力扣550(MySQL)-游戏玩法分析Ⅳ（中等）-526互联

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 
2     player_id, 
3     event_date, 
4     lead ( event_date, 1 ) over (partition by player_id ORDER BY event_date ) AS after_date
5 FROM activity；

②筛选出连续两天登录的用户数；

1 SELECT count( * ) AS nums1 
2 FROM ( 
3       SELECT 
4          player_id, 
5          event_date, 
6          lead ( event_date, 1 ) over (partition by player_id ORDER BY event_date ) AS after_date 
7       FROM activity ) AS temp 
8 WHERE datediff(after_date, event_date ) = 1；

③求出玩家总数，注意去重；

1 SELECT count( DISTINCT player_id ) AS total_num 
2 FROM activity；

④使用笛卡尔积拼接一下，计算结果。

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;