Question

我希望这可以清楚地解释我在寻找什么。我已经搜索了本网站上的一些文章，但是找不到我要找的东西。我也花了将近3个小时试图自己解决这个问题。

我正在尝试计算记录数和SUM工作时间。这是我一直在使用的查询。

SELECT Log.User
 , sum(if(hour(endtime) = 0, 1, 0)) AS Midnight
 , sum(if(hour(endtime) = 1, 1, 0)) AS `1AM`
 , sum(if(hour(endtime) = 2, 1, 0)) AS `2AM`
 , sum(if(hour(endtime) = 3, 1, 0)) AS `3AM`
 , sum(if(hour(endtime) = 4, 1, 0)) AS `4AM`
 , sum(if(hour(endtime) = 5, 1, 0)) AS `5AM`
 , sum(if(hour(endtime) = 6, 1, 0)) AS `6AM`
 , sum(if(hour(endtime) = 7, 1, 0)) AS `7AM`
 , sum(if(hour(endtime) = 8, 1, 0)) AS `8AM`
 , sum(if(hour(endtime) = 9, 1, 0)) AS `9AM`
 , sum(if(hour(endtime) = 10, 1, 0)) AS `10AM`
 , sum(if(hour(endtime) = 11, 1, 0)) AS `11AM`
 , sum(if(hour(endtime) = 12, 1, 0)) AS `12PM`
 , sum(if(hour(endtime) = 13, 1, 0)) AS `1PM`
 , sum(if(hour(endtime) = 14, 1, 0)) AS `2PM`
 , sum(if(hour(endtime) = 15, 1, 0)) AS `3PM`
 , sum(if(hour(endtime) = 16, 1, 0)) AS `4PM`
 , sum(if(hour(endtime) = 17, 1, 0)) AS `5PM`
 , sum(if(hour(endtime) = 18, 1, 0)) AS `6PM`
 , sum(if(hour(endtime) = 19, 1, 0)) AS `7PM`
 , sum(if(hour(endtime) = 20, 1, 0)) AS `8PM`
 , if(hour(endtime) = 20, sec_to_time(sum(time_to_sec(endtime) - time_to_sec(starttime))), 0) AS `8PM Time`
 , sum(if(hour(endtime) = 21, 1, 0)) AS `9PM`
 , sum(if(hour(endtime) = 22, 1, 0)) AS `10PM`
 , sum(if(hour(endtime) = 23, 1, 0)) AS `11PM`
FROM
  (
  SELECT user
   , controlnumber
   , starttime
   , endtime
   , timediff(endtime, starttime) AS Worktime
FROM
   atrtaxcert.ordertimeworked
  ) AS Log
GROUP BY
   Log.User;

这些开始和结束时间仅相隔几分钟。任何指导都非常感谢。这是我在这里的第一篇文章，无法提供任何图像来帮助描述。

Answer 1

如果starttime和endtime是TIME数据类型，则使用TIME_TO_SEC函数并进行减法。总计秒数，然后将总数转换为字符串表示形式。

SELECT `Log`.`User`
     , ...
     , SUM(HOUR(`Log`.endtime)=20) AS `8PM_count` 
     , SUM(IF(HOUR(`Log`.endtime)=20,work_seconds,0) AS `8PM_seconds`
     , SEC_TO_TIME(SUM(IF(HOUR(`Log`.endtime)=20,`Log`.work_seconds,0) AS `8PM_hhhmmss`
     , ...
  FROM ( SELECT
              , TIME_TO_SEC(endtime)-TIME_TO_SEC(starttime) AS work_seconds
       ) `Log`
 GROUP
    BY `Log`.`User`

注意：这个：

SELECT HOUR(endtime)=0 AS foo

简写等同于

SELECT IF(HOUR(endtime) = 0, 1, 0) AS foo

如果starttime和endtime是DATETIME值，您可以使用TIMESTAMPDIFF函数计算以秒为单位的差异：

SELECT `Log`.`User`
     , ...
     , SUM(HOUR(`Log`.endtime)=20) AS `8PM_count` 
     , SUM(IF(HOUR(endtime)=20,TIMESTAMPDIFF(SECOND,`Log`.starttime,`Log`.endtime),0) AS `8PM_seconds`
     , ...
  FROM (
       ) `Log`
 GROUP
    BY `Log`.`User`

（您可能希望忽略在例如starttime = '23：59：00'和endtime = '00：01：00'时返回的值，这将需要另一个条件测试。）

如何在几小时内完成TIMEDIFF的计时

1 个答案: