注：low-cardinality是指该列或者列的组合具有的不同值的个数较少，即该列有很多重复值。high-cardinality是指该列或者列的组合具有不同的值的个数较多，即该列有很少的重复值。
理解每种索引的适用场合将对性能产生重大影响。
传统观念认为位图索引最适用于拥有很少不同值的列 ---- 例如GENDER, MARITAL_STATUS,和RELATION。但是，这种假设是不准确的。实际上，对于大多非频繁更新的并发系统，位图索引也是适用的。事实上，下面将会展示，对以一个具有100%唯一值的列（主键的候选列）来说，位图索引和B树索引一样有效。
本文将提供一些例子以及优化建议，它们对于low-cardinality和high-cardinality上的两种索引都是通用的。这些例子将帮助DBA理解位图索引不是依赖于cardinality而是依赖于程序自身。
索引比较
在唯一列上使用位图索引有一些缺点 --- 其中一个是需要足够的空间(Oracle不推荐使用)。但是，位图索引的大小不但与位图索引列的cardinality有关，还与数据的分布有关。因此，GENDER列上的位图索引比B树索引要小，相反，EMPNO上的位图索引比B树索引大的多。但是，相对于OLTP系统来说，决策支持系统只有很少的用户访问，因而对于这些系统，资源不是问题。
为了阐明这个观点，我创建了两个表，test_normal和test_random。用PL/SQL块在test_normal中插入100万条记录，然后在test_random表中随机插入相同的记录。
1. Create table test_normal (empno number(10), ename varchar2(30), sal number(10));
2.
3. Begin
4. For i in
5. Loop
6. Insert into test_normal
7. values(i, dbms_random.string('U',30), dbms_random.value(1000,7000));
8. If mod(i, 10000) = 0 then
9. Commit;
10. End if;
11. End loop;
12. End;
13. /
14.
15. Create table test_random
16. as
17. select /*+ append */
18.
19. SQL> select count(*) "Total Rows"
20.
21. Total Rows
22. ----------
23. 1000000
24.
25. Elapsed: 00:00:01.09
26.
27. SQL> select count(distinct empno) "Distinct Values"
28.
29. Distinct Values
30. ---------------
31. 1000000
32.
33. Elapsed: 00:00:06.09
34. SQL> select count(*) "Total Rows"
35.
36. Total Rows
37. ----------
38. 1000000
39.
40. Elapsed: 00:00:03.05
41. SQL> select count(distinct empno) "Distinct Values"
42.
43. Distinct Values
44. ---------------
45. 1000000
46.
47. Elapsed: 00:00:12.07
Create table test_normal (empno number(10), ename varchar2(30), sal number(10));
Begin
For i in 1..1000000
Loop
Insert into test_normal
values(i, dbms_random.string('U',30), dbms_random.value(1000,7000));
If mod(i, 10000) = 0 then
Commit;
End if;
End loop;
End;
/
Create table test_random
as
select /*+ append */ * from test_normal order by dbms_random.random;
SQL> select count(*) "Total Rows" from test_normal;
Total Rows
----------
1000000
Elapsed: 00:00:01.09
SQL> select count(distinct empno) "Distinct Values" from test_normal;
Distinct Values
---------------
1000000
Elapsed: 00:00:06.09
SQL> select count(*) "Total Rows" from test_random;
Total Rows
----------
1000000
Elapsed: 00:00:03.05
SQL> select count(distinct empno) "Distinct Values" from test_random;
Distinct Values
---------------
1000000
Elapsed: 00:00:12.07注意，test_normal表是组织良好的，test_random表是随机创建的，因此，其中的数据是无组织的。在上面的表中，EMPNO列上的值完全不同，因此可以作为候选主键。如果你把该列定义为主键，oracle将会建立一个B树索引，因为Oracle不支持主键位图索引。
为了分析这些索引的行为，我们执行下面的步骤：
- 在表test_normal上：
- 在EMPNO列上建立一个位图索引，并执行一些相等性查询。
- 在EMPNO列上建立一个B树索引，执行一些相等性查询，并且比较获得不同结果集所执行的查询需要的物理I/O和逻辑I/O的次数。
- 在表test_random表上：
- 和1.1相同的步骤
- 和1.2相同的步骤
- 在表test_normal上：
- 和1.1相同的步骤，但是执行范围查询。
- 和1.2相同的步骤，但是执行范围查询。比较统计结果。
- 在表test_random表上：
- 和3.1相同的步骤。
- 和3.2相同的步骤
- 在表test_normal上：
- 在SAL列上建立一个位图索引，并且执行一些相等性查询和范围查询。
- 在SAL列上建立一个B树索引，并且执行一些相等性查询和范围查询(和5.1相同的结果集)，比较获取结果执行的I/O次数。
- 在两个表中添加GENDER列，并且把该列更新为3个可能的值：M(女性), F(男性), null(未知)。根据一些条件更新该列的值。
- 在该列上建立一个位图索引并且执行一些相等性查询。
- 在GENDER列上建立一个B树索引并且执行一些相等性查询，和步骤7的结果比较。
步骤1到4涉及一个high-cardinality列（完全不同），步骤5是一个normal-cardinality列，步骤7和8是一个low-cardinality列。
步骤1.1（在表test_normal上）
在该步中，我们在表test_normal上建立一个位图索引，然后检查索引的大小、聚簇因子(clustering factor)和表的大小。然后执行一些相等性查询并且查看使用位图索引时查询需要的I/O次数。
1. SQL> create bitmap index normal_empno_bmx on test_normal(empno);
2.
3. Index created.
4.
5. Elapsed: 00:00:29.06
6. SQL> analyze table test_normal compute statistics for table for all indexes for
7.
8.
9. Table analyzed.
10.
11. Elapsed: 00:00:19.01
12. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
13. 2 from user_segments
14. 3* where segment_name in ('TEST_NORMAL','NORMAL_EMPNO_BMX');
15.
16. SEGMENT_NAME Size in
17. ------------------------------------ ---------------
18. TEST_NORMAL 50
19. NORMAL_EMPNO_BMX 28
20.
