Question

如何计算用户相隔1天的会话数？这就是我到目前为止所拥有的。答案应为46，但此代码仅返回最后排名的记录以及它与第一条记录之间的差异。我想把数字46作为正确的输出。

set @pk1 =''; set @rn1 = 1; set @days = '';

select c.user_id,  c.day_session, datediff(d.day_session, c.day_session)

from 

(select user_id, day_session, rank

FROM

(select user_id,        day_session, 
        @rn1 := if(@pk1=user_id, if(@days=day_session, @rn1, @rn1+1),1) as rank, 
        @pk1 := user_id, 
        @days := @day_session

from

(select user_id, date(reg_utc_timestamp) AS day_session

from mobile_traffic

where user_id = 'abcdxyz'

group by 1,2) a) b) c

inner join (select user_id, day_session, rank

FROM

(select user_id,        day_session, 
        @rn1 := if(@pk1=user_id, if(@days=day_session, @rn1, @rn1+1),1) as rank, 
        @pk1 := user_id, 
        @days := @day_session

from

(select user_id, date(reg_utc_timestamp) AS day_session

from mobile_traffic

where user_id = 'abcdxyz'

group by 1,2) x) y) d on c.user_id = d.user_id and d.rank = c.rank  + 1

当我运行这个片段时，我得到了46条记录。

set @pk1 ='';
set @rn1 = 1;
set @days = '';

select Respondent_ID, day_session, rank

FROM

(select user_id, 
        day_session, 
        @rn1 := if(@pk1=user_id, if(@days=day_session, @rn1, @rn1+1),1) as rank, 
        @pk1 := user_id, 
        @days := @day_session

from

(select user_id, date(reg_utc_timestamp) AS day_session

from mobile_traffic

where user_id = 'abcdxyz'

group by 1,2) a) b

以下是此用户的示例数据：

abcdxyz 2017-11-19  1
abcdxyz 2017-11-20  2
abcdxyz 2017-11-21  3
abcdxyz 2017-11-22  4
abcdxyz 2017-11-23  5
abcdxyz 2017-11-24  6
abcdxyz 2017-11-27  7
abcdxyz 2017-11-28  8
abcdxyz 2017-11-29  9
abcdxyz 2017-11-30  10
abcdxyz 2017-12-01  11
abcdxyz 2017-12-02  12
abcdxyz 2017-12-03  13
abcdxyz 2017-12-04  14
abcdxyz 2017-12-05  15
abcdxyz 2017-12-06  16
abcdxyz 2017-12-07  17
abcdxyz 2017-12-08  18
abcdxyz 2017-12-09  19
abcdxyz 2017-12-10  20
abcdxyz 2017-12-11  21
abcdxyz 2017-12-12  22
abcdxyz 2017-12-13  23
abcdxyz 2017-12-14  24
abcdxyz 2017-12-15  25
abcdxyz 2017-12-16  26
abcdxyz 2017-12-17  27
abcdxyz 2017-12-18  28
abcdxyz 2017-12-19  29
abcdxyz 2017-12-20  30
abcdxyz 2017-12-21  31
abcdxyz 2017-12-22  32
abcdxyz 2017-12-23  33
abcdxyz 2017-12-24  34
abcdxyz 2017-12-25  35
abcdxyz 2017-12-26  36
abcdxyz 2017-12-27  37
abcdxyz 2017-12-28  38
abcdxyz 2017-12-29  39
abcdxyz 2017-12-30  40
abcdxyz 2017-12-31  41
abcdxyz 2018-01-01  42
abcdxyz 2018-01-02  43
abcdxyz 2018-01-03  44
abcdxyz 2018-01-04  45
abcdxyz 2018-01-05  46

Answer 1

我认为解决方案是在相关日期前一天的记录上使用联接。试试这个：

SELECT COUNT(*) FROM mobile_traffic m1
INNER JOIN mobile_traffic m2 ON m1.user_id = m2.user_id
AND DATE(m1.reg_utc_timestamp) =
(DATE(m2.reg_utc_timestamp) + INTERVAL 1 DAY)
WHERE m1.user_id = <some_user_id>

计算用户在MySQL中连续每日会话的数量

1 个答案: