LeetCode MySQL 1308. 不同性别每日分数总计(累加/变量/窗口函数)

文章目录

1. 题目

表: Scores

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| player_name   | varchar |
| gender        | varchar |
| day           | date    |
| score_points  | int     |
+---------------+---------+
(gender, day)是该表的主键

一场比赛是在女队和男队之间举行的
该表的每一行表示一个名叫 (player_name) 性别为 (gender) 的参赛者在某一天获得了 (score_points) 的分数
如果参赛者是女性,那么 gender 列为 ‘F’,
如果参赛者是男性,那么 gender 列为 ‘M’

写一条SQL语句查询每种性别在每一天的总分,并按性别和日期对查询结果排序

下面是查询结果格式的例子:

Scores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           |
+-------------+--------+------------+--------------+
结果表:
+--------+------------+-------+
| 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    |
+--------+------------+-------+
女性队伍:
第一天是 2019-12-30,Priyanka 获得 17 分,队伍的总分是 17 分
第二天是 2019-12-31, Priya 获得 23 分,队伍的总分是 40 分
第三天是 2020-01-01, Aron 获得 17 分,队伍的总分是 57 分
第四天是 2020-01-07, Alice 获得 23 分,队伍的总分是 80 分
男性队伍:
第一天是 2019-12-18, Jose 获得 2 分,队伍的总分是 2 分
第二天是 2019-12-25, Khali 获得 11 分,队伍的总分是 13 分
第三天是 2019-12-30, Slaman 获得 13 分,队伍的总分是 26 分
第四天是 2019-12-31, Joe 获得 3 分,队伍的总分是 29 分
第五天是 2020-01-07, Bajrang 获得 7 分,队伍的总分是 36

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/running-total-for-different-genders
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

2. 解题

类似题目:LeetCode MySQL 1204. 最后一个能进入电梯的人(累加/变量/窗口函数)

# Write your MySQL query statement below
select s1.gender, s1.day, sum(s2.score_points) total
from Scores s1, Scores s2
where s1.gender = s2.gender and s1.day >= s2.day
group by s1.day, s1.gender
order by s1.gender, s1.day

1104 ms

  • 变量法
# Write your MySQL query statement below
select gender, day, 
        case gender when 'F'
            then @prePointF := @prePointF + score_points
            else @prePointM := @prePointM + score_points
        end as total
from Scores, (select @prePointF := 0, @prePointM := 0) temp
order by gender, day

466 ms

  • 窗口函数
# Write your MySQL query statement below
select gender, day, 
        sum(score_points) over (partition by gender order by day) total
from Scores
order by gender, day

549 ms


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转载自blog.csdn.net/qq_21201267/article/details/107452243