21. Elapsed: 00:00:02.00
22. SQL> select index_name, clustering_factor from user_indexes;
23.
24. INDEX_NAME CLUSTERING_FACTOR
25. ------------------------------ ---------------------------------
26. NORMAL_EMPNO_BMX 1000000
27.
28. Elapsed: 00:00:00.00
SQL> create bitmap index normal_empno_bmx on test_normal(empno);
Index created.
Elapsed: 00:00:29.06
SQL> analyze table test_normal compute statistics for table for all indexes for all indexed columns;
Table analyzed.
Elapsed: 00:00:19.01
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3* where segment_name in ('TEST_NORMAL','NORMAL_EMPNO_BMX');
SEGMENT_NAME Size in MB
------------------------------------ ---------------
TEST_NORMAL 50
NORMAL_EMPNO_BMX 28
Elapsed: 00:00:02.00
SQL> select index_name, clustering_factor from user_indexes;
INDEX_NAME CLUSTERING_FACTOR
------------------------------ ---------------------------------
NORMAL_EMPNO_BMX 1000000
Elapsed: 00:00:00.00
可以看到，表上索引的大小是28M并且聚簇因子的大小等于表中的行数。现在我们为不同的结果集执行一些相等性查询：
1. SQL> set
2. SQL> select * from test_normal where empno=&empno;
3. Enter value for
4. old 1: select * from test_normal where empno=&empno
5. new
6.
7. Elapsed: 00:00:00.01
8.
9. Execution Plan
10. ----------------------------------------------------------
11. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
12. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
13. d=1 Bytes=34)
14. 2 1 BITMAP CONVERSION (TO ROWIDS)
15. 3 2 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_EMPNO_BMX'
16.
17. Statistics
18. ----------------------------------------------------------
19. 0 recursive calls
20. 0 db block gets
21. 5 consistent gets
22. 0 physical reads
23. 0 redo size
24. 515 bytes sent via SQL*Net to client
25. 499 bytes received via SQL*Net from client
26. 2 SQL*Net roundtrips to/from client
27. 0 sorts (memory)
28. 0 sorts (disk)
29. 1 rows processed
SQL> set autotrace only
SQL> select * from test_normal where empno=&empno;
Enter value for empno: 1000
old 1: select * from test_normal where empno=&empno
new 1: select * from test_normal where empno=1000
Elapsed: 00:00:00.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=4 Car
d=1 Bytes=34)
2 1 BITMAP CONVERSION (TO ROWIDS)
3 2 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_EMPNO_BMX'
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
5 consistent gets
0 physical reads
0 redo size
515 bytes sent via SQL*Net to client
499 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
步骤1.2（在表test_normal上）现在删除表中EMPNO列上的位图索引并创建一个B树索引。像前面一样我们查看索引的大小、聚簇因子的大小并且执行相同的查询，比较I/O的次数。
1. SQL> drop index NORMAL_EMPNO_BMX;
2.
3. Index dropped.
4.
5. SQL> create index normal_empno_idx on test_normal(empno);
6.
7. Index created.
8.
9. SQL> analyze table test_normal compute statistics for table for all indexes for
10.
11. Table analyzed.
12.
13. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
14. 2 from user_segments
15. 3 where segment_name in ('TEST_NORMAL','NORMAL_EMPNO_IDX');
16.
17. SEGMENT_NAME Size in
18. ---------------------------------- ---------------
19. TEST_NORMAL 50
20. NORMAL_EMPNO_IDX 18
21.
22. SQL> select index_name, clustering_factor from user_indexes;
23.
24. INDEX_NAME CLUSTERING_FACTOR
25. ---------------------------------- ----------------------------------
26. NORMAL_EMPNO_IDX 6210
SQL> drop index NORMAL_EMPNO_BMX;
Index dropped.
SQL> create index normal_empno_idx on test_normal(empno);
Index created.
SQL> analyze table test_normal compute statistics for table for all indexes for all indexed columns;
Table analyzed.
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3 where segment_name in ('TEST_NORMAL','NORMAL_EMPNO_IDX');
SEGMENT_NAME Size in MB
---------------------------------- ---------------
TEST_NORMAL 50
NORMAL_EMPNO_IDX 18
SQL> select index_name, clustering_factor from user_indexes;
INDEX_NAME CLUSTERING_FACTOR
---------------------------------- ----------------------------------
NORMAL_EMPNO_IDX 6210很明显，在该表的EMPNO列上，B树索引比位图索引要小。B树索引上的聚簇因子接近于表中的数据块数；因此B树索引对于范围查询更有效。
现在，我们使用B树索引执行相同的查询。
1. SQL> set
2. SQL> select * from test_normal where empno=&empno;
3. Enter value for
4. old 1: select * from test_normal where empno=&empno
5. new
6.
7. Elapsed: 00:00:00.01
8.
9. Execution Plan
10. ----------------------------------------------------------
11. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
12. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
13. d=1 Bytes=34)
14. 2 1 INDEX (RANGE SCAN) OF 'NORMAL_EMPNO_IDX'
15. ost=3 Card=1)
16.
17. Statistics
18. ----------------------------------------------------------
19. 29 recursive calls
20. 0 db block gets
21. 5 consistent gets
22. 0 physical reads
23. 0 redo size
24. 515 bytes sent via SQL*Net to client
25. 499 bytes received via SQL*Net from client
26. 2 SQL*Net roundtrips to/from client
27. 0 sorts (memory)
28. 0 sorts (disk)
29. 1 rows processed
SQL> set autot trace
SQL> select * from test_normal where empno=&empno;
Enter value for empno: 1000
old 1: select * from test_normal where empno=&empno
new 1: select * from test_normal where empno=1000
Elapsed: 00:00:00.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=4 Car
d=1 Bytes=34)
2 1 INDEX (RANGE SCAN) OF 'NORMAL_EMPNO_IDX' (NON-UNIQUE) (C
ost=3 Card=1)
Statistics
----------------------------------------------------------
29 recursive calls
0 db block gets
5 consistent gets
0 physical reads
0 redo size
515 bytes sent via SQL*Net to client
499 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed可以看到，对于相同的结果集，在唯一列上的位图索引和B树索引需要相同的物理和逻辑读取次数。
BITMAP（位图） | EMPNO | B-TREE（B树） | ||
Consistent Reads | Physical Reads | Consistent Reads | Physical Reads | |
5 | 0 | 1000 | 5 | 0 |
5 | 2 | 2398 | 5 | 2 |
5 | 2 | 8545 | 5 | 2 |
5 | 2 | 98008 | 5 | 2 |
5 | 2 | 85342 | 5 | 2 |
5 | 2 | 128444 | 5 | 2 |
5 | 2 | 858 | 5 | 2 |
步骤2.1（在表test_random上）
现在，在test_random表上执行相同的操作：
1. SQL> create bitmap index random_empno_bmx on test_random(empno);
2.
