Mysql SUM 两次并返回总持续时间为天、小时、分钟

Question

我想从按某个 ID 分组的几行中选择持续时间。该行包含时间 (00:04:25)、(00:05:25)、(12:04:25)(小时、分钟、秒)等。我如何对这些行求和并返回总持续时间为天、小时， 分钟 ？ 我所做的是这个(实际上不是我，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';

这工作不正确。在 r.* 中还存储平均速度、距离，例如此代码返回 2 小时 23 分钟，而它应该返回 2 小时 30 分钟。请帮忙。谢谢!

编辑: 这是一些数据:

 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, Königsbenden, 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, Grüner 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, Hauptstraße, 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, Bahnhofstraße, Langerwehe  27, Königsbenden, Eschweiler   18

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

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;

查询逻辑如下:
1. 对总秒数求和并将其存储在变量 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;

休息是纯数学。
a.)天数:下限(time_in_sec/3600)。
b.)小时数:下限((time_in_sec/3600 - 天)*24)
c.)分钟:Floor(((time_in_sec/3600 - 天)*24 - 小时)*60)