如果我有一个开始时间和结束时间,并且我找到总时间,如果超过24小时,我希望它而不是显示26,显示1天和2小时。怎么办呢?
答案 0 :(得分:2)
你可能想要做一个
IF (MOD(HOUR(TIME) / 24) ) > 24,
SELECT CONCAT(
FLOOR(HOUR(TIME) / 24), ' days ',
MOD(HOUR(TIME), 24), ' hours ',
MINUTE(TIME), ' minutes')
)
使用Timediff
,
FLOOR(HOUR(TIMEDIFF('2010-01-06 08:46', '2010-01-01 12:30')) / 24), ' days ',
答案 1 :(得分:0)
如果您的start_time
和end_time
是日期时间类型,请使用TIMEDIFF功能并应用所需的DATE_FORMAT。
e.g。
select DATE_FORMAT(TIMEDIFF('2012:01:01 00:00:00', '2012:01:01 00:00:00.000001'),
'%d: %H: %m');
答案 2 :(得分:0)
这对我有用:
select
o.id,
o.created,
o.complete_date,
concat(timestampdiff(day, o.created, o.complete_date),
'd ',
lpad(
mod(timestampdiff(hour, o.created, o.complete_date), 24),
2, '0'),
'h'
) as time_to_complete,
from
orders o
;
结果:
+---------+---------------------+---------------------+------------------+
| id | created | complete_date | time_to_complete |
+---------+---------------------+---------------------+------------------+
| 2393721 | 2018-02-20 00:01:44 | 2018-02-21 02:32:08 | 1d 02h |
| 2393722 | 2018-02-20 00:02:56 | 2018-02-21 00:32:08 | 1d 00h |
| 2393723 | 2018-02-20 00:04:12 | 2018-02-20 21:32:07 | 0d 21h |
| 2393724 | 2018-02-20 00:04:50 | 2018-02-20 23:47:05 | 0d 23h |
| 2393725 | 2018-02-20 00:07:07 | 2018-02-20 17:32:03 | 0d 17h |
| 2393726 | 2018-02-20 00:13:41 | 2018-02-21 00:32:08 | 1d 00h |
| 2393727 | 2018-02-20 00:14:25 | 2018-02-21 00:32:08 | 1d 00h |
| 2393728 | 2018-02-20 00:16:47 | 2018-02-20 23:32:04 | 0d 23h |
| 2393729 | 2018-02-20 00:17:19 | 2018-02-20 23:47:09 | 0d 23h |
| 2393730 | 2018-02-20 00:17:59 | 2018-02-21 01:32:08 | 1d 01h |
+---------+---------------------+---------------------+------------------+