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MySQL查询速度很慢

发布时间:2021-04-01 09:30:29 所属栏目:MySql教程 来源:网络整理
导读:我的表格包含以下列: gamelogs_id (auto_increment primary key)player_id (int)player_name (varchar)game_id (int)season_id (int)points (int) 该表具有以下索引 +-----------------+------------+--------------------+--------------+----------------

我的表格包含以下列:

gamelogs_id (auto_increment primary key)
player_id (int)
player_name (varchar)
game_id (int)
season_id (int)
points (int)

该表具有以下索引

+-----------------+------------+--------------------+--------------+--------------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| Table           | Non_unique | Key_name           | Seq_in_index | Column_name        | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | Index_comment |
+-----------------+------------+--------------------+--------------+--------------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| player_gamelogs |          0 | PRIMARY            |            1 | player_gamelogs_id | A         |      371330 |     NULL | NULL   |      | BTREE      |         |               |
| player_gamelogs |          1 | player_name        |            1 | player_name        | A         |        3375 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | points          |            1 | points          | A         |         506 |     NULL | NULL   | YES  | BTREE      |         ## Heading ##|               |
| player_gamelogs |          1 | game_id            |            1 | game_id            | A         |       37133 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | season             |            1 | season             | A         |          30 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | team_abbreviation  |            1 | team_abbreviation  | A         |          70 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | player_id          |            1 | game_id            | A         |       41258 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | player_id          |            2 | player_id          | A         |      371330 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | player_id          |            3 | dk_points          | A         |      371330 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | game_player_season |            1 | game_id            | A         |       41258 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | game_player_season |            2 | player_id          | A         |      371330 |     NULL | NULL   | YES  | BTREE      |         |               |
| player_gamelogs |          1 | game_player_season |            3 | season_id          | A         |      371330 |     NULL | NULL   |      | BTREE      |         |               |
+-----------------+------------+--------------------+--------------+--------------------+-----------+-------------+----------+--------+------+------------+---------+---------------+

我试图在比赛开始之前计算一个赛季和球员的积分平均值.因此,对于本赛季的第3场比赛,avg_points将是游戏1和2的平均值.游戏数量按顺序排列,使得较早的游戏比较晚的游戏少.我也可以选择使用日期字段,但我认为数字比较会更快?

我的查询如下:

SELECT game_id,player_id,player_name,(SELECT avg(points) 
          FROM player_gamelogs t2
         WHERE t2.game_id < t1.game_id 
           AND t1.player_id = t2.player_id 
           AND t1.season_id = t2.season_id) AS avg_points
  FROM player_gamelogs t1
 ORDER BY player_name,game_id;

EXPLAIN生成以下输出:

| id | select_type        | table | type | possible_keys                        | key  | key_len | ref  | rows   | Extra                                           |
+----+--------------------+-------+------+--------------------------------------+------+---------+------+--------+-------------------------------------------------+
|  1 | PRIMARY            | t1    | ALL  | NULL                                 | NULL | NULL    | NULL | 371330 | Using filesort                                  |
|  2 | DEPENDENT SUBQUERY | t2    | ALL  | game_id,game_player_season | NULL | NULL    | NULL | 371330 | Range checked for each record (index map: 0xC8) |

我不确定这是因为涉及的任务的性质还是因为我的查询效率低下.谢谢你的任何建议! 最佳答案 请考虑以下查询:

SELECT t1.season_id,t1.game_id,t1.player_id,t1.player_name,AVG(COALESCE(t2.points,0)) AS average_player_points
FROM player_gamelogs t1
        LEFT JOIN player_gamelogs t2 ON 
                t1.game_id > t2.game_id 
            AND t1.player_id = t2.player_id
            AND t1.season_id = t2.season_id 
GROUP BY
    t1.season_id,t1.player_name
ORDER BY t1.player_name,t1.game_id;

笔记:

>要以最佳方式执行,您需要一个额外的索引(season_id,game_id,player_name)
>更好的是,将播放器表从id中检索名称.对我来说,我们必须从日志表中获取播放器名称,而且如果它在索引中是必需的,这似乎是多余的.
>按已按分组列排序已分组.如果可以,请避免事后订购,因为它会产生无用的开销.正如评论中所述,这不是一种官方行为,并且假设其随时间的一致性的结果应该考虑与突然失去分类的风险.

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