Question

我有这样的数据：

+----+-------------------------+----------+----------+
| ID |      DateReceived       | Quantity | VendorID |
+----+-------------------------+----------+----------+
|  1 | 2010-08-09 06:53:44.783 |        2 |        1 |
|  2 | 2010-08-01 13:31:26.893 |        1 |        1 |
|  3 | 2010-07-26 07:52:29.403 |        2 |        1 |
|  4 | 2011-03-22 13:31:11.000 |        1 |        2 |
|  5 | 2011-03-22 13:31:11.000 |        1 |        2 |
|  6 | 2011-03-22 11:27:01.000 |        1 |        2 |
|  7 | 2011-03-18 09:04:58.000 |        1 |        1 |
|  8 | 2011-12-17 08:21:29.000 |        1 |        3 |
|  9 | 2012-08-10 10:55:20.000 |        9 |        3 |
| 10 | 2012-08-02 20:18:10.000 |        5 |        1 |
| 11 | 2012-07-12 20:44:36.000 |        3 |        1 |
| 12 | 2012-07-05 20:45:29.000 |        1 |        1 |
| 13 | 2013-03-22 13:31:11.000 |        1 |        2 |
| 14 | 2013-03-22 13:31:11.000 |        1 |        2 |
+----+-------------------------+----------+----------+

我希望按DateReceived对数据进行排序，并将Quantity求和。但是，我想将Quantity归为VendorID，只要它们相邻，就像下面的示例输出一样。

+----------+----------+
| VendorID | Quantity |
+----------+----------+
|        1 |        5 |
|        2 |        3 |
|        1 |        1 |
|        3 |       10 |
|        1 |        9 |
|        2 |        2 |
+----------+----------+

我目前正在通过加载所有行并在我的应用程序代码中浏览它来完成此操作。这是我想要消除的软件瓶颈。

什么是MS Sql Server查询以生成所需的输出？

PS。有关更好标题的任何建议吗？

Answer 1

在 SQL Server 2005 + 中，您可以执行以下操作：

with cte as (
    select
        VendorID, Quantity,
        row_number() over(partition by VendorID order by DateReceived) as rn1,
        row_number() over(order by DateReceived) as rn2
    from Table1
)
select
    VendorID, sum(Quantity) as Quantity
from cte 
group by VendorID, rn2 - rn1
order by min(rn2)

<强> sql fiddle demo

在 SQL Server 2012 中，您可以使用lag（）函数：

with cte as (
    select
        VendorID, Quantity, DateReceived,
        case when lag(VendorID) over(order by DateReceived) <> VendorID then 1 else 0 end as rn
    from Table1
), cte2 as (
    select
        VendorID, Quantity, sum(rn) over(order by DateReceived) as s
    from cte 
)
select
    VendorID, sum(Quantity)
from cte2
group by VendorID, s
order by s asc

<强> sql fiddle demo

BTW，看起来你的输出不正确。正确的是：

+----------+----------+
| VendorID | Quantity |
+----------+----------+
|        1 |        6 |
|        2 |        3 |
|        3 |        1 |
|        1 |        9 |
|        3 |        9 |
|        2 |        2 |
+----------+----------+

Answer 2

试试这个解决方案：

SELECT  z.VendorID, z.GroupID,
        MIN(z.DateReceived) AS DateReceivedStart, 
        MAX(z.DateReceived) AS DateReceivedStop, 
        SUM(z.Quantity) AS SumOfQuantity
FROM
(
    SELECT  y.VendorID,
            y.RowNum1 - y.RowNum2 AS GroupID,
            y.DateReceived,
            y.Quantity
    FROM 
    (
        SELECT  *, ROW_NUMBER() OVER(ORDER BY x.DateReceived) AS RowNum1,
                   ROW_NUMBER() OVER(ORDER BY x.VendorID, x.DateReceived) AS RowNum2
        FROM    @MyTable x
    ) y
) z
GROUP BY z.VendorID, z.GroupID
ORDER BY DateReceivedStart

SQL Fiddle demo

Sql查询组由连续的外键值组成

2 个答案: