[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
