SQL Server：如何使用LAG在特定条件下获取上一行的值

Question

我希望新列始终显示Type不为null的上一条记录的Price值。到目前为止，我可以接受当前结果，但最后一行显示的是之前2行的值。最后一行应显示0.666，而不是1.66。

我使用的是我在此处找到的示例，并添加了几行进行测试，以查看它是否始终可以正常运行以达到我的目的。

我的数据大于示例，并且可能有多个记录，其中Type不为null。看来，如果group_nbr碰巧返回与其他组相同的数字，则会引起问题...

我正在使用SQL Server 2017：

DECLARE @a TABLE ( number int, price money, type varchar(2),
                   date date, time time)
INSERT @a VALUES
(23454,1.50, NULL,'2014/02/02','07:00:02'),
(23455,1.60, NULL,'2014/02/02','07:05:02'),
(23456,0.665,'SV','2014/02/02','07:50:48'),
(23457,1.3  ,NULL,'2014/02/02','07:50:45'),
(23658,2.4  ,NULL,'2014/02/02','07:50:47'),
(23660,2.4  ,NULL,'2014/02/02','07:50:49'),
(23465,0.668,'SV','2014/02/02','07:50:46'),
(23467,0.666,'SV','2014/02/02','08:50:40'),
(23668,1.4  ,NULL,'2014/02/02','09:50:49'),
(23466,1.66, NULL,'2014/02/02','08:36:34');

; WITH a AS 
(
     SELECT 
         *,
         ROW_NUMBER() OVER (ORDER BY [date], [time] ) x, 
         ROW_NUMBER() OVER (PARTITION BY CASE WHEN [type] IS NOT NULL THEN 1 ELSE 0 END ORDER BY [date], [time]) y 
     FROM 
         @a
), b AS 
(
     SELECT 
         *,
         x - y as group_nbr,
         ROW_NUMBER() OVER(PARTITION BY x-y ORDER BY x ASC) z1 
     FROM
         a
)
SELECT 
    *,
    CASE 
       WHEN [type] IS NOT NULL 
          THEN LAG(price, z1) OVER (PARTITION BY [type] ORDER BY x) 
          ELSE LAG(price, z1) OVER (ORDER BY x) 
    END
FROM
    b
ORDER BY 
    x

最后一条记录应返回0.666，而不是1.66。

输出应始终返回Type不为null的行的先前价格值。

Answer 1

尝试一下：

WITH DataSource as
(
    SELECT *
      ,COUNT(type) OVER (ORDER BY [date], [time] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)  - MAX(IIF(type is null, 0, 1)) OVER (ORDER BY [date], [time] ROWS BETWEEN 0 FOLLOWING AND 0 FOLLOWING) groupID
    FROM @a
), Groups (groupID, value) AS
(
    SELECT groupID
          ,MAX(IIF(type is not null, price, NULL))
    FROM DataSource
    GROUP BY groupID
)
SELECT *
FROM DataSource A
LEFT JOIN Groups B
    ON A.groupID = B.[groupID] + 1

这个想法是将行分成几组，然后获得该组的价格（这将是type不为空的价格。然后，只需联接表并获得price上一组的人。