 00 00 04 02 02 00
 00 00 00 00
itl=2, kdxlkflg=0x5 sdc=0 indexid=0x1c0008a block=0x01c0008b
(kdxIndexLogicalNonkeyUpdate): index logical nonkey update
ncol: 4  nvec: 2 nnew: 2 size: 0
unique key: (13):  05 41 41 41 41 41 06 01 80 00 c3 00 00
logical nonkey columns operation records:
 column  2:
  atom same length replace: from offset 0 replace 6 bytes:
   01 80 00 db 01 3f
 column  3:
  atom replace: from offset 279 replace 10 bytes with 10 bytes:
   ff 1a ff ff ff ff ff ff ff ff
 
*-----------------------------
* Rec #0x21  slt: 0x05  objn: 77553(0x00012ef1)  objd: 77553  tblspc: 7(0x00000007)
*       Layer:  10 (Index)   opc: 22   rci 0x20   
Undo type:  Regular undo   Last buffer split:  No 
Temp Object:  No 
Tablespace Undo:  No 
rdba: 0x00000000
*-----------------------------
index undo for leaf key operations
KTB Redo 
op: 0x02  ver: 0x01  
compat bit: 4 (post-11) padding: 1
op: C  uba: 0x00c00576.0145.20
Dump kdilk : itl=2, kdxlkflg=0x1 sdc=0 indexid=0x1c0008a block=0x01c0008b
(kdxlpu): purge leaf row
key :(21): 
 05 42 42 42 42 42 06 00 00 00 00 00 00 06 00 00 00 00 00 00 ff
 
*-----------------------------
* Rec #0x22  slt: 0x05  objn: 77553(0x00012ef1)  objd: 77553  tblspc: 7(0x00000007)
*       Layer:  10 (Index)   opc: 37   rci 0x21   
Undo type:  Regular undo   Last buffer split:  No 
Temp Object:  No 
Tablespace Undo:  No 
rdba: 0x00000000
*-----------------------------
index undo for leaf key operations
index change (kdxIndexlogicalNonkeyUpdate): count=6
KTB Redo 
op: 0x02  ver: 0x01  
compat bit: 4 (post-11) padding: 1
op: C  uba: 0x00c00576.0145.21
Dump kdilk : len=29 != sizeof(kdilk)=20 :(29): 
 26 02 05 00 8a 00 c0 01 8b 00 c0 01 00 00 00 00 00 00 00 00 00 04 02 02 fa
 ff 00 00 00
itl=2, kdxlkflg=0x5 sdc=0 indexid=0x1c0008a block=0x01c0008b
(kdxIndexLogicalNonkeyUpdate): index logical nonkey update
ncol: 4  nvec: 2 nnew: 2 size: -6
unique key: (13):  05 42 42 42 42 42 06 00 00 00 00 00 00
logical nonkey columns operation records:
 column  2:
  atom same length replace: from offset 0 replace 6 bytes:
   00 00 00 00 00 00
 column  3:
  atom delete: from offset 0 delete 6 bytes

（10）我们来dump一下id=1和id=10001这两行所在的数据块的数据。

gyj@OCM> select dbms_rowid.rowid_relative_fno(rowid) file#,dbms_rowid.rowid_block_number(rowid) block from gyj_bitmap_t1 where id in(1,10001);

     FILE#      BLOCK
---------- ----------
         6        199
         6        219

(11)转储6号文件199号块

sys@OCM> alter system dump datafile 6 block 199;

System altered.

gyj@OCM> select UTL_RAW.CAST_TO_NUMBER(replace('c1 02',' ')) from dual;

UTL_RAW.CAST_TO_NUMBER(REPLACE('C3020102',''))
----------------------------------------------
                         1

gyj@OCM> select UTL_RAW.CAST_TO_VARCHAR2(REPLACE('42 42 42 42 42',' ')) from dual;

UTL_RAW.CAST_TO_VARCHAR2(REPLACE('4141414141',''))
----------------------------------------------------------------------------------
BBBBB

(11)转储6号文件219号块

gyj@OCM> select UTL_RAW.CAST_TO_NUMBER(replace('c3 02 01 02',' ')) from dual;

UTL_RAW.CAST_TO_NUMBER(REPLACE('C3020102',''))
----------------------------------------------
                                         10001

gyj@OCM> select UTL_RAW.CAST_TO_VARCHAR2(REPLACE('41 41 41 41 41',' ')) from dual;

UTL_RAW.CAST_TO_VARCHAR2(REPLACE('4141414141',''))
----------------------------------------------------------------------------------
AAAAA

具体这个不再说了，需要有一定的基础！！！

QQ：252803295

技术交流QQ群：
DSI&Core Search Ⅰ 群：127149411（2000人技术群:未满）
DSI&Core Search Ⅱ 群：177089463（1000人技术群:未满）
DSI&Core Search Ⅲ 群：284596437（500人技术群:未满）
DSI&Core Search Ⅳ 群：192136702（500人技术群:未满）
DSI&Core Search Ⅴ 群：285030382（500人闲聊群:未满）

MAIL：dbathink@hotmail.com

BLOG： http://blog.csdn.net/guoyjoe

WEIBO：http://weibo.com/guoyJoe0218

ITPUB： http://www.itpub.net/space-uid-28460966.html

OCM： http://education.oracle.com/education/otn/YGuo.HTM

查看全文

相关阅读:
Asp.net 弹出对话框基类 (输出alet警告框)
Request.ServerVariables获取环境变量
 Asp.net 字符串操作基类(安全,替换,分解等)
asp.net 取得远程的IP地址和浏览器类型
 存储过程得到三个值
 弹出无边框网页的Javscrpt代码
 .net打包自动安装数据库!
精妙Sql语句
 关于跨语言站点搜索结果非预期的问题
 SharePoint服务器场环境安装升级补丁的详细步骤

原文地址：https://www.cnblogs.com/riskyer/p/3343353.html