oracle性能调优学习0622

发布时间:2020-08-02 14:21:44 作者:llc018198
来源:网络 阅读:406
  1. 查询数据库中的等待事件:

    SET lines 100
    SET pages 10000
    COLUMN wait_class format a12
    COLUMN event format a30
    COLUMN total_waits format 999999
    COLUMN total_us format 999999999
    COLUMN pct_time format 99.99
    COLUMN avg_us format 999999.99
    SET echo on

    SELECT   wait_class, event, total_waits AS waits,
             ROUND (time_waited_micro / 1000) AS total_ms,
             ROUND (time_waited_micro * 100 / SUM (time_waited_micro) OVER (),
                    2
                   ) AS pct_time,
             ROUND ((time_waited_micro / total_waits) / 1000, 2) AS avg_ms
        FROM v$system_event
       WHERE wait_class <> 'Idle'
    ORDER BY time_waited_micro DESC;

2.整合时间模型和等待接口:

SELECT event,
       total_waits,
       round(time_waited_micro / 1000000) AS time_waited_secs,
       round(time_waited_micro * 100 / SUM(time_waited_micro) over(), 2) AS pct_time
  FROM (SELECT event, total_waits, time_waited_micro
          FROM v$system_event
         WHERE wait_class <> 'Idle'
        UNION
        SELECT stat_name, NULL, VALUE
          FROM v$sys_time_model
         WHERE stat_name IN ('DB CPU', 'backup cpu time'))
 ORDER BY 3 DESC

SELECT sample_seconds,
       stat_name,
       waits_per_second        waits_per_sec,
       microseconds_per_second ms_per_sec,
       pct_of_time             pct
  FROM opsg_delta_report
 WHERE microseconds_per_second > 0;
监控索引的使用情况:

WITH in_plan_objects AS
 (SELECT DISTINCT object_name FROM v$sql_plan WHERE object_owner = 'SCOTT')
SELECT table_name,
       index_name,
       CASE
         WHEN object_name IS NULL THEN
          'NO'
         ELSE
          'YES'
       END AS in_cached_plan
  FROM user_indexes
  LEFT OUTER JOIN in_plan_objects
    ON (index_name = object_name);

4.识别从绑定变量中获益的sql语句:

WITH force_matches AS
       (SELECT force_matching_signature,
               COUNT( * )  matches,
               MAX(sql_id || child_number) max_sql_child,
               DENSE_RANK() OVER (ORDER BY COUNT( * ) DESC)
                  ranking
        FROM v$sql
        WHERE force_matching_signature <> 0
          AND parsing_schema_name <> 'SYS'
        GROUP BY force_matching_signature
        HAVING COUNT( * ) > 5)
SELECT sql_id,  matches, parsing_schema_name schema, sql_text
  FROM    v$sql JOIN force_matches
    ON (sql_id || child_number = max_sql_child)
WHERE ranking <= 10
ORDER BY matches DESC;

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