繁体   English   中英

以最少的时间总和获得最高的用户分数

[英]Get max user scores with min time sums

我有看起来像这样的MySql表

scores
game_id | level | score | time
--------+-------+-------+-----
1       | 1     | 1     | 10
1       | 2     | 0     | 10
1       | 3     | 1     | 20
2       | 1     | 1     | 5
2       | 2     | 1     | 15
2       | 3     | 0     | 10
3       | 1     | 0     | 10
3       | 2     | 0     | 10
3       | 3     | 0     | 10

games
game_id | user_id
--------+--------
1       | 1
2       | 1
3       | 2

而且我需要查询它,以便它总结每场比赛的时间点和时间,所以看起来像这样

game_id | user_id | sumPoints | sumTime
--------+---------+-----------+--------
1       | 1       | 2         | 40
2       | 1       | 2         | 30
3       | 2       | 0         | 30

并且比我需要为每个用户获得最佳分数(最佳分数是指max(sumPoints)和min(sumTime)的位置,因此它看起来像这样:

game_id | user_id | sumPoints | sumTime
--------+---------+-----------+--------
2       | 1       | 2         | 30
3       | 2       | 0         | 30

那就是最终结果。 我可以做点和时间相加的部分,但是我不知道如何获得每个用户的最高分。 我想知道如何通过SQL查询做到这一点? 提前致谢

下面的查询将为您提供部分结果。

SELECT  a.game_ID, 
        b.user_id,
        SUM(a.Score) sumPoint, 
        SUM(a.time) sumTime
FROM    scores a
        INNER JOIN games b
            ON a.game_ID = b.game_ID
GROUP   BY a.game_ID, b.user_id

OUTPUT

╔═════════╦═════════╦══════════╦═════════╗
║ GAME_ID ║ USER_ID ║ SUMPOINT ║ SUMTIME ║
╠═════════╬═════════╬══════════╬═════════╣
║       1 ║       1 ║        2 ║      40 ║
║       2 ║       1 ║        2 ║      30 ║
║       3 ║       2 ║        0 ║      30 ║
╚═════════╩═════════╩══════════╩═════════╝

由于MySQL不像其他RDBMS一样支持Windows功能,因此可能难以获得所需的结果。

SELECT  a.*
FROM
        (
            SELECT  a.game_ID, 
                    b.user_id,
                    SUM(a.Score) sumPoint, 
                    SUM(a.time) sumTime
            FROM    scores a
                    INNER JOIN games b
                        ON a.game_ID = b.game_ID
            GROUP   BY a.game_ID, b.user_id
        ) a
        INNER JOIN
        (
            SELECT  user_ID,
                    MIN(sumTime) sumTime
            FROM
                    (
                        SELECT  a.game_ID, 
                                b.user_id,
                                SUM(a.Score) sumPoint, 
                                SUM(a.time) sumTime
                        FROM    scores a
                                INNER JOIN games b
                                    ON a.game_ID = b.game_ID
                        GROUP   BY a.game_ID, b.user_id
                    ) s
            GROUP   BY user_ID
        ) b ON  a.user_id = b.user_id AND
                a.sumTime = b.sumTime

OUTPUT

╔═════════╦═════════╦══════════╦═════════╗
║ GAME_ID ║ USER_ID ║ SUMPOINT ║ SUMTIME ║
╠═════════╬═════════╬══════════╬═════════╣
║       2 ║       1 ║        2 ║      30 ║
║       3 ║       2 ║        0 ║      30 ║
╚═════════╩═════════╩══════════╩═════════╝

JW的答案与我的基本相同。 我只是认为,如此早地吸引用户似乎不必要地使事情复杂化。 怎么样...

SELECT x.*
     , u.user_id
  FROM 
     ( SELECT game_id
            , SUM(score) ttl_score
            , SUM(time) ttl_time 
         FROM scores 
        GROUP 
           BY game_id
     ) x
  JOIN
     ( SELECT ttl_score
            , MIN(ttl_time) min_ttl_time
         FROM
            ( SELECT game_id
                   , SUM(score) ttl_score
                   , SUM(time) ttl_time 
                FROM scores 
               GROUP 
                  BY game_id 
            ) a
        GROUP
           BY ttl_score
     ) y
    ON y.ttl_score = x.ttl_score
   AND y.min_ttl_time = x.ttl_time
  JOIN games g
    ON g.game_id = x.game_id;

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM