前述

窗口函数相关知识推荐阅读：通俗易懂的学会：SQL窗口函数

题目描述

leetcode 题目：511. 游戏玩法分析 I

思路

方法一： min, group by：分组，排序

1. 不用别名：1242 ms 击败5.08%使用 MySQL 的用户

select player_id, MIN(event_date) as first_login
from Activity
group by player_id;

2. 用别名：473ms击败56.30%使用 MySQL 的用户

select A.player_id, MIN(A.event_date) as first_login
from Activity A
group by A.player_id;

经过验证：当增加新的列名时，使用别名效率更高。

方法二：窗口函数

--错误示例:报错：Unknown column 'rk' in 'where clause' ，解决：得设个临时表
-- select player_id,event_date,
--     row_number() over (partition by player_id order by event_date) as rk
-- from Activity
-- where rk = 1; --正确示范：
select B.player_id, B.event_date as first_login
from (select A.player_id,A.event_date,row_number() over (partition by A.player_id order by A.event_date) as rkfrom Activity A
) B
where B.rk = 1;

方法三： FIRST_VALUE()

-- note: 加上别名A.就不超出时间限制了。可提高效率
select  distinct A.player_id,first_value(A.event_date) over(partition by A.player_id order by A.event_date) as first_login
from Activity A;