Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id,event_date)是此表的主键。
这张表显示了某些游戏的玩家的活动情况。
每一行是一个玩家的记录,他在某一天使用某个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,同时报告每组玩家和日期,以及玩家到目前为止玩了多少游戏。也就是说,在此日期之前玩家所玩的游戏总数。详细情况请查看示例。
查询结果格式如下所示:
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 1 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+------------+---------------------+
| player_id | event_date | 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 |
+-----------+------------+---------------------+
对于 ID 为 1 的玩家,2016-05-02 共玩了 5+6=11 个游戏,2017-06-25 共玩了 5+6+1=12 个游戏。
对于 ID 为 3 的玩家,2018-07-03 共玩了 0+5=5 个游戏。
请注意,对于每个玩家,我们只关心玩家的登录日期。
drop table if EXISTS Activity;
Create table If Not Exists Activity
(player_id int,
device_id int,
event_date date,
games_played int,
primary key (player_id, event_date));
insert into Activity values (1, 2, '2016-03-01', 5);
insert into Activity values (1, 2, '2016-05-02', 6);
insert into Activity values (2, 3, '2017-06-25', 1);
insert into Activity values (3, 1, '2016-03-02', 0);
insert into Activity values (3, 4, '2018-07-03', 5);
开窗
select player_id,event_date ,
sum(games_played) over(partition by player_id order by event_date )games_played_so_far
from Activity
它默认就是 rows between unbounded preceding and current now
自连接【实践】
select
a1.player_id,
a1.event_date,
sum(a2.games_played) games_played_so_far
from Activity a1,Activity a2
where a1.player_id=a2.player_id and //这里就是笛卡尔积了
a1.event_date>=a2.event_date //看似只是一个简单的日期比较缺筛选出了1,2,3然后进行sum即可
group by 1,2; // 个人觉得group by 日期,但是其它的就取不出来了,所以1,2有啥用