Question

考虑这个样本数据：

+----------+------+--------+-----------------+
| username | qnum | qvalue | date            |
+----------+------+--------+-----------------+
| Linda    | 1    | 2      | 11/14/2017 7:25 |
+----------+------+--------+-----------------+
| Fred     | 1    | 1      | 11/23/2017 7:59 |
+----------+------+--------+-----------------+
| Brian    | 5    | 2      | 11/17/2017 7:25 |
+----------+------+--------+-----------------+
| Sandra   | 6    | 1      | 11/25/2017 7:26 |
+----------+------+--------+-----------------+
| Tom      | 6    | 1      | 11/22/2017 7:32 |
+----------+------+--------+-----------------+
| Paul     | 6    | 1      | 11/22/2017 7:36 |
+----------+------+--------+-----------------+
| Andrew   | 7    | 2      | 11/23/2017 7:37 |
+----------+------+--------+-----------------+
| Luke     | 3    | 1      | 11/23/2017 8:03 |
+----------+------+--------+-----------------+
| William  | 8    | 1      | 11/23/2017 8:03 |
+----------+------+--------+-----------------+
| Linda    | 9    | 2      | 11/15/2017 8:03 |
+----------+------+--------+-----------------+
| Brian    | 3    | 2      | 11/17/2017 8:04 |
+----------+------+--------+-----------------+
| Joan     | 9    | 1      | 11/23/2017 8:04 |
+----------+------+--------+-----------------+
| Chris    | 8    | 1      | 11/23/2017 8:04 |
+----------+------+--------+-----------------+
| Kim      | 8    | 1      | 11/15/2017 8:04 |
+----------+------+--------+-----------------+

我试图让上周获得最高q值的人。我可以使用以下SQL获取此信息，但我的问题是，如果多个用户具有最高分，那么它不会显示它们的名称，因为我使用LIMIT函数。有没有办法一起使用max和sum来获得所需的结果？期望的结果将是Linda和Brian列出的结果集，因为上周他们的总和得分为4并且并列。

SELECT username, SUM(qvalue) AS score FROM trivia_scoreboard 
WHERE `date` >= CURDATE() - INTERVAL DAYOFWEEK(CURDATE())+6 DAY AND `date` < CURDATE() - INTERVAL DAYOFWEEK(CURDATE())-1 DAY
GROUP BY username
ORDER BY score DESC
LIMIT 1

Answer 1

您必须使用一个可以获得所有人总数的查询来加入该查询。

SELECT t1.*
FROM (
    SELECT username, SUM(qvalue) AS score FROM trivia_scoreboard 
    WHERE `date` >= CURDATE() - INTERVAL DAYOFWEEK(CURDATE())+6 DAY AND `date` < CURDATE() - INTERVAL DAYOFWEEK(CURDATE())-1 DAY
    GROUP BY username
) AS t1
JOIN (
    SELECT SUM(qvalue) AS score FROM trivia_scoreboard 
    WHERE `date` >= CURDATE() - INTERVAL DAYOFWEEK(CURDATE())+6 DAY AND `date` < CURDATE() - INTERVAL DAYOFWEEK(CURDATE())-1 DAY
    GROUP BY username
    ORDER BY score DESC
    LIMIT 1
) AS t2 ON t1.score = t2.score

DEMO

Answer 2

您也可以使用having：

对此进行说明

SELECT ts.username, SUM(ts.qvalue) AS score
FROM trivia_scoreboard fs
WHERE ts.`date` >= CURDATE() - INTERVAL DAYOFWEEK(CURDATE())+6 DAY AND 
      ts.`date` < CURDATE() - INTERVAL DAYOFWEEK(CURDATE())-1 DAY
GROUP BY username
HAVING score = (SELECT SUM(ts2.qvalue) as score
                FROM trivia_scoreboard ts2
                WHERE ts2.`date` >= CURDATE() - INTERVAL DAYOFWEEK(CURDATE())+6 DAY AND 
                      ts2.`date` < CURDATE() - INTERVAL DAYOFWEEK(CURDATE())-1 DAY
                GROUP BY username
                ORDER BY score DESC
                LIMIT 1
               );

这与使用join的版本之间的区别实际上是一个品味问题。由于有关聚合如何在更大量的数据上扩展的详细信息，这可能会有更好的性能。

获得多个总和值的最大值

2 个答案: