LeetCode 1097. Game Play Analysis V

1、题目描述

We define the install date of a player to be the first login day of that player.
We also define day 1 retention of some date X to be the number of players whose install date is X and they logged back in on the day right after X, divided by the number of players whose install date is X, rounded to 2 decimal places.
Write an SQL query that reports for each install date, the number of players that installed the game on that day and the day 1 retention.
The query result format is in the following example:

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
2	1	2016-03-01	0
3	4	2016-07-03	5

Result table:

install_dt	installs	Day1_retention
2016-03-01	2	0.50
2017-06-25	1	0.00

Player 1 and 3 installed the game on 2016-03-01 but only player 1 logged back in on 2016-03-02 so the day 1 retention of 2016-03-01 is 1 / 2 = 0.50
Player 2 installed the game on 2017-06-25 but didn’t log back in on 2017-06-26 so the day 1 retention of 2017-06-25 is 0 / 1 = 0.00

2、解题思路

建表语句：

 1 CREATE TABLE Activity (
 2   player_id INT (8),
 3   event_date VARCHAR (20)
 4 ) ;
 5 
 6 INSERT INTO Activity (player_id, event_date) 
 7 VALUES
 8   (1, '2018-01-01') ;
 9 
10 INSERT INTO Activity (player_id, event_date) 
11 VALUES
12   (1, '2018-01-02') ;
13 
14 INSERT INTO Activity (player_id, event_date) 
15 VALUES
16   (2, '2018-01-01') ;
17 
18 INSERT INTO Activity (player_id, event_date) 
19 VALUES
20   (2, '2019-03-01') ;
21 
22 INSERT INTO Activity (player_id, event_date) 
23 VALUES
24   (2, '2019-03-02') ;
25 
26 INSERT INTO Activity (player_id, event_date) 
27 VALUES
28   (3, '2019-02-28') ;
29 
30 INSERT INTO Activity (player_id, event_date) 
31 VALUES
32   (3, '2019-03-01') ;
33   
34 INSERT INTO Activity (player_id, event_date) 
35 VALUES
36   (4, '2019-02-25') ;
37 
38 INSERT INTO Activity (player_id, event_date) 
39 VALUES
40   (4, '2019-03-01') ;
41  
42 INSERT INTO Activity (player_id, event_date) 
43 VALUES
44   (5, '2019-03-05') ;

查询每日安装游戏的玩家数：

 1 SELECT 
 2   event_date,
 3   COUNT(1) AS installs 
 4 FROM
 5   (SELECT 
 6     player_id,
 7     event_date 
 8   FROM
 9     Activity 
10   GROUP BY player_id 
11   ORDER BY player_id,
12     event_date) a 
13 GROUP BY event_date;

按照访问日期升序，为每位玩家的事件记录进行编号：

如果你使用的数据库支持ROW_NUMBER()函数，可以更优雅的实现这一步。

 1 SELECT 
 2     player_id,
 3     event_date,
 4     IF(
 5       @id = player_id,
 6       @rank := @rank + 1,
 7       @rank := 1
 8     ) AS rank,
 9     @id := player_id 
10   FROM
11     Activity,
12     (SELECT 
13       @id := NULL,
14       @rank := 0) r 
15   ORDER BY player_id,
16     event_date ;

统计安装游戏后第二日仍然登陆游戏的玩家信息：

　　用上一步的中间表自关联，关联条件见SQL语句。

 1 SELECT 
 2   *
 3 FROM
 4   (SELECT 
 5     player_id,
 6     event_date,
 7     IF(
 8       @id = player_id,
 9       @rank := @rank + 1,
10       @rank := 1
11     ) AS rank,
12     @id := player_id 
13   FROM
14     Activity,
15     (SELECT 
16       @id := NULL,
17       @rank := 0) r 
18   ORDER BY player_id,
19     event_date) t1,
20     (SELECT 
21     player_id,
22     event_date,
23     IF(
24       @id2 = player_id,
25       @rank2 := @rank2 + 1,
26       @rank2 := 1
27     ) AS rank,
28     @id2 := player_id 
29   FROM
30     Activity,
31     (SELECT 
32       @id2 := NULL,
33       @rank2 := 0) r 
34   ORDER BY player_id,
35     event_date) t2  
36 WHERE t1.rank = 1 
37 AND t2.rank =2
38 AND t1.player_id = t2.player_id
39 AND DATEDIFF(t2.event_date,t1.event_date)=1;

完整查询语句：

 1 SELECT 
 2   temp1.event_date AS install_dt,
 3   temp1.installs,
 4   ROUND(
 5     IFNULL(temp2.counts, 0) / temp1.installs,
 6     2
 7   ) AS Day1_retention 
 8 FROM
 9   (SELECT 
10     event_date,
11     COUNT(1) AS installs 
12   FROM
13     (SELECT 
14       player_id,
15       event_date 
16     FROM
17       Activity 
18     GROUP BY player_id 
19     ORDER BY player_id,
20       event_date) a 
21   GROUP BY event_date) temp1 
22   LEFT JOIN 
23     (SELECT 
24       t1.event_date,
25       COUNT(1) AS counts 
26     FROM
27       (SELECT 
28         player_id,
29         event_date,
30         IF(
31           @id = player_id,
32           @rank := @rank + 1,
33           @rank := 1
34         ) AS rank,
35         @id := player_id 
36       FROM
37         Activity,
38         (SELECT 
39           @id := NULL,
40           @rank := 0) r 
41       ORDER BY player_id,
42         event_date) t1,
43       (SELECT 
44         player_id,
45         event_date,
46         IF(
47           @id2 = player_id,
48           @rank2 := @rank2 + 1,
49           @rank2 := 1
50         ) AS rank,
51         @id2 := player_id 
52       FROM
53         Activity,
54         (SELECT 
55           @id2 := NULL,
56           @rank2 := 0) r 
57       ORDER BY player_id,
58         event_date) t2 
59     WHERE t1.rank = 1 
60       AND t2.rank = 2 
61       AND t1.player_id = t2.player_id 
62       AND DATEDIFF(t2.event_date, t1.event_date) = 1 
63     GROUP BY t1.event_date) temp2 
64     ON temp1.event_date = temp2.event_date ;

3、小结

本题知识点有以下几个：

如何实现分组按顺序进行编号。这是一个常见需求，最简单的实现方式是row_number函数，但mysql中没有这个函数，需要自己用sql语句实现。
日期函数的使用。
左连接与空值判定。