sql - 使用 SQL 的日期和时间跨度

Question

给出下表和数据。

 create table prices
(productKey int
,PriceType char(10)
,BeginDate date
,EndDate date
,price decimal(18,2))

insert into prices(productKey, PriceType,BeginDate,EndDate, price)
values
(1,'LIST','1-1-2010','1-15-2010',10),
(1,'LIST','1-16-2010','10-15-2010',20),
(1,'DISCOUNT','1-10-2010','1-15-2010',-5),
(2,'LIST','2-1-2010','10-15-2010',30),
(2,'LIST','10-16-2010','1-1-9999',35),
(2,'DISCOUNT','2-10-2010','10-25-2010',-10),
(2,'LIST','1-1-2010','1-15-2010',10),
(3,'DISCOUNT','1-12-2010','1-1-9999',-5),
(3,'LIST','1-16-2010','1-1-9999',10)

我需要将记录插入到计算每个时间段的实际价格（列表折扣）的同一个表中。

例如，对于产品 1，我应该有以下“实际”记录

Begin     End      Price
1-1-2010  1-9-2010 10
1-10-2010  1-15-2010 5
1-16-2010  10-15-2010 20

我有点想出在标价范围内开始打折的任何事情，但我对其他任何事情都不知所措。

谢谢您的帮助

编辑

每个 ProductKey 可以有多个折扣，但折扣期不会重叠。所以你可以有一个 2010 年的，另一个 2012 年的，但不能有 2 个 2010 年的。

另外，如果有人能想出一个更好的标题，请这样做。在这一点上，我可怜的大脑完全受到挑战。

编辑2

它是 SQL 服务器 2008R2。我喜欢一个漂亮的基于集合的答案（或者让我朝那个方向开始的人），但也会对有效的光标解决方案感到满意。

score 3 · Accepted Answer

聪明的谜题。

您需要重建所有时间跨度。为此，我从价格范围中取出所有日期并重建可能的日期范围。

with alldates as (select d.*, ROW_NUMBER() over (partition by productkey order by thedate) as seqnum
                  from ((select productkey, BeginDate as thedate from prices)
                        union all
                        (select productkey, enddate as thedate from prices)
                       ) d
                 ),
     datepair as (select d1.productkey, d1.thedate as BeginDate, d2.thedate as EndDate
                  from alldates d1 left outer join 
                       alldates d2
                       on d1.seqnum = d2.seqnum - 1 and d1.productKey = d2.productKey
                 )
select dp.productkey, dp.BeginDate, dp.EndDate, SUM(p.price)
from datepair dp join
     prices p
     on dp.productkey = p.productkey and
         dp.BeginDate >= p.BeginDate and
        dp.EndDate <= p.EndDate
group by dp.productkey, dp.BeginDate, dp.EndDate
order by 1, 2, 3

我已经考虑了更多。上面的基本思路是正确的。基本思想是将时间维度分解为区间，其中列表和折扣在整个区间内是恒定的。问题是如何创建这些间隔，它们在 datepairs 别名中。

这些间隔只有几个规则：

日期对间隔可以在任何时间段开始时开始。
日期对间隔可以在任何时间段结束后的一天开始。
日期对间隔可以在任何时间段结束时结束
日期对间隔可以在任何时间段开始前一天结束

一旦我们有了间隔，就可以简单地加入该期间的适当标价和折扣。以下查询使用此逻辑：

with begindates as (select distinct productKey, thedate
                    from ((select productkey, BeginDate as thedate from prices)
                          union all
                          (select productkey, dateadd(d, 1, enddate) as thedate from prices)
                         ) d
                     ),
     enddates as (select distinct productKey, thedate
                  from ((select productkey, DATEADD(d, -1, begindate) as thedate from prices)
                        union all
                        (select productkey, enddate as thedate from prices)
                       ) d
                 ),
     datepair as (select *
                  from (select d1.productkey, d1.thedate as BeginDate,
                               MIN(d2.thedate) as EndDate
                        from begindates d1 left outer join 
                             enddates d2
                             on d1.productKey = d2.productKey and d1.thedate < d2.thedate
                        group by d1.productkey, d1.thedate
                       ) t
                  where BeginDate <> EndDate
                 )
select dp.productkey, dp.BeginDate, dp.EndDate, SUM(p.price)
from datepair dp join
     prices p
     on dp.productkey = p.productkey and
         dp.BeginDate >= p.BeginDate and
        dp.EndDate <= p.EndDate
group by dp.productkey, dp.BeginDate, dp.EndDate
order by 1, 2, 3

