sql_id和hash value的部分转换

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标题：sql_id和hash value的部分转换

作者：惜分飞©版权所有[文章允许转载,但必须以链接方式注明源地址,否则追究法律责任.]

从oracle 10g开始引进了sql_id，在老版本的oralce中,要表明一条sql,一般使用hash value，而在10g及其以后版本中一般建议使用sql_id,从9i的sp和10g的awr中也可以看出.对于Library Cache对象,Oracle使用MD5算法进行哈希,生成一个128位的Hash Value,其中低32位作为HASH VALUE显示,SQL_ID则取了后64位.既然hash value和sql_id之前存在着这样的关系,那么我们就可以通过函数实现两者的部分转换(因为最终取值长度不同,所以不能完全转换)
1.查询sql_id和hash value

SQL> select * from v$version;

BANNER
----------------------------------------------------------------
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
PL/SQL Release 10.2.0.4.0 - Production
CORE    10.2.0.4.0      Production
TNS for Linux: Version 10.2.0.4.0 - Production
NLSRTL Version 10.2.0.4.0 - Production

SQL> select to_char(sysdate,'yyyy-mm-dd hh24:mi:ss')
  2  "www.orasos.com" from dual;

www.orasos.com
-------------------
2012-05-26 01:05:39

SQL> select sql_id,hash_value from v$sql where sql_text like
  2  'select * from dual';

SQL_ID        HASH_VALUE
------------- ----------
a5ks9fhw2v9s1  942515969

2.oracle自带函数转换sql_id to hash value

SQL> select dbms_utility.SQLID_TO_SQLHASH('a5ks9fhw2v9s1') hash_value FROM DUAL;


HASH_VALUE
----------
 942515969

3.自己编写函数sql_id to hash value

SQL> CREATE OR REPLACE FUNCTION sql_id_2_hash_value (sql_id VARCHAR2)
  2     RETURN NUMBER
  3  IS
  4     l_output   NUMBER := 0;
  5  BEGIN
  6         SELECT TRUNC (
  7                   MOD (
  8                      SUM (
  9                         (INSTR ('0123456789abcdfghjkmnpqrstuvwxyz',
 10                                 SUBSTR (LOWER (TRIM (sql_id)), LEVEL, 1))
 11                          - 1)
 12                         * POWER (32, LENGTH (TRIM (sql_id)) - LEVEL)),
 13                      POWER (2, 32)))
 14           INTO l_output
 15           FROM DUAL
 16     CONNECT BY LEVEL <= LENGTH (TRIM (sql_id));
 17     RETURN l_output;
 18  END;
 19  /

函数已创建。

SQL> select sql_id_2_hash_value('a5ks9fhw2v9s1') hash_value FROM DUAL;

HASH_VALUE
----------
 942515969

4.hash value 转换为部分 sql_id

SQL> CREATE OR REPLACE FUNCTION hash_value_2_sql_id (p_hash_value NUMBER)
  2     RETURN VARCHAR2
  3  IS
  4     l_output   VARCHAR2 (8) := '';
  5  BEGIN
  6     FOR i
  7        IN (    SELECT SUBSTR (
  8                          '0123456789abcdfghjkmnpqrstuvwxyz',
  9                          1
 10                          + FLOOR (
 11                               MOD (p_hash_value / (POWER (32, LEVEL - 1)), 32)),
 12                          1)
 13                          sqlidchar
 14                  FROM DUAL
 15            CONNECT BY LEVEL <= LN (p_hash_value) / LN (32)
 16              ORDER BY LEVEL DESC)
 17     LOOP
 18        l_output := l_output || i.sqlidchar;
 19     END LOOP;
 20
 21     RETURN l_output;
 22  END;
 23  /

函数已创建。

SQL> select hash_value_2_sql_id(942515969) from dual;

HASH_VALUE_2_SQL_ID(942515969)
--------------------------------------------------------
2v9s1

参考:http://blog.tanelpoder.com/2009/02/22/sql_id-is-just-a-fancy-representation-of-hash-value/