本文从不绑定变量和绑定变量两种情况讨论直方图的作用
一、不绑定变量
SQL> create table test(name varchar2(10));
表已创建。
SQL> insert into test select ‘A‘ from table1;
已创建25064行。
SQL> insert into test values(‘B‘);
已创建 1 行。
SQL> insert into test values(‘C‘);
已创建 1 行。
SQL> select name,count(1) from test group by name;
NAME COUNT(1)
---------- ----------
A 25064
B 1
C 1
SQL> create index i_test on test(name);
索引已创建。
SQL> analyze table test compute statistics;
表已分析。
SQL> select * from test where name=‘A‘;
已选择25064行。
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 8355 | 8355 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 8355 | 8355 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=‘A‘)
统计信息
----------------------------------------------------------
32 recursive calls
0 db block gets
1720 consistent gets
0 physical reads
0 redo size
337843 bytes sent via SQL*Net to client
18770 bytes received via SQL*Net from client
1672 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
25064 rows processed
SQL> select * from test where name=‘B‘;
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 8355 | 8355 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 8355 | 8355 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=‘B‘)
统计信息
----------------------------------------------------------
1 recursive calls
0 db block gets
47 consistent gets
0 physical reads
0 redo size
407 bytes sent via SQL*Net to client
400 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
只需返回一条数据，但做了全表扫描。
因为，oracle只知道name列有3个不同的值，但不知道每个不同的值分别有多少记录，oracle默认这些数据是完全均匀的，
所以，当用name做条件时，oracle认为会返回总记录的三分之一（从Rows=8355可以看出）
对test表生成直方图后再做同样的查询
SQL> analyze table test compute statistics for table for all indexes for all indexed columns;
表已分析。
SQL> select * from test where name=‘A‘;
已选择25064行。
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 25064 | 25064 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 25064 | 25064 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=‘A‘)
统计信息
----------------------------------------------------------
1 recursive calls
0 db block gets
1717 consistent gets
0 physical reads
0 redo size
337843 bytes sent via SQL*Net to client
18770 bytes received via SQL*Net from client
1672 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
25064 rows processed
SQL> select * from test where name=‘B‘;
执行计划
----------------------------------------------------------
Plan hash value: 3559141341
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 1 | 1 (0)| 00:00:01 |
|* 1 | INDEX RANGE SCAN| I_TEST | 1 | 1 | 1 (0)| 00:00:01 |
---------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - access("NAME"=‘B‘)
统计信息
----------------------------------------------------------
1 recursive calls
0 db block gets
3 consistent gets
0 physical reads
0 redo size
407 bytes sent via SQL*Net to client
400 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
可见，生成了直方图后，oracle会根据数据的实际分布情况选择合适的执行计划。
###############################################################
二、绑定变量的情况下
SQL> analyze table test compute statistics;
表已分析。
SQL> var o varchar2(10)
SQL> exec :o:=‘A‘
PL/SQL 过程已成功完成。
SQL> select * from test where name=:o;
已选择25064行。
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 8355 | 8355 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 8355 | 8355 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=:O)
统计信息
----------------------------------------------------------
1 recursive calls
0 db block gets
1717 consistent gets
0 physical reads
0 redo size
337843 bytes sent via SQL*Net to client
18770 bytes received via SQL*Net from client
1672 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
25064 rows processed
SQL> exec :o:=‘B‘
PL/SQL 过程已成功完成。
SQL> select * from test where name=:o;
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 8355 | 8355 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 8355 | 8355 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=:O)
统计信息
----------------------------------------------------------
0 recursive calls
0 db block gets
47 consistent gets
0 physical reads
0 redo size
407 bytes sent via SQL*Net to client
400 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
从以上测试可以看出，在绑定变量的情况下，如果没有分析直方图，两个查询都使用了相同的执行计划——全表扫描。
在第一次解析SQL的时候，因为oracle不知道数据的具体分布，所以它认为会返回三分之一的数据，所以选择了全表扫描。
