我有一组这样的供应操作：
op_type | time_stamp | product | in | out
----------------------------------------------------
01 | 08:00:00 | p1 | 50 | 0
02 | 08:01:00 | p1 | 0 | 10
02 | 08:02:00 | p1 | 0 | 35
03 | 08:03:00 | p1 | 0 | 5
01 | 08:04:00 | p1 | 60 | 0
02 | 08:09:00 | p1 | 0 | 15
01 | 08:10:00 | p1 | 30 | 0
02 | 08:11:00 | p1 | 0 | 20
01 | 08:00:00 | p2 | 100 | 0
02 | 08:01:00 | p2 | 0 | 20
02 | 08:02:00 | p2 | 0 | 45
03 | 08:03:00 | p2 | 0 | 15
01 | 08:03:10 | p2 | 60 | 0
01 | 08:04:00 | p2 | 5 | 0
02 | 08:09:00 | p2 | 0 | 30
01 | 08:10:00 | p2 | 30 | 0
02 | 08:11:00 | p2 | 0 | 10
我想要的是从给定时间开始选择SUM(in) group by product将涵盖SUM(out) group by product的列表
EX：对于time_stamp>'08:05:00'我有：
SUM(out) for p1=35所以总和
和SUM(out) for p2=40
所以我想要的列表是
op_type | time_stamp | product | in | out
----------------------------------------------------
01 | 08:04:00 | p1 | 60 | 0
02 | 08:09:00 | p1 | 0 | 15
01 | 08:10:00 | p1 | 30 | 0
02 | 08:11:00 | p1 | 0 | 20
01 | 08:03:10 | p2 | 60 | 0
01 | 08:04:00 | p2 | 5 | 0
02 | 08:09:00 | p2 | 0 | 30
01 | 08:10:00 | p2 | 30 | 0
02 | 08:11:00 | p2 | 0 | 10
我所做的就是我在结果中添加了一列，其中一列的总和为out而另一列的累计总和为in
SELECT B.*,C.sum_out FROM (SELECT A.*,SUM(in) OVER (PARTITION BY product ORDER BY time_stamp desc) AS sum_in FROM table A) B
LEFT OUTER JOIN
SELECT C.* FROM (SELECT product,SUM(out) AS sum_out from table GROUP BY product WHERE time_stamp>'08:05:00') C
ON B.product=C.product
所以我得到了：
op_type | time_stamp | product | in | out | sum_in | sum_out
-------------------------------------------------------------------------
01 | 08:00:00 | p1 | 50 | 0 | 140 | 35
02 | 08:01:00 | p1 | 0 | 10 | 90 | 35
02 | 08:02:00 | p1 | 0 | 35 | 90 | 35
03 | 08:03:00 | p1 | 0 | 5 | 90 | 35
01 | 08:04:00 | p1 | 60 | 0 | 90 | 35
02 | 08:09:00 | p1 | 0 | 15 | 30 | 35
01 | 08:10:00 | p1 | 30 | 0 | 30 | 35
02 | 08:11:00 | p1 | 0 | 20 | 0 | 35
01 | 08:00:00 | p2 | 100 | 0 | 195 | 40
02 | 08:01:00 | p2 | 0 | 20 | 95 | 40
02 | 08:02:00 | p2 | 0 | 45 | 95 | 40
03 | 08:03:00 | p2 | 0 | 15 | 95 | 40
01 | 08:03:10 | p2 | 60 | 0 | 95 | 40
01 | 08:04:00 | p2 | 5 | 0 | 35 | 40
02 | 08:09:00 | p2 | 0 | 30 | 30 | 40
01 | 08:10:00 | p2 | 30 | 0 | 30 | 40
02 | 08:11:00 | p2 | 0 | 10 | 0 | 40
如果我添加条款WHERE B.sum_in<=C.sum_out我将能够获得
op_type | time_stamp | product | in | out | sum_in | sum_out
-------------------------------------------------------------------------
02 | 08:09:00 | p1 | 0 | 15 | 30 | 35
01 | 08:10:00 | p1 | 30 | 0 | 30 | 35
02 | 08:11:00 | p1 | 0 | 20 | 0 | 35
01 | 08:04:00 | p2 | 5 | 0 | 35 | 40
02 | 08:09:00 | p2 | 0 | 30 | 30 | 40
01 | 08:10:00 | p2 | 30 | 0 | 30 | 40
02 | 08:11:00 | p2 | 0 | 10 | 0 | 40
所以每个产品还需要一行。
任何想法如何执行？ PS：我正在使用SQL SERVER 2012。
答案 0 :(得分：2)
我起草了一个可能的解决方案，您可以使用子查询来获取所有累积的交易
Select time_stamp,
(Select Sum(SQ.InEntry - SQ.OutEntry) from Table_1 AS SQ where T.time_stamp > SQ.time_stamp AND T.product = SQ.product group by product) +
InEntry - OutEntry
, T.InEntry
, T.OutEntry
, product
from Table_1 T
order by product,time_stamp
为了生成随机数据，我使用了这个语句
declare @FromDate date = '2016-01-01'
declare @ToDate date = '2016-12-31'
declare @product varchar(50)
set @product = ( select top 1 A.Prod from (Select 'A' AS Prod union all select 'B' AS Prod union all select 'C' AS Prod) AS A order by newid())
INSERT INTO [dbo].[Table_1]
([time_stamp]
,[product]
,[InEntry]
,[OutEntry])
Select
dateadd(day,
rand(checksum(newid()))*(1+datediff(day, @FromDate, @ToDate)),
@FromDate)
,@product
,0
,ABS(Checksum(NewID()) % 100)
INSERT INTO [dbo].[Table_1]
([time_stamp]
,[product]
,[InEntry]
,[OutEntry])
Select
dateadd(day,
rand(checksum(newid()))*(1+datediff(day, @FromDate, @ToDate)),
@FromDate)
,@product
,@product ,ABS(Checksum(NewID()) % 100)
,0
GO 40
答案 1 :(得分：0)
您似乎不按产品分组？ 无论如何这可能会有所帮助，你可以使用SUM＆amp; amp; GROUP BY对当时具有适当WHERE子句的内容进行分组。
但要得到你所要求的......
SUM（in）group by product将涵盖SUM（out）
....你可以使用HAVING .....
例如
SELECT Field1, Field2, SUM(IN), SUM(OUT)
FROM Table
WHERE Your_where_clause
GROUP BY Field1, Field2
HAVING SUM(IN) >= SUM(OUT)