使用聚合函数的结果更新mySQL表

时间:2015-07-16 18:41:18

标签: mysql sql join sql-update

我有一个聚合表和一个详细信息表,其中包含许多不同游戏的分数。详细信息表是数百万条记录,聚合表只包含每个游戏以及不同天数的平均分数。

例:
汇总表:

| ID | Name  | AVG_SCORE_7_DAYS | AVG_SCORE_30_DAYS |
-----------------------------------------------------
| 1  | Game 1| 10.3             | 20.3              |
| 2  | Game 2| 14.3             | 26.3              |

详细表:

| ID | Name  | Date                | Score |
--------------------------------------------
| 1  | Game 1| 2015-07-12 01:00:00 | 20    |
| 2  | Game 2| 2015-07-12 01:00:00 | 26    |
| 3  | Game 1| 2015-07-12 01:00:00 | 14    |
| 4  | Game 2| 2015-07-12 01:00:00 | 9     |

我使用以下代码通过存储过程每晚更新我的聚合表:

UPDATE `aggregate` aggr
INNER JOIN
(
    SELECT game_name, avg(score) AS avg_score
    FROM `detail`
    WHERE `date` BETWEEN (CURDATE() + INTERVAL -7 DAY) AND CURDATE()
    GROUP BY `game_name`
) detail7 ON aggr.`game_name` = detail7.`game_name`
INNER JOIN
(
    SELECT game_name, avg(score) AS avg_score
    FROM `detail`
    WHERE `date` BETWEEN (CURDATE() + INTERVAL -30 DAY) AND CURDATE()
    GROUP BY `game_name`
) detail30 ON aggr.`game_name` = detail30.`game_name`

我遇到的问题是,如果7天,30天等某些游戏没有得分,则这些子查询不会返回任何记录,因此如果其中任何一个失败,那么该游戏的任何列都不会得到更新(由于内部联接)。有没有办法可以编写我的查询来更新其他列,即使子查询的结果没有返回任何结果?

1 个答案:

答案 0 :(得分:2)

使用外部联接。您还可以简化逻辑,因此只需要一个聚合:

UPDATE `aggregate` aggr LEFT JOIN
        (SELECT game_name,
                avg(case when `date` BETWEEN (CURDATE() + INTERVAL -7 DAY) AND CURDATE() then score end) AS avg_score_07,
                avg(case when `date` BETWEEN (CURDATE() + INTERVAL -30 DAY) AND CURDATE() then score end) AS avg_score_30
         FROM `detail`
         WHERE `date` BETWEEN (CURDATE() + INTERVAL -30 DAY) AND CURDATE()
         GROUP BY `game_name`
        ) detail
        ON aggr.`game_name` = detail.`game_name`
    SET . . . ;