0

我发现这个问题非常相似，但我仍然遇到一些麻烦。

所以我从名为 Scores 的表开始

id | player | time | scoreA | scoreB |
~~~|~~~~~~~~|~~~~~~|~~~~~~~~|~~~~~~~~|
 1 | John   |  10  |   70   |   80   |
 2 | Bob    |  22  |   75   |   85   |
 3 | John   |  52  |   55   |   75   |
 4 | Ted    |  39  |   60   |   90   |
 5 | John   |  35  |   90   |   90   |
 6 | Bob    |  27  |   65   |   85   |
 7 | John   |  33  |   60   |   80   |

我想选择每个球员的最佳平均得分以及该记录中的信息。澄清一下，最佳平均分数将是 (scoreA + scoreB)/2 的最高值。

结果看起来像这样

id | player | time | scoreA | scoreB | avg_score |
~~~|~~~~~~~~|~~~~~~|~~~~~~~~|~~~~~~~~|~~~~~~~~~~~|
 5 | John   |  35  |   90   |   90   |    90     |
 2 | Bob    |  22  |   75   |   85   |    80     |
 4 | Ted    |  39  |   60   |   90   |    75     |

根据我上面链接的问题，我尝试了这样的查询，

SELECT
   s.*,
   avg_score
FROM
   Scores AS s 
INNER JOIN (
   SELECT
      MAX((scoreA + scoreB)/2) AS avg_score,
      player,
      id
   FROM
      Scores
   GROUP BY
      player
) AS avg_s ON s.id = avg_s.id
ORDER BY
   avg_score DESC,
   s.time ASC

这实际上给我的是，

id | player | time | scoreA | scoreB | avg_score |
~~~|~~~~~~~~|~~~~~~|~~~~~~~~|~~~~~~~~|~~~~~~~~~~~|
 1 | John   |  10  |   70   |   80   |    90     |
 2 | Bob    |  22  |   75   |   85   |    80     |
 4 | Ted    |  39  |   60   |   90   |    75     |

如您所见，它从记录 5 中获得了正确的最大 avg_score，但从另一条记录（记录 1）中获取了其余信息。我错过了什么？如何确保数据全部来自同一记录？我得到了正确的 avg_score 但我想要与该记录关联的其余数据，在本例中为记录 5。

提前致谢！

4

2 回答 2

3

尝试

SELECT s.*,
       q.avg_score
  FROM scores s JOIN
(
  SELECT player, 
         MAX((scoreA + scoreB)/2) AS avg_score
    FROM scores
   GROUP BY player
) q ON s.player = q.player 
   AND (s.scoreA + s.scoreB)/2 = q.avg_score
 ORDER BY q.avg_score DESC, s.time ASC

样本输出：

| 身份证 | 播放器 | 时间 | 斯科雷亚 | 得分 | 平均分数 |
-------------------------------------------------- --
| 5 | 约翰 | 35 | 90 | 90 | 90 |
| 2 | 鲍勃 | 22 | 75 | 85 | 80 |
| 4 | 特德 | 39 | 60 | 90 | 75 |

这是SQLFiddle演示

于 2013-07-10T22:41:13.157 回答
3 
 SELECT x.*
      , (scoreA+scoreB)/2 avg_score 
   FROM scores x 
   JOIN
      ( SELECT player, MAX((scoreA+scoreB)/2) max_avg_score FROM scores GROUP BY player) y 
     ON y.player = x.player 
    AND y.max_avg_score = (scoreA+x.scoreB)/2;
于 2013-07-10T22:49:02.143 回答