3. Index created.
4.
5. SQL> analyze table test_random compute statistics for table for all indexes for
6.
7. Table analyzed.
8.
9. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
10. 2 from user_segments
11. 3* where segment_name in ('TEST_RANDOM','RANDOM_EMPNO_BMX');
12.
13. SEGMENT_NAME Size in
14. ------------------------------------ ---------------
15. TEST_RANDOM 50
16. RANDOM_EMPNO_BMX 28
17.
18. SQL> select index_name, clustering_factor from user_indexes;
19.
20. INDEX_NAME CLUSTERING_FACTOR
21. ------------------------------ ---------------------------------
22. RANDOM_EMPNO_BMX 1000000
SQL> create bitmap index random_empno_bmx on test_random(empno);
Index created.
SQL> analyze table test_random compute statistics for table for all indexes for all indexed columns;
Table analyzed.
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3* where segment_name in ('TEST_RANDOM','RANDOM_EMPNO_BMX');
SEGMENT_NAME Size in MB
------------------------------------ ---------------
TEST_RANDOM 50
RANDOM_EMPNO_BMX 28
SQL> select index_name, clustering_factor from user_indexes;
INDEX_NAME CLUSTERING_FACTOR
------------------------------ ---------------------------------
RANDOM_EMPNO_BMX 1000000再次，索引上的统计结果（大小和聚簇因子）和在表test_normal中是相同的：
1. SQL> select * from test_random where empno=&empno;
2. Enter value for
3. old 1: select * from test_random where empno=&empno
4. new
5.
6. Elapsed: 00:00:00.01
7.
8. Execution Plan
9. ----------------------------------------------------------
10. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
11. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_RANDOM'
12. 2 1 BITMAP CONVERSION (TO ROWIDS)
13. 3 2 BITMAP INDEX (SINGLE VALUE) OF 'RANDOM_EMPNO_BMX'
14.
15. Statistics
16. ----------------------------------------------------------
17. 0 recursive calls
18. 0 db block gets
19. 5 consistent gets
20. 0 physical reads
21. 0 redo size
22. 515 bytes sent via SQL*Net to client
23. 499 bytes received via SQL*Net from client
24. 2 SQL*Net roundtrips to/from client
25. 0 sorts (memory)
26. 0 sorts (disk)
27. 1 rows processed
SQL> select * from test_random where empno=&empno;
Enter value for empno: 1000
old 1: select * from test_random where empno=&empno
new 1: select * from test_random where empno=1000
Elapsed: 00:00:00.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_RANDOM' (Cost=4 Card=1 Bytes=34)
2 1 BITMAP CONVERSION (TO ROWIDS)
3 2 BITMAP INDEX (SINGLE VALUE) OF 'RANDOM_EMPNO_BMX'
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
5 consistent gets
0 physical reads
0 redo size
515 bytes sent via SQL*Net to client
499 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed步骤2.2（在表test_random上）
现在，和步骤1.2一样，我们删除EMPNO列上的位图索引并且创建一个B树索引。
1. SQL> drop index RANDOM_EMPNO_BMX;
2.
3. Index dropped.
4.
5. SQL> create index random_empno_idx on test_random(empno);
6.
7. Index created.
8.
9. SQL> analyze table test_random compute statistics for table for all indexes for
10.
11. Table analyzed.
12.
13. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
14. 2 from user_segments
15. 3 where segment_name in ('TEST_RANDOM','RANDOM_EMPNO_IDX');
16.
17. SEGMENT_NAME Size in
18. ---------------------------------- ---------------
19. TEST_RANDOM 50
20. RANDOM_EMPNO_IDX 18
21.
22. SQL> select index_name, clustering_factor from user_indexes;
23.
24. INDEX_NAME CLUSTERING_FACTOR
25. ---------------------------------- ----------------------------------
26. RANDOM_EMPNO_IDX 999830
SQL> drop index RANDOM_EMPNO_BMX;
Index dropped.
SQL> create index random_empno_idx on test_random(empno);
Index created.
SQL> analyze table test_random compute statistics for table for all indexes for all indexed columns;
Table analyzed.
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3 where segment_name in ('TEST_RANDOM','RANDOM_EMPNO_IDX');
SEGMENT_NAME Size in MB
---------------------------------- ---------------
TEST_RANDOM 50
RANDOM_EMPNO_IDX 18
SQL> select index_name, clustering_factor from user_indexes;
INDEX_NAME CLUSTERING_FACTOR
---------------------------------- ----------------------------------
RANDOM_EMPNO_IDX 999830该表的索引大小和表test_normal是一样的，但是聚簇因子更接近于行数，这就使得该索引对于范围查询不再高效。该聚簇因子不影响相等性查询，因为该列的值是唯一的，每个键对应1行记录。
现在，在相同的结果集上执行相等性查询。
1. SQL> select * from test_random where empno=&empno;
2. Enter value for
3. old 1: select * from test_random where empno=&empno
4. new
5.
6. Elapsed: 00:00:00.01
7.
8. Execution Plan
9. ----------------------------------------------------------
10. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
11. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_RANDOM'
12. 2 1 INDEX (RANGE SCAN) OF 'RANDOM_EMPNO_IDX'
13.
