不同日期的最大值和最小值之差

Question

我会尽力解释这一点。

下面的代码执行以下操作：

从 ServiceLocation table 中查找服务地址。
查找服务类型（电力或水）。
查找过去多少天来提取数据。

一旦有了这个，它就会通过从一天的最小仪表读数中减去一天的最大仪表读数来计算 "daily usage"。

        (MAX(mr.Reading) - MIN(mr.Reading)) AS 'DaytimeUsage'

但是，我缺少的是前一天的最大读数和当天的最小读数。从数学上讲，这应该看起来像这样：

MAX（前一天阅读）- MIN（阅读日期阅读）

基本上，如果它返回 5 天，它应该踢出一个 table，内容如下：

Service Location | Read Date | Usage |

123 Main St | 4/20/15 | 12 |
123 Main St | 4/19/15 | 8 |
123 Main St | 4/18/15 | 6 |
123 Main St | 4/17/15 | 10 |
123 Main St | 4/16/15 | 11 |

其中 "Usage" 是我缺少的 'DaytimeUsage' + 用法（以及上面的问题）。例如，4/18/15 将是下面查询中的 'DaytimeUsage' 加上从 4/17/15 读取的 MAX 和从 4/18/15 读取的 MIN 之间的差异。

我不确定如何完成或是否可能。

SELECT 
    A.ServiceAddress AS 'Service Address', 
    convert(VARCHAR(10),A.ReadDate,101) AS 'Date', 
    SUM(A.[DaytimeUsage]) AS 'Usage' 

FROM 
(
    SELECT
        sl.location_addr AS 'ServiceAddress', 
        convert(VARCHAR(10),mr.read_date,101) AS 'ReadDate', 
        (MAX(mr.Reading) - MIN(mr.Reading)) AS 'DaytimeUsage'
    FROM
        DimServiceLocation AS sl
        INNER JOIN FactBill AS fb ON fb.ServiceLocationKey = sl.ServiceLocationKey
        INNER JOIN FactMeterRead as mr ON mr.ServiceLocationKey = sl.ServiceLocationKey
        INNER JOIN DimCustomer AS c ON c.CustomerKey = fb.CustomerKey
    WHERE 
        c.class_name = 'Tenant'
        AND sl.ServiceLocationKey = @ServiceLocation
        AND mr.meter_type = @ServiceType
    GROUP BY 
        sl.location_addr, 
        convert(VARCHAR(10),
        mr.read_date,101)
) A

WHERE A.ReadDate >= GETDATE()-@Days 
GROUP BY A.ServiceAddress, convert(VARCHAR(10),A.ReadDate,101)
ORDER BY convert(VARCHAR(10),A.ReadDate,101) DESC

Answer 1

我不确定我是否完全理解您的数据库结构，但我可能有解决方案，因此请随时编辑我的答案以适应或更正任何错误。

想法是为 table FactMeterRead 使用两个别名。 mrY（昨天的 Y）和 mrT（今天的 T）。并用 read_date 限制来区分它们。但是，我对您的 table 理解不够，无法编写功能齐全的查询。我希望你能通过这个例子理解这个想法。

SELECT
        sl.location_addr AS 'ServiceAddress', 
        convert(VARCHAR(10),mrT.read_date,101) AS 'ReadDate', 
        (MAX(mrY.Reading) - MIN(mrT.Reading)) AS 'DaytimeUsage'
    FROM
        DimServiceLocation AS sl
          INNER JOIN FactMeterRead as mrY ON mrY.ServiceLocationKey = sl.ServiceLocationKey
          INNER JOIN FactMeterRead as mrT ON mrT.ServiceLocationKey = sl.ServiceLocationKey
    WHERE mrY.read_date=DATE_SUB(mrT.read_date,1 DAY)

Answer 2

如果您使用的是 sql server 2005 以上版本，则可以使用 APPLY 运算符。这是文档的 link。 https://technet.microsoft.com/en-us/library/ms175156(v=sql.105).aspx APPLY 操作有两种形式：OUTER APPLY 和 CROSS APPLY - OUTER 类似于左联接，CROSS 类似于内联接。他们让您运行对返回的每一行进行一次查询。我设置了自己的示例，说明您要尝试做什么，就在这里，希望对您有所帮助。

http://sqlfiddle.com/#!6/fdb3f/1

CREATE TABLE SequencedValues (
  Location varchar(50) NOT NULL,
  CalendarDate datetime NOT NULL,
  Reading int
  )

