1

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

我需要查询它，所以它总结了每场比赛的积分和时间，所以它看起来像这样

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

而且我需要为每个用户获得最好的分数（最好的分数我的意思是 max(sumPoints) 和 min(sumTime)，所以它看起来像这样：

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

这就是最终的结果。我可以做点和时间相加的部分，但我不知道如何获得每个用户的最高分。我想知道，我怎样才能通过 SQL 查询来做到这一点？提前致谢

4

2 回答 2

1

下面的这个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

输出

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

由于MySQL不像任何其他 RDBMS 那样支持 Windows 功能，因此产生您想要的结果可能会很麻烦。

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

输出

╔═════════╦═════════╦══════════╦═════════╗
║ GAME_ID ║ USER_ID ║ SUMPOINT ║ SUMTIME ║
╠═════════╬═════════╬══════════╬═════════╣
║       2 ║       1 ║        2 ║      30 ║
║       3 ║       2 ║        0 ║      30 ║
╚═════════╩═════════╩══════════╩═════════╝
于 2013-05-02T00:27:30.630 回答
0

JW 的回答与我的基本相同。我只是认为这么早让用户加入似乎会使事情变得不必要地复杂化。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;
于 2013-05-02T11:40:45.937 回答