1 
     > Start                    stop                  pinnumber
     > ---------------------------------------------------------
     > 2012-03-14 13:22:17    2012-03-14 15:22:50      2001
     > 2012-03-14 18:11:10    2012-03-14 19:10:10      2001
     > 2012-03-15 07:20:10    2012-03-15 13:20:50      2001
     >**2012-03-16 19:21:55       2012-03-17 02:55:22  2001** //on 16(19:21:55
                                                                     to 23:59:59) and     
                                                                 //on 17(00 to 02:55:22) 
     > 2012-03-17 14:15:05    2012-03-17 17:44:50      2001
     > 2012-03-18 19:11:10    2012-03-18 19:10:10      2002
     > 2012-03-18 10:20:10    2012-03-18 13:20:50      2003
     > 2012-03-18 11:20:10    2012-03-18 15:11:50      2001

问题：

如何计算每个用户每天（'开始'，'停止'）每天的总时间？请参阅上面突出显示的点。假设，如果用户今天“开始”并明天停止，那么今天的时间不同，明天的时间不同？

现在我正在使用以下查询：-

SELECT SEC_TO_TIME( SUM( TIME_TO_SEC(TIMEDIFF( stop, start) ) ) ) 作为时间 1，时钟。* FROM table_name WHERE pin_number= '2001' GROUP BY DATE_FORMAT( start, '%W %M %Y' )

从上面的查询我得到每天的记录，但是当开始日期和停止日期不同时。它计算的总时间不是一天的时间，但我需要每天的时间。

4

1 回答 1

2

I think I finally got there. You first need to get a set of days, which I obtain through a subquery that takes a UNION of the start and stop times (you could filter this for your pinnumber if desired in order to reduce the size of the JOIN).

One then joins each such date with those (start,stop) pairs that encompass that date (i.e. either start during the day, or the start of the day is between the start and stop time).

Finally, one groups by day and takes the sum of the amount of time between the start and end times, cutting off at the day start and end as appropriate (the magic 86400 is the number of seconds in a day = 24*60*60). Sadly this won't play nice with daylight savings, leap seconds, etc...

SELECT FROM_UNIXTIME(unixday, '%d/%m/%Y'), SUM(
    LEAST(   unixday+86400, UNIX_TIMESTAMP(Stop ))
  - GREATEST(unixday      , UNIX_TIMESTAMP(Start))
) AS Seconds
FROM table_name JOIN (

  SELECT UNIX_TIMESTAMP(DATE(Start)) AS unixday FROM table_name
UNION 
  SELECT UNIX_TIMESTAMP(DATE(Stop )) AS unixday FROM table_name

) AS days ON (
      UNIX_TIMESTAMP(Start) BETWEEN unixday AND unixday+86400
  OR (unixday BETWEEN UNIX_TIMESTAMP(Start) AND UNIX_TIMESTAMP(Stop))
)
WHERE pinnumber = 2001
GROUP BY unixday;

See it on sqlfiddle.

于 2012-05-02T11:02:20.613 回答