2

这是我的activities桌子。

activities
+----+---------+----------+-----------------+
| id | user_id | activity |    log_time     |
+----+---------+----------+-----------------+
|  6 |       1 | start    | 12 Oct, 1000hrs |
|  2 |       1 | task     | 12 Oct, 1010hrs |
|  7 |       1 | task     | 12 Oct, 1040hrs |
|  3 |       1 | start    | 12 Oct, 1600hrs |
|  1 |       1 | task     | 12 Oct, 1610hrs |
|  9 |       1 | start    | 14 Oct, 0800hrs |
| 10 |       1 | start    | 16 Oct, 0900hrs |
|  4 |       1 | task     | 16 Oct, 0910hrs |
|  8 |       2 | start    | 12 Oct, 1000hrs |
|  5 |       2 | task     | 12 Oct, 1020hrs |
+----+---------+----------+-----------------+

我需要用户在所有会话中花费的总时间。每个会话在一天内发生，包括一个“开始”和多个“任务”（在下一个会话以“开始”开始之前）。一个会话持续时间 = 最后一个任务 - 开始 [时间戳差异]

output
+---------+------------+------------------------------------------------+
| user_id | total_time |       This is explanation (not a column)       |
+---------+------------+------------------------------------------------+
|       1 |         60 | 12_Oct[40+10] + 14_Oct[0] + 16_Oct[10] = 60min |
|       2 |         20 | 12_Oct[20]  = 20min                            |
+---------+------------+------------------------------------------------+

我无法弄清楚如何获得会话中的最后一个任务。我已经尝试了基本的聚合和连接查询 - 但它不起作用。

作为一种方法，我认为我真正需要的是以某种方式获取最后一列（低于/ session_group），然后我可以聚合并获得最大/最小时间戳之间的差异。

+----+---------+----------+-----------------+---------------+
| id | user_id | activity |    log_time     | session_group |
+----+---------+----------+-----------------+---------------+
|  6 |       1 | start    | 12 Oct, 1000hrs |             1 |
|  2 |       1 | task     | 12 Oct, 1010hrs |             1 |
|  7 |       1 | task     | 12 Oct, 1040hrs |             1 |
|  3 |       1 | start    | 12 Oct, 1600hrs |             2 |
|  1 |       1 | task     | 12 Oct, 1610hrs |             2 |
|  9 |       1 | start    | 14 Oct, 0800hrs |             3 |
| 10 |       1 | start    | 16 Oct, 0900hrs |             4 |
|  4 |       1 | task     | 16 Oct, 0910hrs |             4 |
|  8 |       2 | start    | 12 Oct, 1000hrs |             5 |
|  5 |       2 | task     | 12 Oct, 1020hrs |             5 |
+----+---------+----------+-----------------+---------------+

请让我知道是否甚至可以通过 sql (MySQL) 获得所需的输出以及如何去做？或者是否有必要通过 Javascript 循环遍历数据？

下面是表的 MySQL 查询：

create table activities (
  id INT NOT NULL, 
  user_id INT NULL, 
  activity VARCHAR(45), 
  log_time DATETIME NOT NULL DEFAULT NOW(),
  PRIMARY KEY(id))
 ENGINE = InnoDB;    
 
insert into activities
    (id, user_id, activity, log_time) 
values
    (6,1,'start', '2021-10-12 10:00:00'), 
    (2,1,'task' , '2021-10-12 10:10:00'), 
    (7,1,'task' , '2021-10-12 10:40:00'), 
    (3,1,'start', '2021-10-12 16:00:00'), 
    (1,1,'task',  '2021-10-12 16:10:00'), 
    (9,1,'task',  '2021-10-14 08:00:00'), 
    (10,1,'start','2021-10-16 09:00:00'), 
    (4,1,'task',  '2021-10-16 09:10:00'), 
    (8,2,'start', '2021-10-12 10:00:00'), 
    (5,2,'task',  '2021-10-12 10:20:00');
4

