mysql算术运算（减法）最后一天的第一天（周、月）

我得到了一个MySQL tbl，带有一些列，每5分钟插入一行，其中包含3个值

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

为了在图表中显示此信息，例如每周功耗，我需要选择allready工作的第一个和最后一个，但必须从第一个中减去最后一个，并仅显示星期二结果&werk的日期

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

我花了几天时间寻找解决办法，但没有成功。

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希望您能帮助我。

如果您对您的查询感到满意，那么您可以通过再次嵌套它来进行计算，如下所示：

第二个视图将第一个视图与其自身连接起来，为您提供一周的开始和结束范围：

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

您可以在此处看到结果：。请注意，如果一周尚未结束，我将使用now获取当前日期和时间

从那里，您可以将视图与原始表连接起来，并使用带有分组的“最小”和“最大”函数来获取起始值和结束值“wert”，并根据需要对其进行任何计算

下面是一个例子：

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我希望这会有所帮助。

返回日期平均值的查询对您有用吗？不，不太有用。。。谢谢！非常感谢你。。。第一个和第二个解决方案对我有效@Saschachmitt，太好了。如果您对答案感到满意，请将其标记为已接受。

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

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

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