14. Statistics
15. ----------------------------------------------------------
16. 0 recursive calls
17. 0 db block gets
18. 5 consistent gets
19. 0 physical reads
20. 0 redo size
21. 515 bytes sent via SQL*Net to client
22. 499 bytes received via SQL*Net from client
23. 2 SQL*Net roundtrips to/from client
24. 0 sorts (memory)
25. 0 sorts (disk)
26. 1 rows processed
SQL> select * from test_random where empno=&empno;
Enter value for empno: 1000
old 1: select * from test_random where empno=&empno
new 1: select * from test_random where empno=1000
Elapsed: 00:00:00.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=1 Bytes=34)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_RANDOM' (Cost=4 Card=1 Bytes=34)
2 1 INDEX (RANGE SCAN) OF 'RANDOM_EMPNO_IDX' (NON-UNIQUE) (Cost=3 Card=1)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
5 consistent gets
0 physical reads
0 redo size
515 bytes sent via SQL*Net to client
499 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed再次表明，结果和步骤1.1和1.2几乎相同。对于唯一列来说，数据分布不影响逻辑和物理I/O。
步骤3.1（在表test_normal上）
在该步中，我们将创建一个位图索引。我们知道索引的聚簇因子大小和表中的行数相同。现在我们执行一些范围查询。
1. SQL> select * from test_normal where empno between &range1 and &range2;
2. Enter value for
3. Enter value for
4. old 1: select * from test_normal where empno between &range1 and &range2
5. new
6.
7. 2300 rows selected.
8.
9. Elapsed: 00:00:00.03
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=451 Card=2299 Bytes=78166)
14. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
15. 2 1 BITMAP CONVERSION (TO ROWIDS)
16. 3 2 BITMAP INDEX (RANGE SCAN) OF 'NORMAL_EMPNO_BMX'
17.
18. Statistics
19. ----------------------------------------------------------
20. 0 recursive calls
21. 0 db block gets
22. 331 consistent gets
23. 0 physical reads
24. 0 redo size
25. 111416 bytes sent via SQL*Net to client
26. 2182 bytes received via SQL*Net from client
27. 155 SQL*Net roundtrips to/from client
28. 0 sorts (memory)
29. 0 sorts (disk)
30. 2300 rows processed
SQL> select * from test_normal where empno between &range1 and &range2;
Enter value for range1: 1
Enter value for range2: 2300
old 1: select * from test_normal where empno between &range1 and &range2
new 1: select * from test_normal where empno between 1 and 2300
2300 rows selected.
Elapsed: 00:00:00.03
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=451 Card=2299 Bytes=78166)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=451 Card=2299 Bytes=78166)
2 1 BITMAP CONVERSION (TO ROWIDS)
3 2 BITMAP INDEX (RANGE SCAN) OF 'NORMAL_EMPNO_BMX'
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
331 consistent gets
0 physical reads
0 redo size
111416 bytes sent via SQL*Net to client
2182 bytes received via SQL*Net from client
155 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2300 rows processed步骤3.2（在表test_normal上）
该步中，我们在test_normal的B树索引上执行查询。
1. SQL> select * from test_normal where empno between &range1 and &range2;
2. Enter value for
3. Enter value for
4. old 1: select * from test_normal where empno between &range1 and &range2
5. new
6.
7. 2300 rows selected.
8.
9. Elapsed: 00:00:00.02
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=23 Card=2299 Bytes=78166)
14. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
15. 2 1 INDEX (RANGE SCAN) OF 'NORMAL_EMPNO_IDX'
16.
17. Statistics
18. ----------------------------------------------------------
19. 0 recursive calls
20. 0 db block gets
21. 329 consistent gets
22. 15 physical reads
23. 0 redo size
24. 111416 bytes sent via SQL*Net to client
25. 2182 bytes received via SQL*Net from client
26. 155 SQL*Net roundtrips to/from client
27. 0 sorts (memory)
28. 0 sorts (disk)
29. 2300 rows processed
SQL> select * from test_normal where empno between &range1 and &range2;
Enter value for range1: 1
Enter value for range2: 2300
old 1: select * from test_normal where empno between &range1 and &range2
new 1: select * from test_normal where empno between 1 and 2300
2300 rows selected.
Elapsed: 00:00:00.02
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=23 Card=2299 Bytes=78166)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=23 Card=2299 Bytes=78166)
2 1 INDEX (RANGE SCAN) OF 'NORMAL_EMPNO_IDX' (NON-UNIQUE) (Cost=8 Card=2299)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
329 consistent gets
15 physical reads
0 redo size
111416 bytes sent via SQL*Net to client
2182 bytes received via SQL*Net from client
155 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2300 rows processed在不同的范围上执行的查询结果如下：
BITMAP | EMPNO (Range) | B-TREE | ||
Consistent Reads | Physical Reads | Consistent Reads | Physical Reads | |
331 | 0 | 1-2300 | 329 | 0 |
285 | 0 | 8-1980 | 283 | 0 |
346 | 19 | 1850-4250 | 344 | 16 |
427 | 31 | 28888-31850 | 424 | 28 |
371 | 27 | 82900-85478 | 367 | 23 |
2157 | 149 | 984888-1000000 | 2139 | 35 |
可以看到，在两种索引上需要的逻辑和物理IO基本上是相同的。最后一个范围(984888-1000000)差不多返回了15,000行，是所有范围查询中最大的。当我们执行全表扫描时（通过/*+ full(test_normal) */
步骤4.1（在表test_random上）
在该步中，我们将在表test_random的位图索引上执行范围查询，在这儿，你将看到聚簇因子的影响。
1. SQL>select * from test_random where empno between &range1 and &range2;
2. Enter value for
3. Enter value for
4. old 1: select * from test_random where empno between &range1 and &range2
5. new
6.
7. 2300 rows selected.
8.
9. Elapsed: 00:00:08.01
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=453 Card=2299 Bytes=78166)
14. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_RANDOM'
15. 2 1 BITMAP CONVERSION (TO ROWIDS)
16. 3 2 BITMAP INDEX (RANGE SCAN) OF 'RANDOM_EMPNO_BMX'
17.
