Question

我想从按行ID分组的多行中选择持续时间。该行包含时间（00:04:25），（00:05:25），（12:04:25）（小时，分钟，秒）等。如何将这些行相加并将总持续时间返回为天，小时，分钟？我做的是这个（实际上它不是我，我在stackoverflow上的其他帖子中看到了它）

$sql = 'SELECT r.*,d.*,        CONCAT(
        FLOOR(HOUR(SEC_TO_TIME(SUM(TIME_TO_SEC(r.duration)))) / 24), "' . $this-     >_translator->translate(' days ') . '",
        MOD(HOUR(SEC_TO_TIME(SUM(TIME_TO_SEC(r.duration)))), 24), "' . $this->_translator->translate(' hours ') . '",
        MINUTE(SEC_TO_TIME(SUM(TIME_TO_SEC(r.duration)))), "' . $this->_translator->translate(' minutes ') . '") as total_duration, 
        ROUND(SUM(TIME_TO_SEC(r.duration)/3600)) as total_duration_hours FROM routes r LEFT JOIN drivers d ON d.id = r.driver_id GROUP BY r.driver_id';

这是不正确的。在r。*中也存储了平均速度，距离，例如，此代码返回2小时，23分钟时应返回2小时30分钟。请帮忙。谢谢！

编辑：以下是一些数据：

 Full texts     id  filename    distance    startRoute  endRoute    duration    avgSpeed    car_id  fromLocation    toLocation  driver_id
Edit Edit   Copy Copy   Delete Delete   748     GPSdata20140109073948   5.74    2014-01-09 07:40:23     2014-01-09 07:50:38     00:10:15    33.6    20  22, Ulhaus, Langerwehe  27, KÃ¶nigsbenden, Eschweiler   18
Edit Edit   Copy Copy   Delete Delete   754     GPSdata20140110182521   1.43    2014-01-10 18:25:26     2014-01-10 18:29:31     00:04:05    21.01   18  36, GrÃ¼ner Weg, Aachen     3, Rehmplatz, Aachen    18
Edit Edit   Copy Copy   Delete Delete   758     GPSdata20140112145245   70.55   2014-01-12 14:54:05     2014-01-12 16:54:42     02:00:37    35.09   23  275, HauptstraÃŸe, Langerwehe   , Pescher Weg, Cologne  18
Edit Edit   Copy Copy   Delete Delete   759     GPSdata20140113072802   6.17    2014-01-13 07:28:37     2014-01-13 07:37:02     00:08:25    43.98   19  13, BahnhofstraÃŸe, Langerwehe  27, KÃ¶nigsbenden, Eschweiler   18

5.74 + 1.43 + 70.55 + 6.17 = 83.9距离 33.6 + 21.01 + 35.09 + 43.98 =平均速度33.42 所以距离/平均速度= 2小时30分钟，我得到2小时23分钟。我在哪里做错了？

Answer 1

您可以使用以下查询来获得所需的结果：

select CONCAT(FLOOR(time_in_sec/(60*60*24))," days " ,
FLOOR((time_in_sec/(60*60*24) - FLOOR(time_in_sec/(60*60*24)))*24)," hours ",
FLOOR(((time_in_sec/(60*60*24) - FLOOR(time_in_sec/(60*60*24)))*24 - 
FLOOR((time_in_sec/(60*60*24) - 
FLOOR(time_in_sec/(60*60*24)))*24))*60)," minutes" ) total_duration

FROM (
select SUM(TIME_TO_SEC(r.duration)) time_in_sec 
from routes r LEFT JOIN drivers d ON d.id = r.driver_id 
GROUP BY r.driver_id') T;

查询的逻辑如下：
1.总和秒数并将其存储在变量time_in_sec中。使用此结果来自：

FROM (
    select SUM(TIME_TO_SEC(r.duration)) time_in_sec 
    from routes r LEFT JOIN drivers d ON d.id = r.driver_id 
    GROUP BY r.driver_id') T;

休息是纯粹的数学 a。）连续几天：楼层（time_in_sec / 3600） b。）小时：楼层（（time_in_sec / 3600 - 天）* 24）
c。）分钟：楼层（（（time_in_sec / 3600 - 天）* 24小时）* 60）

Mysql SUM两次，返回总持续时间为天，小时，分钟

1 个答案: