LeetCode-550. 游戏玩法分析 IV，

Table: Activity

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：

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+

Result table:
+-----------+
| fraction |
+-----------+
| 0.33 |
+-----------+
只有 ID 为 1 的玩家在第一天登录后才重新登录，所以答案是 1/3 = 0.33

题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/game-play-analysis-iv
审题：计算首次玩后第二天也登陆的用户数量。然后除以总人数。就是从首次登陆之后连续两天登陆的人数。

思考：首先查出首次登陆日期，然后再给这个日期加一，判断这个日期有没有登陆。

解题：

解题1:

计算每人登陆的时间，使用group by:

(
SELECT player_id, MIN(event_date) AS `date`
FROM Activity
GROUP BY player_id
) AS C

表activity与表C连接，查询首次登陆的行：

SELECT *
FROM Activity AS A
JOIN (
SELECT player_id, MIN(event_date) AS `date`
FROM Activity
GROUP BY player_id
) AS C ON (A.player_id = C.player_id AND A.event_date = C.DATE)

select * from Activity AS A JOIN(
select id,min(date) as 'date' from activity group by id
) AS C on (A.id = C.id and A.date = C.date)

接着再left join表activity ，排除掉首次登录后第二天没有登录的行。这里用left join，并投影处字段A.player_id和B.player_id。部分B.player_id 为NULL，表示首次登录后第二天没有登录: 查询第二天有登陆的：

SELECT DISTINCT A.player_id AS id1,B.player_id AS id2
FROM Activity AS A
JOIN (
SELECT player_id, MIN(event_date) AS `date`
FROM Activity
GROUP BY player_id
) AS C ON (A.player_id = C.player_id AND A.event_date = C.DATE)
LEFT JOIN Activity AS B ON (A.player_id = B.player_id AND DATEDIFF(B.event_date,A.event_date)=1)

##DATEDIFF() 函数返回两个日期之间的天数。
##distinct 去重

基于这样的结果，计算比例：

SELECT ROUND(SUM(IF(id2 IS NOT NULL,1,0)) / COUNT(DISTINCT id1),2) AS `fraction`
FROM (
SELECT DISTINCT A.player_id AS id1,B.player_id AS id2
FROM Activity AS A
JOIN (
SELECT player_id, MIN(event_date) AS `date`
FROM Activity
GROUP BY player_id
) AS C ON (A.player_id = C.player_id AND A.event_date = C.DATE)
LEFT JOIN Activity AS B ON (A.player_id = B.player_id AND DATEDIFF(B.event_date,A.event_date)=1)
) AS D

解法2：

先求出首次登录后第二天又登录的人数。首次登录日期：

SELECT MIN(event_date) FROM Activity WHERE player_id = xxxx

表activity自连接，排除掉日期不是首次登录日的行。并统计人数：

	SELECT COUNT(DISTINCT A.player_id)
		FROM Activity AS A
		JOIN Activity AS B ON A.player_id = B.player_id AND DATEDIFF(B.event_date,A.event_date)=1
		WHERE A.event_date = (
			SELECT MIN(event_date)
			FROM Activity
			WHERE player_id = A.player_id
		)

总人数：

SELECT COUNT(DISTINCT player_id) FROM Activity

两者相除：

SELECT ROUND(
	(
		SELECT COUNT(DISTINCT A.player_id)
		FROM Activity AS A
		JOIN Activity AS B ON A.player_id = B.player_id AND DATEDIFF(B.event_date,A.event_date)=1
		WHERE A.event_date = (
			SELECT MIN(event_date)
			FROM Activity
			WHERE player_id = A.player_id
		)
	)
	/
	(
		SELECT COUNT(DISTINCT player_id)
		FROM Activity
	)
	,
	2
) AS `fraction`

知识点：

distinct：查询的时候去重。

DATEDIFF() ：函数返回两个日期之间的天数。

left join:左连接LEFT JOIN的含义就是求两个表的交集外加左表剩下的数据。

ROUND() ：函数用于把数值字段舍入为指定的小数位数。

count(*) 它返回检索行的数目，不论其是否包含 NULL值。

COUNT(DISTINCT 字段)，返回不同的非NULL值数目；若找不到匹配的项，则COUNT(DISTINCT)返回 0 。

SUM()函数用于计算一组值或表达式的总和，

IS NOT NULL:不为null

红楼终究一场梦

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