18. Statistics
19. ----------------------------------------------------------
20. 0 recursive calls
21. 0 db block gets
22. 2463 consistent gets
23. 1200 physical reads
24. 0 redo size
25. 111416 bytes sent via SQL*Net to client
26. 2182 bytes received via SQL*Net from client
27. 155 SQL*Net roundtrips to/from client
28. 0 sorts (memory)
29. 0 sorts (disk)
30. 2300 rows processed
SQL>select * from test_random where empno between &range1 and &range2;
Enter value for range1: 1
Enter value for range2: 2300
old 1: select * from test_random where empno between &range1 and &range2
new 1: select * from test_random where empno between 1 and 2300
2300 rows selected.
Elapsed: 00:00:08.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=453 Card=2299 Bytes=78166)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_RANDOM' (Cost=453 Card=2299 Bytes=78166)
2 1 BITMAP CONVERSION (TO ROWIDS)
3 2 BITMAP INDEX (RANGE SCAN) OF 'RANDOM_EMPNO_BMX'
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
2463 consistent gets
1200 physical reads
0 redo size
111416 bytes sent via SQL*Net to client
2182 bytes received via SQL*Net from client
155 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2300 rows processed步骤4.2（在表test_random上）
在该步中，我们将在test_random的B树索引上执行范围查询。回想一下，该索引上的聚簇因子接近于表中记录的行数。下面是优化器的输出：
1. SQL> select * from test_random where empno between &range1 and &range2;
2. Enter value for
3. Enter value for
4. old 1: select * from test_random where empno between &range1 and &range2
5. new
6.
7. 2300 rows selected.
8.
9. Elapsed: 00:00:03.04
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=613 Card=2299 Bytes=78166)
14. 1 0 TABLE ACCESS (FULL) OF 'TEST_RANDOM'
15.
16. Statistics
17. ----------------------------------------------------------
18. 0 recursive calls
19. 0 db block gets
20. 6415 consistent gets
21. 4910 physical reads
22. 0 redo size
23. 111416 bytes sent via SQL*Net to client
24. 2182 bytes received via SQL*Net from client
25. 155 SQL*Net roundtrips to/from client
26. 0 sorts (memory)
27. 0 sorts (disk)
28. 2300 rows processed
SQL> select * from test_random where empno between &range1 and &range2;
Enter value for range1: 1
Enter value for range2: 2300
old 1: select * from test_random where empno between &range1 and &range2
new 1: select * from test_random where empno between 1 and 2300
2300 rows selected.
Elapsed: 00:00:03.04
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=613 Card=2299 Bytes=78166)
1 0 TABLE ACCESS (FULL) OF 'TEST_RANDOM' (Cost=613 Card=2299 Bytes=78166)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
6415 consistent gets
4910 physical reads
0 redo size
111416 bytes sent via SQL*Net to client
2182 bytes received via SQL*Net from client
155 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2300 rows processed
因为聚簇因子的缘故，优化器选择了全表扫描而不是使用索引：
BITMAP | EMPNO (Range) | B-TREE | ||
Consistent Reads | Physical Reads | Consistent Reads | Physical Reads | |
2463 | 1200 | 1-2300 | 6415 | 4910 |
2114 | 31 | 8-1980 | 6389 | 4910 |
2572 | 1135 | 1850-4250 | 6418 | 4909 |
3173 | 1620 | 28888-31850 | 6456 | 4909 |
2762 | 1358 | 82900-85478 | 6431 | 4909 |
7254 | 3329 | 984888-1000000 | 7254 | 4909 |
仅对于最后一个范围（984888-1000000），对于位图索引优化器选择了全表扫描。然而，对于B树索引，全部使用全表扫描。引起这种差异的原因是聚簇因子：优化器在产生执行计划时不考虑位图索引的聚簇因子，但是对于B树索引来说，则需要 考虑聚簇因子。在上面的情况中，位图索引比B树索引更有效。
下面的步骤揭示了这些索引更有趣的方面。
步骤5.1（在表test_normal上）
在表test_normal的SAL列上建立一个位图索引，该列拥有普通的cardinality。
1. SQL> create bitmap index normal_sal_bmx on test_normal(sal);
2.
3. Index created.
4.
5. SQL> analyze table test_normal compute statistics for table for all indexes for
6.
7. Table analyzed.
SQL> create bitmap index normal_sal_bmx on test_normal(sal);
Index created.
SQL> analyze table test_normal compute statistics for table for all indexes for all indexed columns;
Table analyzed.
得到索引的大小和聚簇因子：
1. SQL>select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2. 2* from user_segments
3. 3* where segment_name in ('TEST_NORMAL','NORMAL_SAL_BMX');
4.
5. SEGMENT_NAME Size in
6. ------------------------------ --------------
7. TEST_NORMAL 50
8. NORMAL_SAL_BMX 4
9.
10. SQL> select index_name, clustering_factor from user_indexes;
11.
12. INDEX_NAME CLUSTERING_FACTOR
13. ------------------------------ ----------------------------------
14. NORMAL_SAL_BMX 6001
SQL>select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2* from user_segments
3* where segment_name in ('TEST_NORMAL','NORMAL_SAL_BMX');
SEGMENT_NAME Size in MB
------------------------------ --------------
TEST_NORMAL 50
NORMAL_SAL_BMX 4
SQL> select index_name, clustering_factor from user_indexes;
INDEX_NAME CLUSTERING_FACTOR
------------------------------ ----------------------------------
NORMAL_SAL_BMX 6001下面执行查询，首先执行相等性查询：
1. SQL> set
2. SQL> select * from test_normal where sal=&sal;
3. Enter value for
4. old 1: select * from test_normal where sal=&sal
5. new
6.
7. 164 rows selected.
8.
9. Elapsed: 00:00:00.08
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=39 Card=168 Bytes=4032)
14. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
15. 2 1 BITMAP CONVERSION (TO ROWIDS)
16. 3 2 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
17.
