计算用户在平台上花费的时间答案

【问题标题】：Calculate Time spent by users on the platform计算用户在平台上花费的时间
【发布时间】：2020-08-13 10:32:43
【问题描述】：

我正在尝试计算用户每周在我的平台上花费的时间。我有以下 api_request_logs 表（*注意：每一行都是重复的）

time                            user_id       url 
------------------------        -------       ----
January 6,2020, 08:56 PM           14           api\v3\user1
January 6,2020, 08:58 PM           14           api\v3\session 
January 6,2020, 08:59 PM           14           api\v3\notification 
January 6,2020, 09:01 PM           14           api\v3\announcements
January 6,2020, 09:02 PM           14           api\v3\user2 
January 6,2020, 09:28 PM           14           api\v3\session

[enter image description here][1]

我试过这个查询，但是，它似乎没有输出正确的计算。

select  SUM(timespent) timespent, a.user_id as user
from (
    select  timestamp_diff(max(time),min(time),minute) timespent,  DATE_TRUNC(EXTRACT(date from time), day) as Time, a.user_id as user
    from (
    select * , (countif (start) over (order by time)) as  grp from (
    select *, ifnull (lag(url) over(order by time), '') !=url as start from logs.api_request_logs a
    INNER JOIN logs.cohort_members on a.user_id = cohort_members.user_id
    INNER JOIN `logs.cohorts` on cohort_members.cohort_id = `logs.cohorts`.cohort_id
    INNER Join logs.users on cohort_members.user_id = users.user_id
    WHERE a.user_id = 14  and time > timestamp ('2020-01-01 00:00:00') 
    )) group by user, grp)
    group by 2
    order by 2 asc

当前输出：

timespent     users        Time 
---------     -------      -----
32 min         14       January 6,2020, 12:00 AM

期望的输出：

Timespent    users       Time 
----------  --------     -----
  6 min      14          January 6,2020, 12:00 AM

我做错了什么或者你有更好的主意吗？谢谢

【问题讨论】：

(1)请说明计算；不清楚为什么您认为答案应该是“6”。 (2)“每一行都是重复的”是什么意思？
@GordonLinoff 1：用户每天或每周花费的总时间（以分钟为单位）。 2.每行重复10次以上。希望对你有帮助
@GordonLinoff：它应该是“6”，因为用户 (14) 在每一行花费一分钟。
。 .不，他们没有。根据问题中的数据，该用户在 URL“api\v3\user2”上花费了 26 分钟。
@GordonLinoff：我在我的问题中添加了一个快照，请看一下，让我知道你的想法？ [1]：i.stack.imgur.com/gNR3b.png

标签： sql google-bigquery data-analysis metabase

【解决方案1】：

我认为您也只想按您的网址进行分组...尝试以下操作:-)

#standardsql：

SELECT
  SUM(timespent)   timespent,
  a.user_id AS user
FROM (
  SELECT
    TIMESTAMP_DIFF(MAX(time),MIN(time),minute) timespent,
    DATE_TRUNC(EXTRACT(date  FROM  time), day) AS Time,
    a.user_id AS user
    ,url
  FROM (
    SELECT
      *,
      (countif (start) OVER (ORDER BY time)) AS grp
    FROM (
      SELECT
        *,
        ifnull (LAG(url) OVER(ORDER BY time),
          '') !=url AS start
      FROM
        logs.api_request_logs a
      INNER JOIN
        logs.cohort_members
      ON
        a.user_id = cohort_members.user_id
      INNER JOIN
        `logs.cohorts`
      ON
        cohort_members.cohort_id = `logs.cohorts`.cohort_id
      INNER JOIN
        logs.users
      ON
        cohort_members.user_id = users.user_id
      WHERE
        a.user_id = 14
        AND time > timestamp ('2020-01-01 00:00:00') ))
  GROUP BY
    user,
    gr,url)
GROUP BY
  2
ORDER BY
  2 ASC

【讨论】：