计算用户在MySQL中的连续每日会话数

如何计算用户之间间隔1天的会话数？这就是我目前所拥有的。答案应该是46，但这段代码只返回最后一条排名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

当我运行这段代码时，我得到了那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

---

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>

从移动通信m1中选择计数（*）
m1上的内部加入移动通信量m2.user\u id=m2.user\u id
和日期（m1.reg\u utc\u时间戳）=
（日期（m2.reg\U utc\U时间戳）+间隔1天）
其中m1.user_id=

“到目前为止，这就是我所拥有的。答案应该是46”没有示例数据的好的反向工程SQL很难做到。换句话说，提供示例数据和所需结果。答案应该是添加示例数据，TIA