LeetCode 534. 游戏玩法分析 III

Question

下面是leetcode题

桌子：活动

列名	类型
player_id	int
device_id	int
事件日期	约会
玩过的游戏	int

(player_id, event_date) 是这个表的主键。 此表显示了某些游戏的玩家活动。 每一行都是一个玩家的记录，该玩家在某天使用某种设备注销之前登录并玩了很多游戏（可能是 0）。

编写一个 SQL 查询来报告每个玩家和日期，玩家到目前为止玩了多少场比赛。也就是说，玩家在该日期之前玩的游戏总数。检查示例是否清晰。

按任意顺序返回结果表

查询结果格式如下例

Player_Id	Device_ID	活动日期	Games_Played
1	2	2016-03-01	5
1	2	2016-03-02	6
1	3	2017-06-25	1
3	1	2016-03-02	0
3	4	2018-07-03	5

我已经用Windows函数解决了。我的代码如下

select player_id,event_date,sum(games_played) over(partition by player_id order by event_date) as games_played_so_far from activity

和输出

player_id	事件日期	games_played_So_Far
1	2016-03-01	5
1	2016-05-02	11
1	2017-06-25	12
3	2016-03-02	0
3	2018-07-03	5

但是在使用 JOIN 解决它时，我无法理解，为什么我们需要对 a2.games_played 而不是 a1.games_played 求和。代码如下

SELECT a1.player_id, a1.event_date ,SUM(a2.games_played) AS games_played_so_far
FROM activity a1, activity a2
WHERE a1.player_id = a2.player_id
AND a1.event_date >=a2.event_date
GROUP BY a1.player_id, a1.event_date
ORDER BY a1.player_id, a1.event_date;

并且，我编写了以下代码并得到了以下结果。如果看到下面的输出，则 a1_played 看起来对齐，而 a2_Played 仅包含 0 和 5 值。我不明白为什么我们要总结 a2_played

SELECT a1.player_id, a1.event_date as a1_Date, a2.event_date as a2_Date,a1.games_played as a1played,a2.games_played as a2played,
SUM(a1.games_played) AS sum_a1,SUM(a2.games_played) AS sum_a2
FROM activity a1, activity a2
WHERE a1.player_id = a2.player_id
AND a1.event_date >=a2.event_date
GROUP BY a1.player_id, a1.event_date
ORDER BY a1.player_id, a1.event_date;

Player_Id	a1_日期	a2_日期	a1_Played	a2_Played	sum_a1_Played	Sum_a2_Played
1	2016-03-01	2016-03-01	5	5	5	5
1	2016-05-02	2016-03-01	6	5	12	11
1	2017-06-25	2016-03-01	1	5	3	12
3	2016-03-02	2016-03-02	0	0	0	0
3	2018-07-03	2016-03-02	5	0	10	5

Answer 1

如果您使用您父亲的 SQL 而不是您祖父的 SQL 重写您的查询，即使用显式 JOIN，答案将跃然纸上。 （很抱歉，我很尖刻，但我个人在 1994 年转向了显式 JOIN，并且再也没有回头。）

SELECT a1.player_id, a1.event_date, SUM(a2.games_played) AS games_so_far
  FROM activity a1
  JOIN activity a2     ON a1.player_id = a2.player_id
                      AND a1.event_date >=a2.event_date
 GROUP BY a1.player_id, a1.event_date
 ORDER BY a1.player_id, a1.event_date

ON 子句第二部分中的

>=

不等式可以解决问题。来自 a1 的每个连续的

event_date

连接到不断增加的行数。然后他们得到总结。

Answer 2

select a.*, sum(a.games_played) over(partition by player_id order by a.event_date rows between unbounded preceding and current row) as games_played
 from Activity a;

在 player_id 的窗口上使用窗口函数和滚动求和