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

SQL> select * from th where owner='SYS';

已选择36216行。

[@more@]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=881329)

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

SQL> 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=881329)

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

SQL> 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=2643768)

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

SQL> 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

SQL> analyze table th compute statistics;

表已分析。

SQL> var o varchar2(20)SQL> exec :o:='SYS'

PL/SQL 过程已成功完成。

SQL> 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=783363)

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

SQL> exec :o:='SUK'

PL/SQL 过程已成功完成。

SQL> 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=783363)

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:

SQL> alter system flush shared_pool;

系统已更改。

SQL> analyze table th delete statistics;

表已分析。

SQL> analyze table th compute statistics for table for all indexes for all indexed columns;

表已分析。

SQL> exec :o:='SYS'

PL/SQL 过程已成功完成。

SQL> 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=890673)

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

SQL> alter system flush shared_pool;

系统已更改。

SQL> exec :o:='SUK'

PL/SQL 过程已成功完成。

SQL> 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=890673)

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

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