[英]How to add or substract minutes from a timediff result in mysql
我正在尝试进行查询,以显示所有工作时间,天数和人员。 我有那种快感。
假设我有一个叫做uren的表:
rec_id | user_id | start (datetime) | eind (datetime)
我有一个叫用户的表
user_id | name |
在下面的查询中,我几乎拥有了我想要的所有信息。
select users.name, sec_to_time(SUM(TIME_TO_SEC(TIMEDIFF(uren.eind, uren.start)))),count(distinct(date(start))) as dagen
from uren, users
where date(uren.start) between CAST('2017-10-04 00:00:00' as Date) and CAST('2017-11-04 00:00:00' as DATE) and
uren.user_id = users.user_id
group by uren.user_id
ORDER BY name
这告诉我这个
Piet (name) 230 (hours total) 24(days worked)
现在是真正的问题:
我想减去每天工作少于5小时的30分钟。
我无能为力的atm。 有人可以帮忙吗
假设每个用户每天一行:
select users.name,
(sec_to_time(sum(time_to_sec(timediff(uren.eind, uren.start))) -
30 * 60 * sum(time_to_sec(timediff(uren.eind, uren.start)) < 5*60*60)
)
),
count(distinct(date(start))) as dagen
from uren join users
on uren.user_id = users.user_id
where date(uren.start) between '2017-10-04' and '2017-11-04'
group by uren.user_id
order by name;
如果某天有多个班次,则需要先按天汇总:
select u.name,
(sec_to_time(sum(day_secs) -
30 * 60 * sum(day_secs < 5*60*60)
)
),
count(*) as dagen
from (select uren.user_id, uren.name, date(uren.start),
sum(time_to_sec(timediff(uren.eind, uren.start))) as day_secs
from uren join
users
on uren.user_id = users.user_id
where uren.start >= '2017-10-04' and uren.start < '2017-11-05'
group by uren.user_id, date(uren.start)
) u
group by name
order by name
您没有线索是有充分的理由的。 这是因为您要问的内容很复杂。
您需要接受的第一件事(我的意思是绝望地接受)是您的功能取决于天数和用户,因此,首先需要对两者进行分组,然后在最终结果中仅按用户分组。 为此,在父级可以按用户分组之前,您将需要一个按天和用户分组的子查询。
这是我想出的...
SELECT
users.name,
sec_to_time(SUM(ur.timeWorked)) as tWorked,
SUM(ur.dagen) as Dagen
FROM users
INNER JOIN (
SELECT
user_id,
SUM(TIME_TO_SEC(TIMEDIFF(`eind`, `start`))) - (1800 *
IF(SUM(TIME_TO_SEC(TIMEDIFF(`eind`, `start`))) < 18000,1,0))
as `timeWorked`,
count(distinct(date(`start`))) as `dagen`
FROM uren
WHERE date(`start`)
BETWEEN CAST('2017-10-04 00:00:00' as Date)
AND CAST('2017-11-04 00:00:00' as DATE)
GROUP BY user_id, DAYOFYEAR(`start`)
) as ur ON ur.user_id = users.user_id
GROUP BY ur.user_id
ORDER BY name
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