MySQL：总和（案例时（...））+ 分组。为什么即使使用 GROUP BY 也无法显示数据集的详细信息？

Question

Credit: Leetcode 1308. Running Total for different Genders

样本分数表：

+-------------+--------+------------+--------------+
| player_name | gender | day        | score_points |
+-------------+--------+------------+--------------+
| Aron        | F      | 2020-01-01 | 17           |
| Alice       | F      | 2020-01-07 | 23           |
| Bajrang     | M      | 2020-01-07 | 7            |
| Khali       | M      | 2019-12-25 | 11           |
| Slaman      | M      | 2019-12-30 | 13           |
| Joe         | M      | 2019-12-31 | 3            |
| Jose        | M      | 2019-12-18 | 2            |
| Priya       | F      | 2019-12-31 | 23           |
| Priyanka    | F      | 2019-12-30 | 17           |
+-------------+--------+------------+--------------+

要求：编写一个 SQL 查询来查找每个性别每天的（累积）总分。 按性别和日期对结果表进行排序。示例结果表如下：

+--------+------------+-------+
| gender | day        | total |
+--------+------------+-------+
| F      | 2019-12-30 | 17    |
| F      | 2019-12-31 | 40    |
| F      | 2020-01-01 | 57    |
| F      | 2020-01-07 | 80    |
| M      | 2019-12-18 | 2     |
| M      | 2019-12-25 | 13    |
| M      | 2019-12-30 | 26    |
| M      | 2019-12-31 | 29    |
| M      | 2020-01-07 | 36    |
+--------+------------+-------+

我的代码如下。我不明白为什么它只返回一行[{"headers": ["gender", "day", "total"], "values": [["F", "2019-12-30", 184] ]}]，即使使用 GROUP BY。

SELECT s1.gender, s1.day, 
SUM(CASE WHEN s1.day < s2.day AND s1.gender = s2.gender THEN s1.score_points ELSE 0 END) AS total
FROM Scores s1, Scores s2
GROUP BY s1.gender AND s1.day
ORDER BY s1.gender AND s1.day

如果有人可以帮助我，非常感谢！

Answer 1

您应该使用ON 子句进行正确连接，并为GROUP BY 和ORDER BY 使用有效语法：

select s1.gender, s1.day, sum(s2.score_points) total
from scores s1 inner join scores s2
on s2.gender = s1.gender and s2.day <= s1.day
group by s1.gender, s1.day
order by s1.gender, s1.day

请参阅demo。
结果：

| gender | day        | total |
| ------ | -----------| ----- |
| F      | 2019-12-30 | 17    |
| F      | 2019-12-31 | 40    |
| F      | 2020-01-01 | 57    |
| F      | 2020-01-07 | 80    |
| M      | 2019-12-18 | 2     |
| M      | 2019-12-25 | 13    |
| M      | 2019-12-30 | 26    |
| M      | 2019-12-31 | 29    |
| M      | 2020-01-07 | 36    |

Answer 2

我不确定这是否有效，但它会通过：

SELECT T1.gender, T1.day, SUM(T2.score_points) AS total
FROM Scores T1
JOIN Scores T2
WHERE T1.gender = T2.gender AND T2.day <= T1.day
GROUP BY T1.gender, t1.day
ORDER BY t1.gender, t1.day;

---

参考文献

• 有关其他详细信息，您可以查看Discussion Board。有很多公认的解决方案，有各种languages 和解释、高效的算法，以及渐近的time/space 复杂性分析^{1, 2}。

Answer 3

AND 是返回真/假 (1/0) 的布尔运算。您似乎想要条件聚合：

SELECT s.day, 
       SUM(CASE WHEN s.gender = 'M' THEN s.score_points ELSE 0 END) AS total_male,
       SUM(CASE WHEN s.gender = 'F' THEN s.score_points ELSE 0 END) AS total_female
FROM Scores s
GROUP BY s.day
ORDER BY s.day;

GROUP BY 键确定结果集的外观。每组唯一值在结果集中都有一行，其余列汇总以适合该行。

在这种情况下，您需要每天一行，所以这应该是 GROUP BY 中的内容。

那么对于累计量使用窗口函数：

SELECT s.day, 
       SUM(CASE WHEN s.gender = 'M' THEN s.score_points ELSE 0 END) AS total_male,
       SUM(CASE WHEN s.gender = 'F' THEN s.score_points ELSE 0 END) AS total_female,
       SUM(SUM(CASE WHEN s.gender = 'M' THEN s.score_points ELSE 0 END)) OVER (ORDER BY s.day) AS running_male,
       SUM(CASE WHEN s.gender = 'F' THEN s.score_points ELSE 0 END) OVER (ORDER BY s.day) AS running_female
FROM Scores s
GROUP BY s.day
ORDER BY s.day;