18. Statistics
19. ----------------------------------------------------------
20. 0 recursive calls
21. 0 db block gets
22. 165 consistent gets
23. 0 physical reads
24. 0 redo size
25. 8461 bytes sent via SQL*Net to client
26. 609 bytes received via SQL*Net from client
27. 12 SQL*Net roundtrips to/from client
28. 0 sorts (memory)
29. 0 sorts (disk)
30. 164 rows processed
SQL> set autot trace
SQL> select * from test_normal where sal=&sal;
Enter value for sal: 1869
old 1: select * from test_normal where sal=&sal
new 1: select * from test_normal where sal=1869
164 rows selected.
Elapsed: 00:00:00.08
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=39 Card=168 Bytes=4032)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=39 Card=168 Bytes=4032)
2 1 BITMAP CONVERSION (TO ROWIDS)
3 2 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
165 consistent gets
0 physical reads
0 redo size
8461 bytes sent via SQL*Net to client
609 bytes received via SQL*Net from client
12 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
164 rows processed
接下来是范围查询：
1. SQL> select * from test_normal where sal between &sal1 and &sal2;
2. Enter value for
3. Enter value for
4. old 1: select * from test_normal where sal between &sal1 and &sal2
5. new
6.
7. 83743 rows selected.
8.
9. Elapsed: 00:00:05.00
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=83376 Bytes
14. =2001024)
15. 1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL'
16. Bytes=2001024)
17.
18. Statistics
19. ----------------------------------------------------------
20. 0 recursive calls
21. 0 db block gets
22. 11778 consistent gets
23. 5850 physical reads
24. 0 redo size
25. 4123553 bytes sent via SQL*Net to client
26. 61901 bytes received via SQL*Net from client
27. 5584 SQL*Net roundtrips to/from client
28. 0 sorts (memory)
29. 0 sorts (disk)
30. 83743 rows processed
SQL> select * from test_normal where sal between &sal1 and &sal2;
Enter value for sal1: 1500
Enter value for sal2: 2000
old 1: select * from test_normal where sal between &sal1 and &sal2
new 1: select * from test_normal where sal between 1500 and 2000
83743 rows selected.
Elapsed: 00:00:05.00
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=83376 Bytes
=2001024)
1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL' (Cost=601 Card=83376
Bytes=2001024)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
11778 consistent gets
5850 physical reads
0 redo size
4123553 bytes sent via SQL*Net to client
61901 bytes received via SQL*Net from client
5584 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
83743 rows processed
现在，删除test_normal上的位图索引并且建立一个B树索引。
1. SQL> create index normal_sal_idx on test_normal(sal);
2.
3. Index created.
4.
5. SQL> analyze table test_normal compute statistics for table for all indexes for
6.
7. Table analyzed.
SQL> create index normal_sal_idx on test_normal(sal);
Index created.
SQL> analyze table test_normal compute statistics for table for all indexes for all indexed columns;
Table analyzed.
查看索引大小和聚簇因子：
1. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2. 2 from user_segments
3. 3 where segment_name in ('TEST_NORMAL','NORMAL_SAL_IDX');
4.
5. SEGMENT_NAME Size in
6. ------------------------------ ---------------
7. TEST_NORMAL 50
8. NORMAL_SAL_IDX 17
9.
10. SQL> select index_name, clustering_factor from user_indexes;
11.
12. INDEX_NAME CLUSTERING_FACTOR
13. ------------------------------ ----------------------------------
14. NORMAL_SAL_IDX 986778
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3 where segment_name in ('TEST_NORMAL','NORMAL_SAL_IDX');
SEGMENT_NAME Size in MB
------------------------------ ---------------
TEST_NORMAL 50
NORMAL_SAL_IDX 17
SQL> select index_name, clustering_factor from user_indexes;
INDEX_NAME CLUSTERING_FACTOR
------------------------------ ----------------------------------
NORMAL_SAL_IDX 986778从上表可以看出，B树索引大于相同列上的位图索引，它的聚簇因子接近于表中的行数。
现在，先执行相等性查询：
1. SQL> set
2. SQL> select * from test_normal where sal=&sal;
3. Enter value for
4. old 1: select * from test_normal where sal=&sal
5. new
6.
7. 164 rows selected.
8.
9. Elapsed: 00:00:00.01
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=169 Card=168 Bytes=4032)
14. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
15. 2 1 INDEX (RANGE SCAN) OF 'NORMAL_SAL_IDX'
16.
17. Statistics
18. ----------------------------------------------------------
19. 0 recursive calls
20. 0 db block gets
21. 177 consistent gets
22. 0 physical reads
23. 0 redo size
24. 8461 bytes sent via SQL*Net to client
25. 609 bytes received via SQL*Net from client
26. 12 SQL*Net roundtrips to/from client
27. 0 sorts (memory)
28. 0 sorts (disk)
29. 164 rows processed
SQL> set autot trace
SQL> select * from test_normal where sal=&sal;
Enter value for sal: 1869
old 1: select * from test_normal where sal=&sal
new 1: select * from test_normal where sal=1869
164 rows selected.
Elapsed: 00:00:00.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=169 Card=168 Bytes=4032)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=169 Card=168 Bytes=4032)
2 1 INDEX (RANGE SCAN) OF 'NORMAL_SAL_IDX' (NON-UNIQUE) (Cost=3 Card=168)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
177 consistent gets
0 physical reads
0 redo size
8461 bytes sent via SQL*Net to client
609 bytes received via SQL*Net from client
12 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
164 rows processed接下来是范围查询：
1. SQL> select * from test_normal where sal between &sal1 and &sal2;
2. Enter value for
3. Enter value for
4. old 1: select * from test_normal where sal between &sal1 and &sal2
5. new
6.
7. 83743 rows selected.
8.
9. Elapsed: 00:00:04.03
10.
11. Execution Plan
12. ----------------------------------------------------------
13. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=83376 Bytes
14. =2001024)
15. 1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL'
16. Bytes=2001024)
17.
18.
