oracle中约束(constraints)是如何影响查询计划的
来源:程序员人生 发布时间:2015-01-05 08:52:05 阅读次数:3994次
原文:http://www.oracle.com/technetwork/issue-archive/2009/09-may/o39asktom-096149.html
oracle中束缚(constraints)是如何影响查询计划的
通常人们认为束缚只是和数据完全性有关,没问题。但是束缚也被优化器使用来优化履行计划。
优化器会拿以下资源最为输入inputs:
1)待优化的查询
2)所有
数据库对象统计
3)系统统计,如果可以获得的话(CPU速度、单块I/O速度等等作为物理硬件的衡量尺度)
4)初始化参数
5)束缚
我常常听到人们在数据仓库/报表系统中疏忽束缚的使用,理由是:数据本身OK,并且我们做了数据清洗(如果让束缚enable,他们反而不高兴),但事实证明为了取得更好的履行计划他们的确需要数据完全性束缚。数据仓库中1个差的履行计划可能耗时几小时乃至几天才能履行完。下面我们通过实例来讲明束缚对履行计划的重要性:
1. 来看NOT NULL束缚如何影响履行计划:
code1: 创建数据隔离的表和视图
drop table t1;
drop table t2;
drop view v;
create table t1
as
select * from all_objects
where object_type in ('TABLE','VIEW');
alter table t1 modify object_type not null;
alter table t1 add constraint t1_check_otype check (object_type in('TABLE','VIEW'));
create table t2
as
select * from all_objects
where object_type in ( 'SYNONYM', 'PROCEDURE' );
alter table t2 modify object_type not null;
alter table t2 add constraint t2_check_otype
check (object_type in ('SYNONYM', 'PROCEDURE'));
create or replace view v
as
select * from t1
union all
select * from t2;
code2: 优化掉1个表
set autotrace traceonly explain
select * from v where object_type = 'TABLE';
Execution Plan
----------------------------------------------------------------------------
Plan hash value: 3982894595
-----------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-----------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 40 | 6320 | 151 (1)| 00:00:02 |
| 1 | VIEW | V | 40 | 6320 | (1)| 00:00:02 |
| 2 | UNION-ALL | | | | | |
|* 3 | TABLE ACCESS FULL | T1 | 3083 | 475K| 31 (0)| 00:00:01 |
|* 4 | FILTER | | | | | |
|* 5 | TABLE ACCESS FULL| T2 | 5 | 790 | 12 (1)| 00:00:02 |
-----------------------------------------------------------------------------
Predicate Information (identified by operation id):
-------------------------------
3 - filter("OBJECT_TYPE"='TABLE')
4 - filter(NULL IS NOT NULL)
5 - filter("OBJECT_TYPE"='TABLE')
奇怪的是:我们查的数据只在T1中存在根本不关T2的事!而且看filter4:NULL IS NOT NULL这个条件我们没有指定,是履行计划加的!
这个条件始终未FALSE。
继续,为了说明NOT NULL束缚是如何作用的再来看1个例子:
code3:
drop table t;
create table t
as
select * from all_objects;
create index t_idx on t(object_type);
exec dbms_stats.gather_table_stats( user, 'T' );
select count(*) from t;
Execution Plan
----------------------------------------
Plan hash value: 2966233522
-------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
-------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 283 (1)| 00:00:04 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| T | 68437 | 283 (1)| 00:00:04 |
-------------------------------------------------------------------
这个计划没有使用我们创建的索引。缘由是object_type列是nullable,索引其实不包括所有NULL值,所以没法根据索引键值进行
COUNT操作。如果我们告知
数据库:OBJECT_TYPE IS NOT NULL,履行计划将立马转变!
code4:
alter table t modify object_type NOT NULL;
select count(*) from t;
Execution Plan
------------------------------------------
Plan hash value: 1058879072
------------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 54 (2)| 00:00:01 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | INDEX FAST FULL SCAN| T_IDX | 68437 | 54 (2)| 00:00:01 |
------------------------------------------------------------------------
some kind of 奇异吧!
问题是加入该object_type列可以为NULL,又该如何解决?答案是我们可以创建多列索引(组合索引)固然object_type比在其中。
例如:
code5:
drop index t_idx;
create index t_idx on t (object_type, 0);
code6:
select * from t where object_type is null;
Execution Plan
-----------------------------
Plan hash value: 470836197
--------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 101 | 1 (0)| 00:00:01 |
| 1 | TABLE ACCESS BY INDEX ROWID | T | 1 | 101 | 1 (0)| 00:00:01 |
|* 2 | INDEX RANGE SCAN | T_IDX | 1 | | 1 (0)| 00:00:01 |
--------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 -
access("OBJECT_TYPE" IS NULL)
2、来看主外键束缚如何影响履行计划:
code7:
drop table emp;
drop table dept;
drop view emp_dept;
create table emp
as
select *
from scott.emp;
create table dept
as
select *
from scott.dept;
create or replace view emp_dept
as
select emp.ename, dept.dname
from emp, dept
where emp.deptno = dept.deptno;
--我们伪装EMP和DEPT两个表示大表!