INSERT INTO SequencedValues (
  Location,
  CalendarDate,
  Reading
  )
 SELECT
   'Address1',
   '4/20/2015',
   10
UNION SELECT
   'Address1',
   '4/19/2015',
   9
UNION SELECT
   'Address1',
   '4/19/2015',
   20
UNION SELECT
   'Address1',
   '4/19/2015',
   25
UNION SELECT
   'Address1',
   '4/18/2015',
   8
UNION SELECT
   'Address1',
   '4/17/2015',
   7
UNION SELECT
   'Address2',
   '4/20/2015',
   100
UNION SELECT
   'Address2',
   '4/20/2015',
   111
UNION SELECT
   'Address2',
   '4/19/2015',
   50
UNION SELECT
   'Address2',
   '4/19/2015',
   65


SELECT DISTINCT 
    sv.Location,
    sv.CalendarDate,
    sv_dayof.MINDayOfReading,
    sv_daybefore.MAXDayBeforeReading
FROM SequencedValues sv
OUTER APPLY (
    SELECT MIN(sv_dayof_inside.Reading) AS MINDayOfReading
    FROM SequencedValues sv_dayof_inside
    WHERE sv.Location = sv_dayof_inside.Location
    AND sv.CalendarDate = sv_dayof_inside.CalendarDate
    ) sv_dayof
OUTER APPLY (
    SELECT MAX(sv_daybefore_max.Reading) AS MAXDayBeforeReading
    FROM SequencedValues sv_daybefore_max
    WHERE sv.Location = sv_daybefore_max.Location
    AND sv_daybefore_max.CalendarDate IN (
        SELECT TOP 1 sv_daybefore_inside.CalendarDate
        FROM SequencedValues sv_daybefore_inside
        WHERE sv.Location = sv_daybefore_inside.Location
        AND sv.CalendarDate > sv_daybefore_inside.CalendarDate
        ORDER BY sv_daybefore_inside.CalendarDate DESC
        )
    ) sv_daybefore
ORDER BY
    sv.Location,
    sv.CalendarDate DESC

Answer 3

您似乎可以通过计算昨天和今天的 MAX 之间的差异来解决这个问题，但是这就是我的处理方式。相对于任何给定日期的前一天再次加入相同的 table，并且在您的内部查询中也加入 select Max/Min。此外，如果您将日期放在内部查询 where 子句中，您的数据集 return 会更快更小。

SELECT 
A.ServiceAddress AS 'Service Address', 
  convert(VARCHAR(10),A.ReadDate,101) AS 'Date', 
  SUM(A.[TodayMax]) - SUM(A.[TodayMin]) AS 'Usage',
  SUM(A.[TodayMax]) - SUM(A.[YesterdayMax]) AS 'Usage with extra bit you want' 
FROM 
(
SELECT
    sl.location_addr AS 'ServiceAddress', 
    convert(VARCHAR(10),mr.read_date,101) AS 'ReadDate', 
    MAX(mrT.Reading) AS 'TodayMax',
    MIN(mrT.Reading) AS 'TodayMin',
    MAX(mrY.Reading) AS 'YesterdayMax',
    MIN(mrY.Reading) AS 'YesterdayMin',        
FROM
    DimServiceLocation AS sl
    INNER JOIN FactBill AS fb ON fb.ServiceLocationKey = sl.ServiceLocationKey
    INNER JOIN FactMeterRead as mrT ON mrT.ServiceLocationKey = sl.ServiceLocationKey
    INNER JOIN FactMeterRead as mrY ON mrY.ServiceLocationKey = s1.ServiceLocationKey
                                    AND mrY.read_date = mrT.read_date -1)
    INNER JOIN DimCustomer AS c ON c.CustomerKey = fb.CustomerKey
WHERE 
    c.class_name = 'Tenant'
    AND sl.ServiceLocationKey = @ServiceLocation
    AND mr.meter_type = @ServiceType
    AND convert(VARCHAR(10), mrT.read_date,101) >= GETDATE()-@Days 
GROUP BY 
    sl.location_addr, 
    convert(VARCHAR(10),
    mr.read_date,101)

) A
GROUP BY A.ServiceAddress, convert(VARCHAR(10),A.ReadDate,101)
ORDER BY convert(VARCHAR(10),A.ReadDate,101) DESC

不同日期的最大值和最小值之差

Difference Max and Min from Different Dates

sql

sql-server

datetime

max

min

Service Location | Read Date | Usage |