ORACLE柱状图与执行计划(转)

2008-11-12 13:22
本文从不绑定变量与绑定变量两种情况讨论柱状图的作用。
 
一、不绑定变量的情况：
考虑以下数据
SQL> select owner,count(1) from th group by owner;
OWNER COUNT(1)
------------------------------ ----------
SUK 1
SYS 36216
SYSTEM 1
其中，在表的OWENR上建立有一个索引
做普通分析后，执行查询
SQL> analyze table th compute statistics;
Table analyzed
suk@ORACLE9I> select * from th where owner='SYS';
已选择36216行。
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=50 Card=12073 Bytes=
881329)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=50 Card=12073 Bytes=8813
29)
 
 
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
2894 consistent gets
0 physical reads
0 redo size
2045535 bytes sent via SQL*Net to client
27057 bytes received via SQL*Net from client
2416 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
36216 rows processed
suk@ORACLE9I> select * from th where owner='SUK';
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=50 Card=12073 Bytes=
881329)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=50 Card=12073 Bytes=8813
29)
 
 
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
513 consistent gets
0 physical reads
0 redo size
1133 bytes sent via SQL*Net to client
503 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，这是不合理的执行计划。因为，它只是知道owner列有三个不同的值，但oracle不知道每个不同的owner分别有多少记录，oracle默认为这些数据的分布是完全均匀的，所以，当用owner作条件时，oracle会认为会返回总记录的三分之一（从执行计划中的Card=12073可以看出来）
对表TH生成柱状图后在做同样的查询
SQL> analyze table th compute statistics for table for all indexes for all indexed columns;
Table analyzed
suk@ORACLE9I> select * from th where owner='SYS';
已选择36216行。
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=50 Card=36216 Bytes=
2643768)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=50 Card=36216 Bytes=2643
768)
 
 
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
2894 consistent gets
0 physical reads
0 redo size
2045535 bytes sent via SQL*Net to client
27057 bytes received via SQL*Net from client
2416 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
36216 rows processed
suk@ORACLE9I> select * from th where owner='SUK';
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=2 Card=1 Bytes=73)
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TH' (Cost=2 Card=1 Bytes
=73)
2 1 INDEX (RANGE SCAN) OF 'IDX_TH' (NON-UNIQUE) (Cost=1 Card
=1)
 
 
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
4 consistent gets
0 physical reads
0 redo size
1133 bytes sent via SQL*Net to client
503 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，生成了柱状图后，oracle会根据数据的实际分布情况选择合适的执行计划，提高性能。
----------------------------------------------------------------------
二、绑定变量的情况下
下面看看在绑定变量的情况下，执行同样的操作，会发生什么事情
suk@ORACLE9I> analyze table th compute statistics;
表已分析。
suk@ORACLE9I> var o varchar2(20)suk@ORACLE9I> exec :o:='SYS'
PL/SQL 过程已成功完成。
suk@ORACLE9I> select * from th where owner=:o;
已选择32192行。
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=75 Card=10731 Bytes=
783363)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=75 Card=10731 Bytes=7833
63)
 
 
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
2886 consistent gets
0 physical reads
0 redo size
1818406 bytes sent via SQL*Net to client
24109 bytes received via SQL*Net from client
2148 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
32192 rows processed
suk@ORACLE9I> exec :o:='SUK'
PL/SQL 过程已成功完成。
suk@ORACLE9I> select * from th where owner=:o;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=75 Card=10731 Bytes=
783363)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=75 Card=10731 Bytes=7833
63)
 
 
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
770 consistent gets
0 physical reads
0 redo size
1151 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
从以上测试可以看到，在绑定变量的情况下，如果没有分析柱状图，两个查询都使用了相同的执行计划--全表扫描。这也很容易理解，在第一次解析SQL的时候，会根据:o的绑定的值去窥视表数据，因为oracle不知道连接列的数据的具体分布，所以它会以为会返回三分之一的数据，所以选择了全表扫描。在以后的执行同样的SQL时会重用该SQL，都会使用第一次解析生成的执行计划了。在本例中，由于没有做柱状图，索引第一次执行select * from th where owner=:o时，无论:0是'SYS'还是'SUK'，都会使用全表扫描。那么，我们是否可以得出这样的一个结论：如果对表做了柱状图，那么如果第一次硬解析SQL时:o的值是'SUK'时，这个sql将会使用索引扫描；如果第一次硬解析时:o的值是'SYS'时，SQL将会使用全表扫描呢？看如下的测试例子：
suk@ORACLE9I> alter system flush shared_pool;
系统已更改。
suk@ORACLE9I> analyze table th delete statistics;
表已分析。
suk@ORACLE9I> analyze table th compute statistics for table for all indexes for all indexed columns;
表已分析。
suk@ORACLE9I> exec :o:='SYS'
PL/SQL 过程已成功完成。
suk@ORACLE9I> select * from th where owner=:o;
已选择32192行。
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=75 Card=10731 Bytes=
890673)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=75 Card=10731 Bytes=8906
73)
 
 
Statistics
----------------------------------------------------------
271 recursive calls
0 db block gets
2900 consistent gets
0 physical reads
0 redo size
1818406 bytes sent via SQL*Net to client
24109 bytes received via SQL*Net from client
2148 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
32192 rows processed
suk@ORACLE9I> alter system flush shared_pool;
系统已更改。
suk@ORACLE9I> exec :o:='SUK'
PL/SQL 过程已成功完成。
suk@ORACLE9I> select * from th where owner=:o;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=75 Card=10731 Bytes=
890673)
1 0 TABLE ACCESS (FULL) OF 'TH' (Cost=75 Card=10731 Bytes=8906
73)
 
 
Statistics
----------------------------------------------------------
529 recursive calls
0 db block gets
51 consistent gets
0 physical reads
0 redo size
1151 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
7 sorts (memory)
0 sorts (disk)
1 rows processed
从这个结果可以看到，分析了柱状图后，无论:o的值是'SYS'还是'SUK'，第一次执行该sql时，使用的都是全表扫描，这与刚才我们的推论不一致了，如果真是这样的话，使用绑定变量对表做柱状图还有什么意义呢？其实这应该算是ORACLE的一个BUG，在这里AUTOTRACE的结果是不对的，我们可以用10046看
第一次执行
select *
from
th where owner=:o
当:o:='SYS'时
call count cpu elapsed disk query current rows
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.00 0.01 0 0 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 2148 0.09 0.17 0 2886 0 32192
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 2150 0.09 0.18 0 2886 0 32192
Misses in library cache during parse: 1
Optimizer goal: CHOOSE
Parsing user id: 23
Rows Row Source Operation
------- ---------------------------------------------------
32192 TABLE ACCESS FULL TH
 
第一次执行
select *
from
th where owner=:o
当:o:='SUK'时
call count cpu elapsed disk query current rows
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.01 0.01 0 31 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 2 0.00 0.00 0 4 0 1
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 4 0.01 0.01 0 35 0 1
Misses in library cache during parse: 1
Optimizer goal: CHOOSE
Parsing user id: 23
Rows Row Source Operation
------- ---------------------------------------------------
1 TABLE ACCESS BY INDEX ROWID TH
1 INDEX RANGE SCAN IDX_TH (object id 7248)
从9i开始，oracle在对sql第一次硬解析时，会对绑定的变量值进行窥视，从而根据变量值和数据的分布决定sql的执行计划。从以上的例子可以证明这点。
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
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