一次IO利用率100%，数据库大量全表扫描有关问题

2012-09-27

一次IO利用率100%，数据库大量全表扫描问题现象描述 1，具体什么业务受到影响不清楚，但从系统测看，主机IO资

一次IO利用率100%，数据库大量全表扫描问题
现象描述
1，具体什么业务受到影响不清楚，但从系统测看，主机IO资源比较紧张（HPUX 11.31 +oracle 9i）
HP-UX crmdb3 B.11.23 U ia64    09/19/12

11:09:42    %usr    %sys    %wio   %idle
11:09:45      28       5      64       3
11:09:48      28       2      61       9
11:09:51      28       2      67       3
11:09:54      33       2      57       7
11:09:57      31       2      59       7

glance看IO已接近100%

2，数据库侧看，大量db file scattered read IO相关等待事件

SQL>  select OPTIMIZER_MODE from  v$sqlarea where HASH_VALUE='125827763';OPTIMIZER_MODE---------------------------------------RULETable                   Number                 Empty Average    Chain Average Global User           Sample DateName                   of Rows   Blocks       Blocks   Space    Count Row Len Stats  Stats            Size MM-DD-YYYY--------------- -------------- -------- ------------ ------- -------- ------- ------ ------ -------------- ----------RM_A_X                                                                    NO     NOSQL>  select CREATED,OBJECT_NAME from  dba_objects where object_name ='RM_A_X' and owner='ai';CREATED                                                                                              OBJECT_NAME---------------------------------------------------------------- --------------------2012-06-26 09:52:48                                                                                  RM_A_XSQL> show parameter optimizer_modeNAME                                 TYPE                              VALUE------------------------------------ --------------------------------- ------------------------------optimizer_mode                       string                            RULE

3，rbo的执行顺序如下，安装如下文章所述，IN应该属于rank 10,如果把in改成单条件=则直接走相关索引

sing the RBO, the optimizer chooses an execution plan based on the access paths available and the ranks of these access paths. Oracle's ranking of the access paths is heuristic. If there is more than one way to execute a SQL statement, then the RBO always uses the operation with the lower rank. Usually, operations of lower rank execute faster than those associated with constructs of higher rank.
The list shows access paths and their ranking:
RBO Path 1: Single Row by Rowid
RBO Path 2: Single Row by Cluster Join
RBO Path 3: Single Row by Hash Cluster Key with Unique or Primary Key
RBO Path 4: Single Row by Unique or Primary Key
RBO Path 5: Clustered Join
RBO Path 6: Hash Cluster Key
RBO Path 7: Indexed Cluster Key
RBO Path 8: Composite Index
RBO Path 9: Single-Column Indexes
RBO Path 10: Bounded Range Search on Indexed Columns
RBO Path 11: Unbounded Range Search on Indexed Columns
RBO Path 12: Sort Merge Join
RBO Path 13: MAX or MIN of Indexed Column
RBO Path 14: ORDER BY on Indexed Column
RBO Path 15: Full Table Scan

解决方案：

由于问题比较紧急，需要尽快解决，因此使用了最简单有效的解决办法用hint，使用此方法修改sql后问题很快解决。
SQL> explain plan for
2 select /*+index(a RM_A_M1) */ a.serial_num, x.c_attr_1 crypto, to_char(a.expirt_dt, 'yyyy-mm-dd') expirt_dt, k.key_info     from ai.RM_A_key_info k ,ai.RM_A_x x ,ai.RM_A
3   a where a.row_id = x.row_id (+)
4 and k.row_id(+) = x.n_attr_1        and serial_num in ( '12475014246300079', '12475014246300099');

Explained.

SQL> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------------------

----------------------------------------------------
| Id | Operation                      | Name                 | Rows | Bytes | Cost |
----------------------------------------------------
|   0 | SELECT STATEMENT               |                       |     2 | 6204 |     9 |
|   1 | NESTED LOOPS OUTER            |                       |     2 | 6204 |     9 |
|   2 |   NESTED LOOPS OUTER           |                       |     2 | 4126 |     7 |
|   3 |    INLIST ITERATOR             |                       |       |       |       |
|   4 |     TABLE ACCESS BY INDEX ROWID| RM_A              |     2 |    70 |     5 |
|* 5 |      INDEX RANGE SCAN          | RM_A_M1           |     2 |       |     4 |
|   6 |    TABLE ACCESS BY INDEX ROWID | RM_A_X            |     1 | 2028 |     1 |

PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------------------
|* 7 |     INDEX UNIQUE SCAN          | RM_A_X_P1         |     1 |       |       |
|   8 |   TABLE ACCESS BY INDEX ROWID | RM_A_KEY_INFO     |     1 | 1039 |     1 |
|* 9 |    INDEX UNIQUE SCAN           | RM_A_KEY_INFO_P1 |     1 |       |       |
----------------------------------------------------

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

   5 - access("A"."SERIAL_NUM"='12475014246300079' OR
              "A"."SERIAL_NUM"='12475014246300099')
   7 - access("A"."ROW_ID"="X"."ROW_ID"(+))

PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------------------
   9 - access("K"."ROW_ID"(+)="X"."N_ATTR_1")

Note: cpu costing is off

25 rows selected.

参考资料

http://docs.oracle.com/cd/B10501_01/server.920/a96533/rbo.htm#39232

热点排行

其他数据库

一次IO利用率100%，数据库大量全表扫描有关问题