begin
dbms_stats.set_table_stats
( user, 'EMP', numrows=>1000000, numblks=>100000 );
dbms_stats.set_table_stats
( user, 'DEPT', numrows=>100000, numblks=>10000 );
end;
/
code8:
SQL> select ename from emp_dept;
Execution Plan
-----------------------------
Plan hash value: 615168685
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc | Cost (%CPU) | Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1000K| 31M| | 31515 (1)| 00:06:19 |
|* 1 | HASH JOIN | | 1000K| 31M| 2448K| 31515 (1)| 00:06:19 |
| 2 | TABLE ACCESS FULL| DEPT | 100K| 1269K| | 2716 (1)| 00:00:33 |
| 3 | TABLE ACCESS FULL| EMP | 1000K| 19M| | 27151 (1)| 00:05:26 |
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 -
access("EMP"."DEPTNO"="DEPT"."DEPTNO")
此处我们只查询EMP表的ename列,DEPT表现得没啥必要。但是DEPTNO是DEPT表的主键,EMP表的外键!这样就致使EMP表中DEPTNO列是非空的。但是我们没有指明这层关系,
ORACLE自然不知道,所以我们这么做:
alter table dept add constraint dept_pk primary key(deptno);
alter table emp add constraint emp_fk_dept foreign key(deptno) references dept(deptno);
select ename from emp_dept;
Execution Plan
------------------------------
Plan hash value: 3956160932
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 50000 | 976K| 27152 (1)| 00:05:26 |
|* 1 | TABLE ACCESS FULL| EMP | 50000 | 976K| 27152 (1)| 00:05:26 |
--------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("EMP"."DEPTNO" IS NOT NULL)
起作用了!加了个filter。
3、来看1个NOT NULL和主外键束缚搭配物化视图查询是如何影响履行计划的
我常常把物化视图作为”数据仓库索引“使用,其最重要的用处是作为“preanswer”,将针对特定表的复杂和长时间运行的结果保存在1个永久表中。
说白了就是加速查询速度。
alter table emp drop constraint emp_fk_dept;
alter table dept drop constraint dept_pk;
code10:
create materialized view mv enable query rewrite
as
select dept.deptno, dept.dname, count (*) from emp, dept
where emp.deptno = dept.deptno
group by dept.deptno, dept.dname;
begin
dbms_stats.set_table_stats
( user, 'MV', numrows=>100000, numblks=>10000 );
end;
/
code11: 1个查询使用物化视图
select dept.dname, count (*) from emp, dept
where emp.deptno = dept.deptno and dept.dname = 'SALES'
group by dept.dname;
Execution Plan
------------------------------
Plan hash value: 1703036361
--------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes| Cost (%CPU) | Time |
--------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1000 | 22000 | 2716 (1)| 00:00:33 |
| 1 | SORT GROUP BY NOSORT | | 1000 | 22000 | 2716 (1)| 00:00:33 |
|* 2 | MAT_VIEW REWRITE ACCESS FULL | MV | 1000 | 22000 | 2716 (1)| 00:00:33 |
--------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - filter("MV"."DNAME"='SALES')
code12: 1个查询没有使用物化视图
SQL> select count(*) from emp;
COUNT(*)
--------
14
SQL> select * from table(dbms_xplan.display_cursor);
PLAN_TABLE_OUTPUT
-----------------------
SQL_ID g59vz2u4cu404, child number 1
-----------------------
select count(*) from emp
Plan hash value: 2083865914
-------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
-------------------------------------------------------------------
| 0 | SELECT STATEMENT | | | 27142 (100)| |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| EMP | 1000K| 27142 (1)| 00:05:26 |
-------------------------------------------------------------------
14 rows selected.
很明显,更有的履行计划应当是查询物化视图中实现计算好的数据。
为此,我们需要告知
数据库以下内容:
DEPTNO in DEPT is a primary key
DEPTNO in EMP is a foreign key
DEPTNO in EMP is a NOT NULL column
alter table dept add constraint dept_pk primary key(deptno);
alter table emp add constraint emp_fk_dept foreign key(deptno) references dept(deptno);
alter table emp modify deptno NOT NULL;
code13:
SQL> select count(*) from emp;
COUNT(*)
-------
14
SQL> select * from table(dbms_xplan.display_cursor);
PLAN_TABLE_OUTPUT
-----------------------
SQL_ID g59vz2u4cu404, child number 2
-----------------------
select count (*) from emp
Plan hash value: 1747602359
--------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | | | 2716 (100)| |
| 1 | SORT AGGREGATE | | 1 | 13 | | |
| 2 | MAT_VIEW REWRITE ACCESS FULL| MV | 100K| 1269K| 2716 (1)| 00:00:33 |
--------------------------------------------------------------------------------------
关于物化视图的查询重写特性参考:http://blog.itpub.net/28719055/viewspace⑴258720/
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