2

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 |
+----+-------------------------+----------+----------+

我想对数据进行排序DateReceived并求和Quantity。但是，我想对Quantity分组求和VendorID，只要它们相邻，如下面的示例输出。

+----------+----------+
| 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遍历它们来做到这一点。这是目前我想消除的软件瓶颈。

生成所需输出的 ​​MS Sql Server 查询是什么？

PS。对更好的标题有什么建议吗？

4

2 回答 2

4

在SQL Server 2005+ 中，您可以这样做：

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)

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

sql fiddle demo

顺便说一句，看起来你的输出不正确。正确的一个是：

+----------+----------+
| VendorID | Quantity |
+----------+----------+
|        1 |        6 |
|        2 |        3 |
|        3 |        1 |
|        1 |        9 |
|        3 |        9 |
|        2 |        2 |
+----------+----------+
于 2013-10-24T19:02:34.157 回答
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 演示

于 2013-10-24T19:17:24.757 回答