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我有一个包含三列的 SQL Server 2008+ 表：

Delta (float)
Rate (float)
and Date (datetime)

我需要能够为给定日期的所有增量生成一个费率表，并在必要时进行插值。我可以对 Delta 的单个值执行此操作，但同时为所有 Delta 执行此操作让我感到难过。

一些示例数据是

Rate    ForwardDate             Delta
1.3528  2013-09-30 00:00:00.000 -0.9
1.3528  2013-09-30 00:00:00.000 -0.75
1.3528  2013-09-30 00:00:00.000 -0.5
1.3528  2013-09-30 00:00:00.000 -0.25
1.3528  2013-09-30 00:00:00.000 -0.1
7.121   2013-10-30 00:00:00.000 -0.9
7.152   2013-10-30 00:00:00.000 -0.75
7.387   2013-10-30 00:00:00.000 -0.5
7.972   2013-10-30 00:00:00.000 -0.25
8.564   2013-10-30 00:00:00.000 -0.1
6.972   2013-12-30 00:00:00.000 -0.9
7.035   2013-12-30 00:00:00.000 -0.75
7.405   2013-12-30 00:00:00.000 -0.5
8.275   2013-12-30 00:00:00.000 -0.25
9.187   2013-12-30 00:00:00.000 -0.1
7.185   2014-03-30 00:00:00.000 -0.9
7.321   2014-03-30 00:00:00.000 -0.75
7.82    2014-03-30 00:00:00.000 -0.5
8.919   2014-03-30 00:00:00.000 -0.25
10.075  2014-03-30 00:00:00.000 -0.1
7.755   2014-09-30 00:00:00.000 -0.9
7.92    2014-09-30 00:00:00.000 -0.75
8.485   2014-09-30 00:00:00.000 -0.5
9.76    2014-09-30 00:00:00.000 -0.25
11.175  2014-09-30 00:00:00.000 -0.1

到目前为止，我想出的最好的是：

(
SELECT Delta, ( DATEDIFF(d, @PreviousDate, @ForwardDate) * NextRate 
                   + DATEDIFF(d, @ForwardDate, @NextDate) * PreviousRate
                   ) / DATEDIFF(d, @PreviousDate, @NextDate) AS Rate

FROM 
 (SELECT Main.Delta AS Delta, PR.Rate AS PreviousRate, NR.Rate AS NextRate
  FROM @RatesTable Main
  INNER JOIN @RatesTable PR on PR.Delta = Main.Delta AND PR.ForwardDate = @PreviousDate
  INNER JOIN @RatesTable NR on NR.Delta = Main.Delta AND NR.ForwardDate = @NextDate) AS PrevNextRateTable);

Where@NextDate和@PreviousDateare 从较早的计算中确定表中最接近测试日期的日期 ( @ForwardDate)。这是必需的，因为我们需要确保所有插值都在相同的两个日期之间（如果其中一个日期缺少一个值，则给出 NULL）。我不想在每个 delta 值的不同日期之间进行插值。

这给了我似乎是结果的交叉连接

Delta   Rate
-0.9    7.742609
-0.75   7.906979
-0.5    8.470543
-0.25   9.741717
-0.1    11.15109
-0.9    7.742609
-0.75   7.906979
-0.5    8.470543
-0.25   9.741717
-0.1    11.15109
-0.9    7.742609
-0.75   7.906979
-0.5    8.470543
-0.25   9.741717
-0.1    11.15109
-0.9    7.742609
-0.75   7.906979
-0.5    8.470543
-0.25   9.741717
-0.1    11.15109
-0.9    7.742609
-0.75   7.906979
-0.5    8.470543
-0.25   9.741717
-0.1    11.15109

每个增量有 5 个结果。如有必要，我可以SELECT DISTINCT这样做，但不禁认为这是作弊，我做错了什么。有没有更好的方法来消除对 的需要SELECT DISTINCT？

4

1 回答 1

1

尝试这个：

SELECT Delta, ( DATEDIFF(d, @PreviousDate, @ForwardDate) * NextRate 
                   + DATEDIFF(d, @ForwardDate, @NextDate) * PreviousRate
                   ) / DATEDIFF(d, @PreviousDate, @NextDate) AS Rate

FROM 
 (SELECT Main.Delta AS Delta, PR.Rate AS PreviousRate, NR.Rate AS NextRate
  FROM @RatesTable Main
  INNER JOIN @RatesTable PR on PR.Delta = Main.Delta AND PR.ForwardDate = @PreviousDate
  INNER JOIN @RatesTable NR on NR.Delta = Main.Delta AND NR.ForwardDate = @NextDate) AS PrevNextRateTable
GROUP BY Delta,
        ( DATEDIFF (d, @PreviousDate, @ForwardDate) * NextRate 
                   + DATEDIFF(d, @ForwardDate, @NextDate) * PreviousRate
                   ) / DATEDIFF(d, @PreviousDate, @NextDate) ;
于 2014-04-03T11:03:53.870 回答