力扣550(MySQL)-游戏玩法分析Ⅳ（中等）

题目：

需求：编写一个 SQL 查询，报告在首次登录的第二天再次登录的玩家的分数，四舍五入到小数点后两位。换句话说，您需要计算从首次登录日期开始至少连续两天登录的玩家的数量，然后除以玩家总数。

查询结果格式如下所示：

 解题思路：

①先将登录日期上移一行，方便筛选出连续登录数；

1  select player_id ,event_date,
2             lead(event_date,1) over(partition by player_id order by event_date asc) as date2
3  from Activity

②再以player_id分组，筛选出连续两天登录的用户信息；

1 select  player_id , min(event_date) as date1,date2
2     from(
3          select player_id ,event_date,
4             lead(event_date,1) over(partition by player_id order by event_date asc)  as date2
5          from Activity
6     ) t1
7 group by player_id
8 having datediff(date2 ,date1)=1

③求出连续登录两个的玩家数和玩家总数，进行计算

 1 select round(count(1)/(
 2     select count(distinct player_id)
 3     from Activity
 4 ),2) fraction
 5 from (
 6     select  player_id , min(event_date) as date1,date2
 7     from(
 8          select player_id ,event_date,
 9             lead(event_date,1) over(partition by player_id order by event_date asc)  as date2
10          from Activity
11     ) t1
12     group by player_id
13     having datediff(date2 ,date1)=1
14 ) t2

方法二：

①先求出所有玩家的第一次登录时间

1 SELECT
2    player_id,
3    MIN(event_date) AS first_date
4 FROM Activity
5 GROUP BY player_id；

②将原表与第一步求出来的临时表进行左连接；

1 select *
2 from activity a
3 left join (SELECT
4        player_id,
5        MIN(event_date) AS first_date
6      FROM Activity
7       GROUP BY player_id) as b
8 on a.player_id = b.player_id and datediff(event_date,first_date) = 1;

③统计出左右两个表的数量，进行计算；

1 SELECT round(count(b.player_id)/count(DISTINCT a.player_id) , 2) AS fraction
2 FROM activity a
3 LEFT JOIN (
4     SELECT
5        player_id,
6        MIN(event_date) AS first_date
7     FROM Activity
8     GROUP BY player_id) AS b
9 ON a.player_id = b.player_id AND datediff(event_date,first_date) = 1;

小知识：

①lead() over：

1.点击这里可以看语法

2.这里也可以

②左右内连接：跳转这里看笔记

③笛卡儿积：两个集合相乘的结果。

假设集合A={a, b}，集合B={0, 1,2}，则两个集合的笛卡尔积为{(a, 0),(a, 1), (a, 2), (b, 0), (b, 1), (b, 2)}。