19. Statistics
20. ----------------------------------------------------------
21. 0 recursive calls
22. 0 db block gets
23. 11778 consistent gets
24. 3891 physical reads
25. 0 redo size
26. 4123553 bytes sent via SQL*Net to client
27. 61901 bytes received via SQL*Net from client
28. 5584 SQL*Net roundtrips to/from client
29. 0 sorts (memory)
30. 0 sorts (disk)
31. 83743 rows processed
SQL> select * from test_normal where sal between &sal1 and &sal2;
Enter value for sal1: 1500
Enter value for sal2: 2000
old 1: select * from test_normal where sal between &sal1 and &sal2
new 1: select * from test_normal where sal between 1500 and 2000
83743 rows selected.
Elapsed: 00:00:04.03
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=83376 Bytes
=2001024)
1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL' (Cost=601 Card=83376
Bytes=2001024)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
11778 consistent gets
3891 physical reads
0 redo size
4123553 bytes sent via SQL*Net to client
61901 bytes received via SQL*Net from client
5584 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
83743 rows processed在不同的数据集上执行查询的结果如下，可以看出逻辑和物理I/O的次数基本上是相同的。
BITMAP | SAL (Equality) | B-TREE | Rows Fetched | ||
Consistent Reads | Physical Reads | Consistent Reads | Physical Reads | ||
165 | 0 | 1869 | 177 | 164 | |
169 | 163 | 3548 | 181 | 167 | |
174 | 166 | 6500 | 187 | 172 | |
75 | 69 | 7000 | 81 | 73 | |
177 | 163 | 2500 | 190 | 175 | |
BITMAP | SAL (Range) | B-TREE | Rows Fetched | ||
Consistent Reads | Physical Reads | Consistent Reads | Physical Reads | ||
11778 | 5850 | 1500-2000 | 11778 | 3891 | 83743 |
11765 | 5468 | 2000-2500 | 11765 | 3879 | 83328 |
11753 | 5471 | 2500-3000 | 11753 | 3884 | 83318 |
17309 | 5472 | 3000-4000 | 17309 | 3892 | 166999 |
39398 | 5454 | 4000-7000 | 39398 | 3973 | 500520 |
对于范围查询，优化器选择了全表扫描，根本没有使用索引。但是对于相等性查询，优化器使用了索引。再次，逻辑和物理I/O是相同的。
因此，可以得出结论，对于一个具有normal-cardinality的列来说，优化器对于两种类型的索引的选择是相同的，并且没有明显的I/O差异。
步骤6（增加GENDER列）
在测试low-cardinality列之前，我们先增加一个GENDER列并且把它的值更新成M,F或者null。
1. SQL> alter table test_normal add GENDER varchar2(1);
2.
3. Table altered.
4.
5. SQL> select GENDER, count(*) from test_normal group by GENDER;
6.
7. S COUNT(*)
8. - ----------
9. F 333769
10. M 499921
11. 166310
12.
13. 3 rows selected.
SQL> alter table test_normal add GENDER varchar2(1);
Table altered.
SQL> select GENDER, count(*) from test_normal group by GENDER;
S COUNT(*)
- ----------
F 333769
M 499921
166310
3 rows selected.该列上位图索引的大小大约为570KB，如下表所示：
1. SQL> create bitmap index normal_GENDER_bmx on test_normal(GENDER);
2.
3. Index created.
4.
5. Elapsed: 00:00:02.08
6.
7. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
8. 2 from user_segments
9. 3 where segment_name in ('TEST_NORMAL','NORMAL_GENDER_BMX');
10.
11. SEGMENT_NAME Size in
12. ------------------------------ ---------------
13. TEST_NORMAL 50
14. NORMAL_GENDER_BMX .5625
15.
16. 2 rows selected.
SQL> create bitmap index normal_GENDER_bmx on test_normal(GENDER);
Index created.
Elapsed: 00:00:02.08
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3 where segment_name in ('TEST_NORMAL','NORMAL_GENDER_BMX');
SEGMENT_NAME Size in MB
------------------------------ ---------------
TEST_NORMAL 50
NORMAL_GENDER_BMX .5625
2 rows selected.
相对而言，改列上的B树索引的大小为13M，比位图索引大的多。
1. SQL> create index normal_GENDER_idx on test_normal(GENDER);
2.
3. Index created.
4.
5. SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
6. 2 from user_segments
7. 3 where segment_name in ('TEST_NORMAL','NORMAL_GENDER_IDX');
8.
9. SEGMENT_NAME Size in
10. ------------------------------ ---------------
11. TEST_NORMAL 50
12. NORMAL_GENDER_IDX 13
13.
14. 2 rows selected.
SQL> create index normal_GENDER_idx on test_normal(GENDER);
Index created.
SQL> select substr(segment_name,1,30) segment_name, bytes/1024/1024 "Size in MB"
2 from user_segments
3 where segment_name in ('TEST_NORMAL','NORMAL_GENDER_IDX');
SEGMENT_NAME Size in MB
------------------------------ ---------------
TEST_NORMAL 50
NORMAL_GENDER_IDX 13
2 rows selected.现在，执行相等性查询，优化器将不使用该索引，不论是位图索引还是B树索引，它将使用全部扫描。
1. SQL> select * from test_normal where GENDER is null;
2.
3. 166310 rows selected.
4.
5. Elapsed: 00:00:06.08
6.
7. Execution Plan
8. ----------------------------------------------------------
9. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=166310 Bytes=4157750)
10. 1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL'
11.
12. SQL> select * from test_normal where GENDER='M';
13.
14. 499921 rows selected.
15.
16. Elapsed: 00:00:16.07
17.
18. Execution Plan
19. ----------------------------------------------------------
20. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=499921 Bytes=12498025)
21. 1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL'
22.
23. SQL>select * from test_normal where GENDER='F'
24. /
25.
26. 333769 rows selected.
27.
28. Elapsed: 00:00:12.02
29.
30. Execution Plan
31. ----------------------------------------------------------
32. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=333769 Byte
33. s=8344225)
34. 1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL'
35. Bytes=8344225)
SQL> select * from test_normal where GENDER is null;
166310 rows selected.
