sql语句的哈希值

时间:2013-04-15 09:39:17

标签: sql oracle sql-execution-plan

当我们在Oracle中执行任何sql语句时,会为该sql语句分配一个哈希值并将其存储到库缓存中。因此,稍后,如果另一个用户请求相同的查询,则Oracle会找到哈希值并执行相同的执行计划。但是,我对哈希值有一个疑问。我的意思是,如何生成哈希值?,我的意思是,Oracle服务器是使用某些算法还是只是将sql字符串转换为某个数值。

因为,我正在阅读 Pro Oracle SQL 一书,其中写有,

select * from employees where department_id = 60;

SELECT * FROM EMPLOYEES WHERE DEPARTMENT_ID = 60;

select /* a_comment */ * from employees where department_id = 60;

将返回不同的哈希值,因为当执行sql语句时,Oracle首先将字符串转换为哈希值。但是,当我尝试这个时,它会返回相同的哈希值。

SQL> select * from boats where bid=10;

no rows selected


Execution Plan
----------------------------------------------------------
Plan hash value: 2799518614

-------------------------------------------------------------------------------------
| Id  | Operation                   | Name  | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |       |     1 |    16 |     1   (0)| 00:00:01 |
|   1 |  TABLE ACCESS BY INDEX ROWID| BOATS |     1 |    16 |     1   (0)| 00:00:01 |
|*  2 |   INDEX UNIQUE SCAN         | B_PK  |     1 |       |     0   (0)| 00:00:01 |
-------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   2 - access("BID"=10)

SQL> SELECT * FROM BOATS WHERE BID=10;

no rows selected


Execution Plan
----------------------------------------------------------
Plan hash value: 2799518614

-------------------------------------------------------------------------------------
| Id  | Operation                   | Name  | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |       |     1 |    16 |     1   (0)| 00:00:01 |
|   1 |  TABLE ACCESS BY INDEX ROWID| BOATS |     1 |    16 |     1   (0)| 00:00:01 |
|*  2 |   INDEX UNIQUE SCAN         | B_PK  |     1 |       |     0   (0)| 00:00:01 |
-------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   2 - access("BID"=10)

3 个答案:

答案 0 :(得分:9)

在您的问题文本中,您似乎在描述sql_id和/或hash_value。这是SQL语句文本的哈希值,也是Oracle用于确定共享池中是否已存在特定SQL语句的哈希值。但是,您在示例中显示的是plan_hash_value,它是为SQL语句生成的计划的哈希值。两者之间可能存在多对多的关系。单个SQL语句(sql_id / hash_value)可以有多个不同的计划(plan_hash_value),并且多个不同的SQL语句可以共享相同的计划。

因此,例如,如果我编写两个不同的SQL语句来查询EMP表中的特定行,我将得到相同的plan_hash_value

SQL> set autotrace traceonly;
SQL> select * from emp where ename = 'BOB';

no rows selected


Execution Plan
----------------------------------------------------------
Plan hash value: 3956160932

--------------------------------------------------------------------------
| Id  | Operation         | Name | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |      |     1 |    39 |     3   (0)| 00:00:01 |
|*  1 |  TABLE ACCESS FULL| EMP  |     1 |    39 |     3   (0)| 00:00:01 |
--------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - filter("ENAME"='BOB')


SQL> ed
Wrote file afiedt.buf

  1* select * FROM emp WHERE ename = 'BOB'
SQL> /

no rows selected


Execution Plan
----------------------------------------------------------
Plan hash value: 3956160932

--------------------------------------------------------------------------
| Id  | Operation         | Name | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |      |     1 |    39 |     3   (0)| 00:00:01 |
|*  1 |  TABLE ACCESS FULL| EMP  |     1 |    39 |     3   (0)| 00:00:01 |
--------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - filter("ENAME"='BOB')

但是,如果我查看v$sql,我会看到生成了两个不同的sql_idhash_value

SQL> set autotrace off;
SQL> ed
Wrote file afiedt.buf

  1  select sql_id, sql_text, hash_value, plan_hash_value
  2    from v$sql
  3   where sql_text like 'select%BOB%'
  4*    and length(sql_text) < 50
SQL> /

SQL_ID        SQL_TEXT                                 HASH_VALUE PLAN_HASH_VALUE
------------- ---------------------------------------- ---------- ---------------
161v96c0v9c0n select * FROM emp WHERE ename = 'BOB'      28618772      3956160932
cvs1krtgzfr78 select * from emp where ename = 'BOB'    1610046696      3956160932

Oracle认识到这两个语句是具有不同sql_idhash_value哈希值的不同查询。但他们都碰巧生成了相同的计划,所以他们最终得到了相同的plan_hash_value

答案 1 :(得分:0)

我会说你在这种情况下证明了这本书是错的。理论上,让哈希识别概念性SQL语句而不是随机大写的字符串似乎更好......我希望在生成哈希时也会忽略注释。 ; - )

答案 2 :(得分:0)

设置行300 col BEGIN_INTERVAL_TIME为a30 从dba_hist_snapshot a,dba_hist_sqlstat b中选择a.snap_id,a.begin_interval_time,b.plan_hash_value,其中a.snap_id = b.snap_id和b.sql_id ='&amp; sql_id'顺序为1;