score 0 · Accepted Answer

我相信这应该给你你正在寻找的东西。关键是计算单个产品的所有唯一日期范围而不重叠。

-- Get all end dates
SELECT ROW_NUMBER() OVER (PARTITION BY ProductKey ORDER BY EndDate) RowNum, ProductKey, EndDate 
INTO #EndDates 
FROM (SELECT ProductKey, EndDate
    FROM prices
    UNION
    SELECT  ProductKey, DATEADD(d, -1, BeginDate) AS EndDate
    FROM prices) endDates
ORDER BY EndDate

-- Get all unique date ranges with no overlap
SELECT a.ProductKey, DATEADD(d, 1, a.EndDate) BeginDate, b.EndDate
INTO #DateRange
FROM #EndDates a
    INNER JOIN #EndDates b
        ON a.RowNum = b.RowNum - 1
        AND a.ProductKey = b.ProductKey
ORDER BY productkey, enddate

-- Get actual price
SELECT d.ProductKey, d.BeginDate, d.EndDate, SUM(Price) ActualPrice 
FROM prices p
    INNER JOIN #DateRange d
        ON p.ProductKey = d.ProductKey
        AND p.BeginDate <= d.EndDate
        AND p.EndDate >= d.BeginDate
GROUP BY d.ProductKey, d.BeginDate, d.EndDate
ORDER BY d.ProductKey, d.BeginDate, d.EndDate

-- Clean up
DROP TABLE #EndDates
DROP TABLE #DateRange

score 0 · Accepted Answer

为了让这更有趣，我添加了另一个与之前的部分价格相同的部分，以确保我将它们分开：

INSERT INTO prices(productKey, PriceType,BeginDate,EndDate, price)
VALUES (1, 'DISCOUNT', '5-2-2010', '5-8-2010', -15)

我还有一个名为的表tblDates，由以下内容填充：

INSERT dbo.tblDates (date1)
SELECT TOP(65536) ROW_NUMBER()OVER(ORDER BY v1.number)-1
FROM master.dbo.spt_values v1 CROSS APPLY master.dbo.spt_values v2 WHERE v1.type='p' and v2.type='p'
GO

我在这里给出的脚本不需要那个。但是拥有它不会影响速度，也不会真正占用那么多空间。这是我的答案：

DECLARE @InitialBeginDate DATE
    , @FinalEndDate DATE;

SELECT
    @InitialBeginDate = MIN(BeginDate)
    , @FinalEndDate = MAX(EndDate)
FROM prices
WHERE productKey = @ProductKey


CREATE TABLE #Dates
(
    DateValue DATE NOT NULL
)

INSERT #Dates (DateValue)
SELECT
    DATEADD(DAY, V.number, @InitialBeginDate)
FROM master..spt_values V
WHERE V.[type] = 'P'
    AND V.number BETWEEN 0 AND DATEDIFF(DAY, @InitialBeginDate, @FinalEndDate)

;WITH MergedDays AS
(
    SELECT
        ListedDates.DateValue
        , SUM(P.price) PriceOnDate
        , DATEDIFF(DAY, @InitialBeginDate, ListedDates.DateValue) - DENSE_RANK() OVER(PARTITION BY SUM(P.price) ORDER BY ListedDates.DateValue, SUM(P.price)) DateGroup
    FROM #Dates ListedDates
    INNER JOIN prices P
        ON P.BeginDate <= ListedDates.DateValue
        AND P.EndDate >= ListedDates.DateValue
        AND P.productKey = @ProductKey
    GROUP BY
        ListedDates.DateValue
)
SELECT
    MIN(DateValue) AS SegmentBeginDate
    , MAX(DateValue) AS SegmentEndDate
    , MAX(PriceOnDate) AS SegmentPrice -- This is just to collapse it, it'll be the same for all records.
FROM MergedDays
GROUP BY DateGroup
ORDER BY SegmentBeginDate

DROP TABLE #Dates

There are a couple other answers now, so this is just another way of doing things; there are many.

sql - 使用 SQL 的日期和时间跨度

3 回答 3

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