Question

我需要从以下数据集中确定开始日期和结束日期：

item        date              cost         
12345    01/01/15             2.00
12345    01/02/15             2.00
12345    01/03/15             2.00
12345    01/04/15             2.00
12345    01/05/15             2.00
12345    01/06/15             2.00
12345    01/07/15             1.50
12345    01/08/15             1.50
12345    01/09/15             1.50
12345    01/10/15             1.50
12345    01/11/15             1.50
12345    01/12/15             1.50
12345    01/13/15             1.50
12345    01/14/15             2.00
12345    01/15/15             2.00
12345    01/16/15             2.00
12345    01/17/15             2.00
12345    01/18/15             2.00
12345    01/19/15             2.00
12345    01/20/15             2.00
12345    01/26/15             2.00
12345    01/27/15             2.00
12345    01/28/15             2.00
12345    01/29/15             2.00

如果可能，我想要的是以下输出：

item        start              end                  cost
12345    01/01/15             01/06/15             2.00
12345    01/07/15             01/13/15             1.50
12345    01/14/15             01/20/15             2.00
12345    01/26/15             01/29/15             2.00

基本上，任何时候更改，或任何给定成本的日期之间有超过4天的差距。

感谢。

Answer 1

这是你可以做到的一种方式：

DECLARE @myTable TABLE (item INT, date DATE, cost DECIMAL(5, 2));
INSERT @myTable VALUES (12345, '01/01/15', 2.00), (12345, '01/02/15', 2.00)
, (12345, '01/03/15', 2.00), (12345, '01/04/15', 2.00), (12345, '01/05/15', 2.00)
, (12345, '01/06/15', 2.00), (12345, '01/07/15', 1.50), (12345, '01/08/15', 1.50)
, (12345, '01/09/15', 1.50), (12345, '01/10/15', 1.50), (12345, '01/11/15', 1.50)
, (12345, '01/12/15', 1.50), (12345, '01/13/15', 1.50), (12345, '01/14/15', 2.00)
, (12345, '01/15/15', 2.00), (12345, '01/16/15', 2.00), (12345, '01/17/15', 2.00)
, (12345, '01/18/15', 2.00), (12345, '01/19/15', 2.00), (12345, '01/20/15', 2.00)
, (12345, '01/26/15', 2.00), (12345, '01/27/15', 2.00), (12345, '01/28/15', 2.00)
, (12345, '01/29/15', 2.00);

WITH CTE1 AS (
    SELECT item, date, cost, ROW_NUMBER() OVER (PARTITION BY item ORDER BY date) RN
    FROM @myTable)
, CTE2 AS (
    SELECT item, T.date TDate, T.cost Tcost, prevRow.date SDate, MD.maxDate, ROW_NUMBER() OVER (PARTITION BY item ORDER BY T.date) RN2
    FROM CTE1 T
    OUTER APPLY (
        SELECT date, cost
        FROM CTE1
        WHERE RN = T.RN - 1
        AND item = T.item
        ) prevRow
    OUTER APPLY (
        SELECT MAX(date) FROM CTE1
        ) MD(maxDate)
    WHERE CASE WHEN DATEDIFF(DAY, prevRow.date, T.date) <= 4 THEN T.date END IS NULL
    OR CASE WHEN prevRow.cost = T.cost THEN T.cost END IS NULL)
SELECT item, TDate startDate, ISNULL(nextRow.SDate, C.maxDate) endDate, TCost cost
FROM CTE2 C
OUTER APPLY (
    SELECT SDate
    FROM CTE2
    WHERE RN2 = C.RN2 + 1
    AND item = C.item) nextRow;

基本上，您希望将每一行与下一行进行比较，以检查是否存在成本差异或超过4天的日期差异。在较新版本的SQL Server（2012年起）中，您可以使用LEAD / LAG窗口函数，但我认为在SQL Server 2008中最简单的方法是为每行分配一个行号并使用apply。

CTE1指定行号。 CTE2用于与下一行进行比较。最后的select语句会拉出每个期间的开始和结束日期，其中费用会发生变化，或者日期跳过大于4天。

Answer 2

以下是@ZLK提供的变体，但仅使用连接而不应用。我们再次使用行号。第二个CTE获取第一行和最后一行以及每行有变化的行。最终查询负责选择正确的最终日期。

--set up data
declare @table table (item int, tdate date, cost decimal(5,2));
insert @table values
(12345,'01/01/15',2.00),
(12345,'01/02/15',2.00),
(12345,'01/03/15',2.00),
(12345,'01/04/15',2.00),
(12345,'01/05/15',2.00),
(12345,'01/06/15',2.00),
(12345,'01/07/15',1.50),
(12345,'01/08/15',1.50),
(12345,'01/09/15',1.50),
(12345,'01/10/15',1.50),
(12345,'01/11/15',1.50),
(12345,'01/12/15',1.50),
(12345,'01/13/15',1.50),
(12345,'01/14/15',2.00),
(12345,'01/15/15',2.00),
(12345,'01/16/15',2.00),
(12345,'01/17/15',2.00),
(12345,'01/18/15',2.00),
(12345,'01/19/15',2.00),
(12345,'01/20/15',2.00),
(12345,'01/26/15',2.00),
(12345,'01/27/15',2.00),
(12345,'01/28/15',2.00),
(12345,'01/29/15',2.00);
--select * from @table order by tdate;

--query
with cte as (
        select  *, row_number() over (partition by item order by tdate) row_num
        from    @table
),
cte2 as (
        select  c.item, c.tdate, c_pre.tdate pre_date, c_post.tdate post_date, c.cost, row_number() over (partition by c.item order by c.tdate) row_num
        from    cte c
        left    join cte c_pre
                on  c_pre.row_num = c.row_num - 1
        left    join cte c_post
                on  c_post.row_num = c.row_num + 1
        where   c_pre.row_num is null   --first row
        or      c_post.row_num is null  --last row
        or      c_pre.cost != c.cost    --cost difference
        or      datediff(d, c_pre.tdate, c.tdate) > 4   --more than 4 days
)
select  c.item, c.tdate start, 
        case when c_post.post_date is null then c_post.tdate else c_post.pre_date end [end],    --adjustment for last row
        c.cost
from    cte2 c
left    join cte2 c_post
        on  c_post.row_num = c.row_num + 1
where   c.post_date is not null
order   by c.tdate;

如何从单个列

2 个答案: