来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problemset/database/
SQL结构：
Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)
Truncate table Activity
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-05-02', '6')
insert into Activity (player_id, device_id, event_date, games_played) values ('2', '3', '2017-06-25', '1')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')
活动表 Activity：
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
表的主键是 (player_id, event_date)。
这张表展示了一些游戏玩家在游戏平台上的行为活动。
每行数据记录了一名玩家在退出平台之前，当天使用同一台设备登录平台后打开的游戏的数目（可能是 0 个）。
511. 游戏玩法分析 I（简单）
需求：写一条 SQL 查询语句，获取每位玩家第一次登陆平台的日期。
查询结果的格式如下所示：
Activity table：
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table：
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
+-----------+-------------+
-- 解法一： 根据player_id分组，取最小日期
SELECT
player_id,
MIN(event_date) AS first_login
FROM Activity
GROUP BY player_id;
512. 游戏玩法分析 II（简单）
需求：请编写一个 SQL 查询，描述每位玩家首次登陆的设备名称
查询结果格式在以下示例中：
Activity table：
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+-----------+
| player_id | device_id |
+-----------+-----------+
| 1 | 2 |
| 2 | 3 |
| 3 | 1 |
+-----------+-----------+
-- 解法一： 每位玩家首次登陆 + 内连接 主键是(player_id, event_date)
SELECT
a.player_id,
a.device_id
FROM Activity AS a
INNER JOIN
(
SELECT
player_id,
MIN(event_date) AS first_login
FROM Activity
GROUP BY player_id
) AS b
ON a.player_id = b.player_id AND a.event_date = b.first_login;
)
-- 解法二： 联合查询，根据联合主键，用in匹配每位玩家首次登陆
SELECT
player_id,
device_id
FROM Activity
WHERE (player_id, event_date) IN (
SELECT
player_id,
MIN(event_date) event_date
FROM Activity
GROUP BY player_id
)
-- 解法三： 利用窗口函数rank()系列，取排行的第一个
SELECT
t1.player_id,
t1.device_id
FROM (
SELECT
player_id,
device_id,
dense_rank() OVER (PARTITION BY player_id ORDER BY event_date) rk
FROM Activity
) AS t1
WHERE t1.rk = 1
534. 游戏玩法分析 III（中等）
需求：请编写一个 SQL 查询，同时报告每组玩家和日期，以及玩家到目前为止玩了多少游戏。也就是说，在此日期之前玩家所玩的游戏总数。详细情况请查看示例。
查询结果格式在以下示例中：
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 1 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+------------+---------------------+
| player_id | event_date | games_played_so_far |
+-----------+------------+---------------------+
| 1 | 2016-03-01 | 5 |
| 1 | 2016-05-02 | 11 |
| 1 | 2017-06-25 | 12 |
| 3 | 2016-03-02 | 0 |
| 3 | 2018-07-03 | 5 |
+-----------+------------+---------------------+
对于 ID 为 1 的玩家，2016-05-02 共玩了 5+6=11 个游戏，2017-06-25 共玩了 5+6+1=12 个游戏。
对于 ID 为 3 的玩家，2018-07-03 共玩了 0+5=5 个游戏。
请注意，对于每个玩家，我们只关心玩家的登录日期。
-- 解法一： 通过内连接，过滤出时间在自身前面的数据，再对联合主键进行group by，配合sum(games_played)
SELECT
a.player_id,
a.event_date,
SUM(b.games_played) AS games_played_so_far
FROM
Activity AS a INNER JOIN Activity AS b
ON a.player_id = b.player_id AND a.event_date >= b.event_date
GROUP BY
a.player_id, a.event_date
-- 解法二： 窗口函数
SELECT
player_id,
event_date,
SUM(games_played) OVER (PARTITION BY player_id ORDER BY event_date) AS games_played_so_far
FROM Activity
550. 游戏玩法分析 IV（中等）
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：
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
-- 解法一：1.先求出每位玩家下一次登录的时间，并且按照登录时间排行
-- 2.再过滤出首次登录，并且event_date + 1 = nextDate，才说明是连续两天登录
SELECT
ROUND(SUM(CASE
WHEN event_date + 1 = nextDate THEN 1
ELSE 0
END) / COUNT(*), 2) AS fraction
FROM (
SELECT
player_id,
event_date,
lead(event_date, 1) OVER (PARTITION BY player_id ORDER BY event_date) AS nextDate,
row_number() OVER (PARTITION BY player_id ORDER BY event_date) AS rk
FROM Activity
) AS t1
WHERE t1.rk = 1
-- 解法二：1.先求出玩家第一次登陆的数据，然后把这个作为临时表（b），
-- 2.然后用原始表（a）和临时表（b）左连接进行比较，如果日期相差1天，即为符合条件的玩家，
-- 用这部分的玩家数量除以所有玩家的数量，然后保留两位小数就可以得出结果了
SELECT
ROUND(COUNT(DISTINCT (b.player_id)) / (SELECT COUNT(DISTINCT (player_id)) FROM Activity), 2) AS fraction
FROM Activity a
LEFT JOIN
(
SELECT
player_id,
MIN(event_date) AS first_date
FROM Activity
GROUP BY player_id
) AS b
ON a.player_id = b.player_id AND datediff(a.event_date, b.first_date) = 1;