4 回答 4

1

这可能会奏效。找出所有开始-任务对并以分钟为单位取最大差异，然后总结每个用户的分钟数。

select user_id, sum(minutes) minutes
from (
    select a.user_id, a.id, max(timestampdiff(minute, a.log_time, b.log_time)) minutes
    from activities a 
    join activities b on a.user_id = b.user_id and a.log_time < b.log_time 
    where a.activity = 'start'
    and b.activity = 'task'
    and date(a.log_time) = date(b.log_time)
    and not exists (
      select 1
      from activities c 
      where c.user_id = a.user_id
      and a.activity = c.activity
      and c.log_time > a.log_time
      and c.log_time < b.log_time
    )
    group by a.user_id, a.id   
) f
group by user_id

或使用窗口函数

with combo as
(
select user_id, activity, log_time,
 lag(activity) over( partition by user_id  order by log_time) last_activity,
 lag(log_time) over( partition by user_id  order by log_time) last_log_time
from activities
)
select user_id, sum(timestampdiff(minute, last_log_time, log_time))
from combo
where activity = 'task'
and date(log_time) = date(last_log_time)
group by user_id
于 2021-10-31T14:17:57.820 回答
1

您可以使用SUM()窗口函数为每个会话分配一个数字，然后聚合：

SELECT DISTINCT user_id,
       SUM(TIMESTAMPDIFF(MINUTE, MIN(log_time), MAX(log_time))) OVER (PARTITION BY user_id) total_time 
FROM (
  SELECT *, SUM(activity = 'start') OVER (PARTITION BY user_id, DATE(log_time) ORDER BY log_time) grp
  FROM activities
) t
WHERE grp > 0
GROUP BY user_id, DATE(log_time), grp;

请参阅演示。

于 2021-10-31T15:02:18.327 回答
1

模式和插入语句：

 create table activities (
   id INT NOT NULL, 
   user_id INT NULL, 
   activity VARCHAR(45), 
   log_time DATETIME NOT NULL DEFAULT NOW(),
   PRIMARY KEY(id))
  ENGINE = InnoDB;    
  
 insert into activities
     (id, user_id, activity, log_time) 
 values
     (6,1,'start', '2021-10-12 10:00:00'), 
     (2,1,'task' , '2021-10-12 10:10:00'), 
     (7,1,'task' , '2021-10-12 10:40:00'), 
     (3,1,'start', '2021-10-12 16:00:00'), 
     (1,1,'task',  '2021-10-12 16:10:00'), 
     (9,1,'start',  '2021-10-14 08:00:00'), 
     (10,1,'start','2021-10-16 09:00:00'), 
     (4,1,'task',  '2021-10-16 09:10:00'), 
     (8,2,'start', '2021-10-12 10:00:00'), 
     (5,2,'task',  '2021-10-12 10:20:00');

询问：

 with tasks as
 (
 SELECT
   user_id, partition_condition ,TIMESTAMPDIFF(minute,min(log_time),max(log_time))time_diff
 FROM (
   SELECT
     id, user_id, activity, log_time,
     sum(case when activity='start' then 1 else 0 end) over (partition by user_id order by log_time) as partition_condition
 
   FROM activities
 
 ) as tasks
 group by user_id, partition_condition
 )
 select user_id,sum(time_diff)total_time from tasks
 group by user_id

输出：

用户身份 总时间
1 60
2 20

db<>在这里摆弄

于 2021-10-31T17:33:55.387 回答
1

您可以使用LAG()窗口函数为每个用户和日期进行操作，以便计算具有任务活动的每一行的微小差异，例如

SELECT user_id, 
       SUM( TIMESTAMPDIFF(MINUTE, COALESCE( lg, log_time ), log_time ) ) AS total_time
  FROM (SELECT LAG(log_time) OVER (PARTITION BY user_id, DATE(log_time) 
                                      ORDER BY log_time) AS lg, 
               a.*
          FROM activities AS a
         ORDER BY log_time) AS aa
  WHERE activity != 'start'      
  GROUP BY user_id

Demo

于 2021-10-31T22:08:01.220 回答