Elapsed: 00:00:06.08
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=166310 Bytes=4157750)
1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL' (Cost=601 Card=166310 Bytes=4157750)
SQL> select * from test_normal where GENDER='M';
499921 rows selected.
Elapsed: 00:00:16.07
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=499921 Bytes=12498025)
1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL' (Cost=601 Card=499921Bytes=12498025)
SQL>select * from test_normal where GENDER='F'
/
333769 rows selected.
Elapsed: 00:00:12.02
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=333769 Byte
s=8344225)
1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL' (Cost=601 Card=333769
Bytes=8344225)结论
现在我们了解了优化器对于这些技术做出的反应，现在我们来看对于位图索引和B树索引最适合的程序。
保持GENDER列上的位图索引，在SAL列上再建立一个位图索引然后执行一些查询。对这些列上的B树索引执行同样的查询。
在表test_normal中，你需要所有工资等于下列值的所有女性雇员的雇员号码：
1000
1500
2000
2500
3000
3500
4000
4500因此：
1. SQL>select * from test_normal
2. where sal in (1000,1500,2000,2500,3000,3500,4000,4500,5000) and GENDER='M';
SQL>select * from test_normal
where sal in (1000,1500,2000,2500,3000,3500,4000,4500,5000) and GENDER='M';这是一个典型的数据仓库查询，绝对不要在OLTP系统中执行该查询。下面是两列上具有位图索引时的结果：
1. SQL>select * from test_normal
2. where sal in (1000,1500,2000,2500,3000,3500,4000,4500,5000) and GENDER='M';
3.
4. 1453 rows selected.
5.
6. Elapsed: 00:00:02.03
7.
8. Execution Plan
9. ----------------------------------------------------------
10. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=198 Card=754 Bytes=18850)
11. 1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL'
12. 2 1 BITMAP CONVERSION (TO ROWIDS)
13. 3 2 BITMAP AND
14. 4 3 BITMAP OR
15. 5 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
16. 6 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
17. 7 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
18. 8 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
19. 9 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
20. 10 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
21. 11 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
22. 12 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
23. 13 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
24. 14 3 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_GENDER_BMX'
25.
26. Statistics
27. ----------------------------------------------------------
28. 0 recursive calls
29. 0 db block gets
30. 1353 consistent gets
31. 920 physical reads
32. 0 redo size
33. 75604 bytes sent via SQL*Net to client
34. 1555 bytes received via SQL*Net from client
35. 98 SQL*Net roundtrips to/from client
36. 0 sorts (memory)
37. 0 sorts (disk)
38. 1453 rows processed
SQL>select * from test_normal
where sal in (1000,1500,2000,2500,3000,3500,4000,4500,5000) and GENDER='M';
1453 rows selected.
Elapsed: 00:00:02.03
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=198 Card=754 Bytes=18850)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_NORMAL' (Cost=198 Card=754 Bytes=18850)
2 1 BITMAP CONVERSION (TO ROWIDS)
3 2 BITMAP AND
4 3 BITMAP OR
5 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
6 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
7 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
8 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
9 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
10 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
11 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
12 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
13 4 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_SAL_BMX'
14 3 BITMAP INDEX (SINGLE VALUE) OF 'NORMAL_GENDER_BMX'
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
1353 consistent gets
920 physical reads
0 redo size
75604 bytes sent via SQL*Net to client
1555 bytes received via SQL*Net from client
98 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1453 rows processed
下面是B树索引时的结果：
1. SQL>select * from test_normal
2. where sal in (1000,1500,2000,2500,3000,3500,4000,4500,5000) and GENDER='M';
3.
4. 1453 rows selected.
5.
6. Elapsed: 00:00:03.01
7.
8. Execution Plan
9. ----------------------------------------------------------
10. 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=754 Bytes=18850)
11. 1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL'
12.
13. Statistics
14. ----------------------------------------------------------
15. 0 recursive calls
16. 0 db block gets
17. 6333 consistent gets
18. 4412 physical reads
19. 0 redo size
20. 75604 bytes sent via SQL*Net to client
21. 1555 bytes received via SQL*Net from client
22. 98 SQL*Net roundtrips to/from client
23. 0 sorts (memory)
24. 0 sorts (disk)
25. 1453 rows processed
SQL>select * from test_normal
where sal in (1000,1500,2000,2500,3000,3500,4000,4500,5000) and GENDER='M';
1453 rows selected.
Elapsed: 00:00:03.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=601 Card=754 Bytes=18850)
1 0 TABLE ACCESS (FULL) OF 'TEST_NORMAL' (Cost=601 Card=754 Bytes=18850)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
6333 consistent gets
4412 physical reads
0 redo size
75604 bytes sent via SQL*Net to client
1555 bytes received via SQL*Net from client
98 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1453 rows processed
可以看出，如果使用B树索引，优化器使用全表扫描；而对于位图索引，则使用索引。从获取结果需要的I/O次数可以推断出性能。
总之，基于如下的原因，位图索引适用于决策支持系统，而不管cardinality的高低：
- 使用位图索引，优化器可以高效地执行包含AND，OR或者XOR的查询。
- 使用位图索引，优化器可以回答对null的查询和计数。null值在位图索引时同样被加上索引（不像B树索引）。
- 最重要的是，在决策支持系统中，位图索引支持特殊的查询，但B树索引则不能。具体来说，如果你有一个包含50列的表，用户经常查询其中的10列 ---- 10列的组合或者有时是其中一列，创建B树索引会比价困难。如果你在这些列上建立10个位图索引，这些查询都可以通过索引回答，不论你查询的是全部10列，还是10列中的4列或者6列，或者其中的一列。
相反，B树索引非常适合于OLTP系统，其用户执行的都是常规的查询。因为在OLTP系统中，数据会频繁地更新和删除，如果使用位图索引将会引起严重的锁定性能问题。
两种索引都有一个共同的目的：尽快地得到结果。但是你应该依据程序的类型来选择其中之一，而不是根据cardinality水平。
本文章为转载内容，我们尊重原作者对文章享有的著作权。如有内容错误或侵权问题，欢迎原作者联系我们进行内容更正或删除文章。
