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我有一个表，我有多个用户在工作，我需要根据声明的状态计算每个用户花费的总时间

 ISSUEID    Time             User       date     status
10101   2019-08-30 19:51:43 peter   2019-08-30  CLAIMED
10101   2019-08-30 19:51:46 peter   2019-08-30  CLAIMED
10101   2019-08-30 20:10:39 miller  2019-08-30  CLAIMED
10101   2019-08-30 21:21:14 miller  2019-08-30  CLAIMED
10101   2019-08-30 21:22:33 Renus   2019-08-30  CLAIMED
10101   2019-08-31 1:31:58  Renus   2019-08-31  CLAIMED
10101   2019-08-31 1:32:08  peter   2019-08-31  CLAIMED
10101   2019-08-31 1:32:12  peter   2019-08-31  CLAIMED
10101   2019-08-31 1:32:58  miller  2019-08-31  CLAIMED
10101   2019-09-04 23:16:24 peter   2019-09-04  CLAIMED
10101   2019-09-04 23:16:26 peter   2019-09-04  CLAIMED
10101   2019-09-04 23:16:50 peter   2019-09-04  CLAIMED
10101   2019-09-04 23:17:01 Renus   2019-09-04  CLAIMED
10101   2019-09-04 23:17:06 Renus   2019-09-04  CLAIMED

期望的结果

IssueID user    timespent
10101   Peter   33
10101   Miller  4285
10101   Renus   30
4

2 回答 2

1

使用DATETIME_DIFF()and 在此之前使用将日期时间转换为日期DATETIME_TRUNC

SELECT
  ISSUEID,
  User,
  DATETIME_DIFF(
    min(DATETIME_TRUNC(Time, DAY)), 
    max(day), 
    day)
FROM table 
GROUP BY ISSUEID, User
于 2019-09-24T12:16:56.510 回答
1

以下是 BigQuery 标准 SQL 

#standardSQL
SELECT ISSUEID, user, SUM(timespent) timespent
FROM (
  SELECT ISSUEID, user, TIMESTAMP_DIFF(MAX(time), MIN(time), SECOND) timespent
  FROM (
    SELECT *, COUNTIF(start) OVER(ORDER BY time) grp
    FROM (
      SELECT *, IFNULL(LAG(User) OVER(ORDER BY time), '') != User AS start 
      FROM `project.dataset.table`
    )
  )
  GROUP BY ISSUEID, user, grp
)
GROUP BY ISSUEID, user
于 2019-09-24T12:44:31.387 回答