在以后执行同样的SQL时会重用该SQL，都会使用第一次解析生成的执行计划。
在本例中，无论:o是‘A‘还是‘B‘，都会使用全表扫描，那么，我们是否可以得出这样一个结论：
如果分析了直方图，那么如果第一次硬解析SQL时:o是‘A‘时，会使用全表扫描；:o是‘B‘时，会使用索引扫描呢？看如下的测试：
SQL> alter system flush shared_pool;
系统已更改。
SQL> analyze table test delete statistics;
表已分析。
SQL> analyze table test compute statistics for table for all indexes for all indexed columns;
表已分析。
SQL> exec :o:=‘A‘
PL/SQL 过程已成功完成。
SQL> select * from test where name=:o;
已选择25064行。
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 8355 | 8355 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 8355 | 8355 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=:O)
统计信息
----------------------------------------------------------
32 recursive calls
0 db block gets
1720 consistent gets
0 physical reads
0 redo size
337843 bytes sent via SQL*Net to client
18770 bytes received via SQL*Net from client
1672 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
25064 rows processed
SQL> alter system flush shared_pool;
系统已更改。
SQL> analyze table test delete statistics;
表已分析。
SQL> analyze table test compute statistics for table for all indexes for all indexed columns;
表已分析。
SQL> exec :o:=‘B‘
PL/SQL 过程已成功完成。
SQL> select * from test where name=:o;
执行计划
----------------------------------------------------------
Plan hash value: 1357081020
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 8355 | 8355 | 14 (8)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| TEST | 8355 | 8355 | 14 (8)| 00:00:01 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("NAME"=:O)
统计信息
----------------------------------------------------------
32 recursive calls
0 db block gets
6 consistent gets
0 physical reads
0 redo size
407 bytes sent via SQL*Net to client
400 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
从这个结果可以看出，分析了直方图后，无论:o的值是‘A‘还是‘B‘，第一次执行该sql时，使用的都是全表扫描，这与刚才的推论不一致了。
如果真是这样的话，使用绑定变量对表做直方图还有什么意义呢？其实这应该算是oracl的一个bug，在这里autotrace的结果是不对的，我们可以用10046看
SQL> alter system flush shared_pool;
SQL> analyze table test delete statistics;
SQL> analyze table test compute statistics for table for all indexes for all indexed columns;
SQL> exec :o:=‘A‘
SQL> ALTER SESSION SET EVENTS ‘10046 trace name context forever, level 12‘;
SQL> select * from test where name=:o;
SQL> ALTER SESSION SET EVENTS ‘10046 trace name context off‘;
将C:\oracle\product\10.2.0\admin\orcl\udump下的最新trc文件copy至桌面
C:\Users\LEE\Desktop> tkprof orcl_ora_4516.trc orcla.sql
查看orcla.sql
select *
from
test where name=:o
call count cpu elapsed disk query current rows
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 2 0.04 0.11 0 3 0 0
Execute 2 0.00 0.02 0 0 0 0
Fetch 1674 0.14 0.14 0 1720 0 25065
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 1678 0.18 0.28 0 1723 0 25065
Misses in library cache during parse: 2
Optimizer mode: ALL_ROWS
Parsing user id: 58
Rows Row Source Operation
------- ---------------------------------------------------
25064 TABLE ACCESS FULL TEST (cr=1717 pr=0 pw=0 time=100367 us)
——————————————————
SQL> alter system flush shared_pool;
SQL> analyze table test delete statistics;
SQL> analyze table test compute statistics for table for all indexes for all indexed columns;
SQL> exec :o:=‘B‘
SQL> ALTER SESSION SET EVENTS ‘10046 trace name context forever, level 12‘;
SQL> select * from test where name=:o;
SQL> ALTER SESSION SET EVENTS ‘10046 trace name context off‘;
将C:\oracle\product\10.2.0\admin\orcl\udump下的最新trc文件copy至桌面
C:\Users\LEE\Desktop> tkprof orcl_ora_4516.trc orclb.sql
查看orclb.sql
select *
from
test where name=:o
call count cpu elapsed disk query current rows
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.00 0.04 0 0 0 0
Execute 1 0.00 0.01 0 0 0 0
Fetch 2 0.00 0.00 0 3 0 1
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 4 0.00 0.05 0 3 0 1
Misses in library cache during parse: 1
Optimizer mode: ALL_ROWS
Parsing user id: 58
Rows Row Source Operation
------- ---------------------------------------------------
1 INDEX RANGE SCAN I_TEST (cr=3 pr=0 pw=0 time=51 us)(object id 57877)
到此为止可以可以得出如下结论：
1、无论是否绑定变量，对数据分布不均的情况下柱状图都是很有效的。假如数据是均衡的，没有必要使用直方图。
2、对数据分布不均匀的情况下，使用绑定变量可能会造成恶果，就算对表做了柱状图也一样
3、使用绑定变量，sql第一次执行决定了以后同样的sql执行的执行计划
4、AUTOTRACE的信息不一定准确，必要时要用10046查看需要的信息
本文转自：http://blog.csdn.net/narutobing/article/details/7881082