MySQLbest/first分差查询优化

MySQL best/first score difference query optimisation

谁能帮我优化这个查询?我有下表:

cdu_user_progress:
--------------------------------------------------------------
|id    |uid     |lesson_id    |game_id    |date    |score    |
--------------------------------------------------------------

对于每个用户,我试图获得特定 game_id 和特定 lesson_id 的最佳分数和第一分数之间的差异,并根据该差异对结果进行排序('progress' 在我的查询中):

SELECT ms.uid AS id, ms.max_score - fs.first_score AS progress
FROM (
    SELECT up.uid, MAX(CASE WHEN game_id = 3 THEN score ELSE NULL END) AS max_score
    FROM cdu_user_progress up
    WHERE  (up.uid IN  ('1671', '1672', '1673', '1674', '1675', '1676', '1679', '1716', '1725',         '1726', '1937', '1964', '1996', '2062', '2065', '2066', '2085', '2086')) AND (up.lesson_id = '65') AND (up.score > '-1')
GROUP BY up.uid
) ms
LEFT JOIN (
    SELECT up.uid, up.score AS first_score 
    FROM cdu_user_progress up
    INNER JOIN (
        SELECT up.uid, MIN(CASE WHEN game_id = 3 THEN date ELSE NULL END) AS first_date
        FROM cdu_user_progress up
        WHERE  (up.uid IN  ('1671', '1672', '1673', '1674', '1675', '1676', '1679', '1716', '1725', '1726', '1937', '1964', '1996', '2062', '2065', '2066', '2085', '2086')) AND (up.lesson_id = '65') AND (up.score > '-1') 
        GROUP BY up.uid
    ) fd ON fd.uid = up.uid AND fd.first_date = up.date
) fs ON fs.uid = ms.uid
ORDER BY progress DESC

任何帮助将不胜感激!

缺少任何 EXPLAIN 输出或索引定义,我们无法提出任何建议。 (我在评论中指出,如果我们不能保证 cdu_user_progress 中的 (uid,date) 元组的唯一性,那么看起来有些连接谓词丢失了......我们有可能获得行用于不同的 lesson_id 或不大于 '-1'.

的分数

在查询文本中,紧接在 ) fs 之前,我将添加

        AND up.lesson_id = '65'
        AND up.score > '-1'
      GROUP BY up.uid

我还将 up.score 列(在 fd 视图的 SELECT 列表中)包装在聚合函数中,MIN()MAX(),为了符合 ANSI 标准(尽管当 SQL_MODE 不包含 ONLY_FULL_GROUP_BY 时 MySQL 并不要求这样做)


如果我没有定义 suitable 索引,我会考虑添加一个索引:

... ON cdu_user_progress (lesson_id, uid, score, game_id, date)

派生的 tables(具体化内联视图)有一些开销,那些派生的 tables 不会有索引(在 MySQL 5.5 和较早。)但是每个内联视图中的 GROUP BY 确保我们的行数少于 20,因此这不会成为问题。

因此,如果存在性能问题,则出现在视图查询中。同样,我们确实需要查看 EXPLAIN 的输出和索引定义,以及一些基数估计,以便提出建议。


跟进

鉴于 (uid,date) 没有唯一约束,我将在 fs 视图查询中添加这些谓词。我还会在查询中使用唯一的 table 别名(对于 cdu_user_progress 的每个引用)以使语句和 EXPLAIN 输出更易于阅读。此外,在 fd 视图中添加 GROUP BY 子句和聚合函数...我将这样编写查询:

SELECT ms.uid AS id
     , ms.max_score - fs.first_score AS progress
  FROM ( SELECT up.uid
              , MAX(CASE WHEN up.game_id = 3 THEN up.score ELSE NULL END) AS max_score
           FROM cdu_user_progress up
          WHERE up.uid IN ('1671','1672','1673','1674','1675','1676','1679','1716','1725','1726','1937','1964','1996','2062','2065','2066','2085','2086')
            AND up.lesson_id = '65'
            AND up.score > '-1'
          GROUP BY up.uid
       ) ms
  LEFT
  JOIN ( SELECT uo.uid
              , MIN(uo.score) AS first_score
           FROM ( SELECT un.uid
                       , MIN(CASE WHEN un.game_id = 3 THEN un.date ELSE NULL END) AS first_date
                    FROM cdu_user_progress un
                   WHERE un.uid IN ('1671','1672','1673','1674','1675','1676','1679','1716','1725','1726','1937','1964','1996','2062','2065','2066','2085','2086')
                     AND un.lesson_id = '65' 
                     AND un.score > '-1' 
                   GROUP BY un.uid
                ) fd
           JOIN cdu_user_progress uo
             ON uo.uid = fd.uid
            AND uo.date = fd.first_date
            AND uo.lesson_id = '65'
            AND uo.score > '-1'
          GROUP BY uo.uid
       ) fs
    ON fs.uid = ms.uid
 ORDER BY progress DESC

而且我相信这将使我在上面推荐的索引成为 suitable 用于所有对 cdu_user_progress.

的引用