Question

我有一个MySQL tbl，有一些colums，每5分钟一次。插入一个带有3个值的新行

1. Auto inc. curent Date Unix timestamp --> date
2. power consumption absolut --> wert01
3. Power Generation absolut --> wert02

要在图表中显示此信息，对于每周功耗的示例，我需要选择已经完成的第一个和最后一个，但是必须从第一个中显示最后一个并且仅显示第二个结果＆amp;这个混蛋的日子。

SELECT
  (SELECT wert01
   FROM sml_splitt
   WHERE date >= curdate() - INTERVAL DAYOFWEEK(curdate())+6 DAY
     AND date < curdate() - INTERVAL DAYOFWEEK(curdate()) DAY
   ORDER BY date DESC LIMIT 1) AS 'last',

  (SELECT wert01
   FROM sml_splitt
   WHERE date >= curdate() - INTERVAL DAYOFWEEK(curdate())+6 DAY
     AND date < curdate() - INTERVAL DAYOFWEEK(curdate()) DAY
   ORDER BY date LIMIT 1) AS 'lirst

我正在寻找一些寻找解决方案的日子，但没有成功。 Hopfuly，你可以帮助我。

Answer 1

如果您对查询感到满意，那么您可以再次将其嵌套一次来进行数学计算：http://sqlfiddle.com/#!9/515ef/1

select t1.last, t1.first, t1.last - t1.first as result
from (
  select (
    select wert01
    from sml_splitt
    where dt >= curdate() - interval dayofweek(curdate()) + 6 day
    and dt < curdate() - interval dayofweek(curdate()) day
    order by dt desc limit 1
  ) as 'last',
  (
    select wert01
    from sml_splitt
    where dt >= curdate() - interval dayofweek(curdate()) + 6 day
    and dt < curdate() - interval dayofweek(curdate()) day
    order by dt limit 1
  ) as 'first'
) t1
;

如果您确实希望按周使用此数据进行报告，请让我提出一些观点。第一个将为您提供所有明确的周日开始日期：

create view v1 as
select date(dt) as week_begins
from sml_splitt
where dayofweek(dt) = 1
group by week_begins

第二个视图将第一个视图与自身连接起来，为您提供一周开始和一周结束范围：

create view v2 as
select t1.week_begins, coalesce(t2.week_begins,now()) as week_ends
from v1 t1
left join v1 t2
on t2.week_begins = t1.week_begins + interval 7 day

您可以在此处查看结果：http://sqlfiddle.com/#!9/a4d1b3/2。请注意，如果本周尚未结束，我会使用now()来获取当前日期和时间。

在那里，您可以将自己的观点与原始表格结合使用，并使用min()和max()功能进行分组，以获得开始和结束时间#t;＆＃39;值和你喜欢的任何计算。

以下是一个例子：http://sqlfiddle.com/#!9/a4d1b3/6

select week_begins, week_ends,
min(wert01) as start_wert01,
max(wert01) as end_wert01,
max(wert01) - min(wert01) as power_consumed,
min(wert02) as start_wert02,
max(wert02) as end_wert02,
max(wert02) - min(wert02) as power_generated,
(max(wert02) - min(wert02)) - (max(wert01) - min(wert01)) as net_generated
from v2
inner join sml_splitt
on sml_splitt.dt >= v2.week_begins
and sml_splitt.dt < v2.week_ends
group by week_begins

我希望有所帮助。

mysql arithmetik操作（减法）last - 当天的第一天（周，月）

1 个答案: