Sql 从一类不完整的数据集中获取最大值

许多设备返回一个值。只有在发生更改时，此值才会存储在表中：

Device  Value  Date
B       5      2017-07-01
C       2      2017-07-01
A       3      2017-07-02
C       1      2017-07-04
A       6      2017-07-04

值可以在任何日期进入表格，即日期不会连续递增。多个设备可能在同一日期存储其值

请注意，尽管表中每个日期通常只有几个设备，但所有设备在该日期实际上都有一个值：它是在此之前存储的最新设备。例如，在2017-07-02上，只有设备A存储了一个值。该日期的B和C值为2017-07-01存储的值；这些在-02上仍然有效，只是没有更改

为了检索给定日期（例如2017-07-04）上所有设备的值，我使用以下方法：

---

select device，value from data internal join select device，maxdate as date from data where date我认为最通用的方法是对每个日期使用单独的查询。根据数据库的不同，肯定有更简单的方法。但获得一个适用于SQLite、MariaDB和Postgres的应用程序不会使用任何复杂的功能：

select '2017-07-01' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-01' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date
union all
select '2017-07-02' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-02' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date
select '2017-07-03' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-03' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date
select '2017-07-04' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-04' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date;

---

这应该是你问题的解决方案。 它应该是跨数据库的，因为大多数数据库都支持OVER子句。 您应该使用查询中的所有datesALL_DATE创建一个表，否则每个数据库都有一种特定的方法在没有表的情况下完成此操作

WITH GROUPED_BY_DATE_DEVICE AS (
  SELECT DATE, DEVICE, SUM(VALUE) AS VALUE FROM DEVICE_INFO
  GROUP BY DATE, DEVICE
), GROUPED_BY_DATE AS (
  SELECT A.DATE, MAX(VALUE) AS VALUE
  FROM ALL_DATE A
  LEFT JOIN GROUPED_BY_DATE_DEVICE B
  ON A.DATE = B.DATE
  GROUP BY A.DATE
)
SELECT DATE, MAX(VALUE) OVER (ORDER BY DATE) AS MAX_VALUE 
FROM GROUPED_BY_DATE
ORDER BY DATE;

---

提示：与日历表进行外部联接以包括缺少的日期。

Date        max(value)
2017-07-01  5
2017-07-02  5
2017-07-03  5
2017-07-04  6

select '2017-07-01' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-01' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date
union all
select '2017-07-02' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-02' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date
select '2017-07-03' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-03' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date
select '2017-07-04' as date, max(data.value)
from data inner join
     (select device, max(date) as date 
      from data
      where date <= '2017-07-04' group by device
     ) latestdate
     on data.device = latestdate.device and data.date = latestdate.date;

WITH GROUPED_BY_DATE_DEVICE AS (
  SELECT DATE, DEVICE, SUM(VALUE) AS VALUE FROM DEVICE_INFO
  GROUP BY DATE, DEVICE
), GROUPED_BY_DATE AS (
  SELECT A.DATE, MAX(VALUE) AS VALUE
  FROM ALL_DATE A
  LEFT JOIN GROUPED_BY_DATE_DEVICE B
  ON A.DATE = B.DATE
  GROUP BY A.DATE
)
SELECT DATE, MAX(VALUE) OVER (ORDER BY DATE) AS MAX_VALUE 
FROM GROUPED_BY_DATE
ORDER BY DATE;