MYSQL 'SELECT' 和 COUNT 用户事件表中一天的“活动”用户

Question

+----+--------------+------------+--------+
| id | user    | date(data) | type        |
+----+--------------+------------+--------+
|  3 | 5458848 | 2013-12-19 | SUBSCRIBE   |
|  4 | 5458848 | 2013-12-19 | UNSUBSCRIBE |
|  5 | 5458848 | 2013-12-20 | SUBSCRIBE   |
|  7 | 5458848 | 2013-12-20 | UNSUBSCRIBE |
|  8 | 7883870 | 2013-12-20 | SUBSCRIBE   |
|  9 | 7883870 | 2013-12-23 | UNSUBSCRIBE |
| 10 | 7883870 | 2013-12-24 | SUBSCRIBE   |
| 11 | 7883870 | 2013-12-24 | UNSUBSCRIBE |
| 12 | 7883870 | 2013-12-24 | SUBSCRIBE   |
+----+---------+------------+-------------+                                                 

你好，我想知道如何在mysql中查询每天有多少用户活跃和不活跃，

例如：如果一个用户在同一天订阅和取消订阅，这意味着在那一天，我有一个用户不活跃

+------------+-------------+-------------+
| date(data) | active      | inactive    |
+------------+-------------+-------------+
| 2013-12-19 | 0           | 1           |
| 2013-12-20 | 1           | 1           |
| 2013-12-23 | 0           | 2           |
| 2013-12-24 | 1           | 1           |
+------------+-------------+-------------+

Answer 1

set @ac:=null;
DROP TABLE IF EXISTS ___active_nums;

CREATE TABLE ___active_nums(
  event_id integer,
  event_user_id integer,
  event_date  Date,
  number_of_active_users integer,
  INDEX USING BTREE (event_date)
)
ENGINE=MEMORY 
;

INSERT INTO ___active_nums

SELECT

  events.id,
  events.user,
  events.date,
  ifnull( @ac:= ( @ac + if( type='SUBSCRIBE', 1, -1) ), @ac:=if( type='SUBSCRIBE', 1, -1) ) AS active_num

FROM 
(
  SELECT * FROM events ORDER BY id
)
AS events
;

SELECT

event_date as date,
number_of_active_users,

( SELECT count(DISTINCT event_user_id) FROM ___active_nums WHERE ___active_nums.event_date <= outer_active_nums.event_date )
- number_of_active_users
as number_of_inactive_users

FROM ___active_nums  AS outer_active_nums

WHERE event_id IN ( SELECT max(event_id) FROM ___active_nums GROUP BY event_date );

Answer 2

试试这个...

SELECT SUM(CASE WHEN type='SUBSCRIBE' THEN 1 ELSE 0 END) as active,
       SUM(CASE WHEN type='UNSUBSCRIBE' THEN 1 ELSE 0 END) as inactive
WHERE (date>='2013-12-19' and date<='2013-12-24');

Answer 3

您需要每个日期的累计订阅数和取消订阅数之间的差值来计算是否有人订阅。例如，最后一个用户7883870 在同一天发生了三个事件后，状态为“订阅”。

一种方法是使用相关子查询：

select t.user, t.date,
       (select sum(type = 'SUBSCRIBE') - sum(type = 'UNSUBSCRIBE')
        from table t2
        where t2.user = t.user and
              t2.date <= t.date
       ) netsubs
from table t
group by t.user, t.date;

有了这些信息，我们就可以确定某人在给定日期所处的状态。

select date, 
       sum(case when netsubs > 0 then 1 else 0 end) as active,
       sum(case when netsubs = 0 then 1 else 0 end) as inactive
from (select t.user, t.date,
             (select sum(type = 'SUBSCRIBE') - sum(type = 'UNSUBSCRIBE')
              from table t2
              where t2.user = t.user and
                    t2.date <= t.date
             ) netsubs
      from table t
      group by t.user, t.date
     ) t
group by t.date;

编辑：

我误解了这个问题。在任何一天，您都需要在该点之前处于活动状态或已取消订阅的用户数。

select date, numacive, numusers - numactive as numinactive
from (select date,
             (select count(distinct user)
              from table t2
              where t2.date <= t.date
             ) as numusers,
             (select sum(type = 'SUBSCRIBE') - sum(type = 'UNSUBSCRIBE')
              from table t2
              where t2.date <= t.date
             ) as numactive
      from (select distinct date
            from table t
           ) t
     ) t;