在刷leetcode534. 游戏玩法分析 III 时再解完题后发现还有另一种解法，用到的关键字是 sum over partition by 函数
# method 1
SELECT a1.player_id, a1.event_date, sum(a2.games_played) AS games_played_so_far
FROM Activity a1
LEFT 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 ASC
# method 2
SELECT player_id, event_date, sum(games_played) OVER (PARTITION BY player_id ORDER BY event_date) AS games_played_so_far
FROM Activity
ORDER BY player_id, event_date DESC
0、select * from wmg_test; ---测试数据
1、select v1,v2,sum(v2) over(order by v2) as sum --按照 v2排序，累计n+n-1+....+1
from wmg_test;
2、select v1,v2,sum(v2) over(partition by v1 order by v2) as sum --先分组，组内在进行 1 中的操作
from wmg_test;
3、select v1,v2,sum(v2) over(partition by v1 order by v1) as sum ---稳定排序
from wmg_test;
4、select v1,v2,sum(v2) over(partition by v1) as sum --相同key的进行回填处理
from wmg_test;
5、select distinct v1,sum_01 --取一条
from (
select v1,sum(v2) over(partition by v1) as sum_01
from wmg_test
) a;
6、当然也可以逆序累加，只需order by desc 即可
总结区别：group by 和partition by的区别
group 单纯分组
partition 也能分组，但还具备累计的功能
