我试图通过平均每天的结束数量，在参数@StartDate的日期范围内找到库存的平均数量。我有三个表：零件表，零件交易表和仓库表，在下面模拟。
PartNum | PartNum TranDate TranQty | PartNum OnHandQty
---------- | ------------------------------------ | --------------------
P1 | P1 6/28/2016 5 | P1 30
P2 | P1 6/26/2016 3 | P2 2
| P1 6/26/2016 -1 |
| P1 6/15/2016 2 |
| P2 6/15/2016 1 |
如果今天是2016年6月30日和@StartDate = 2016年6月1日，我希望得到如下结果：
PartNum AverageOnHand
------------------------
P1 22.9
P2 1.5
但是，我不知道什么功能最能让我得到一个合适的加权和，我可以除以日期的差异。是否有SumProduct功能或类似功能，我可以在这里使用？到目前为止，我的代码如下：
select
[Part].[PartNum] as [Part_PartNum],
(max(PartWhse.OnHandQty)*datediff(day,max(PartTran.TranDate),Constants.Today)) as [Calculated_WeightedSum],
(WeightedSum/DATEDIFF(day, @StartDate, Constants.Today)) as [Calculated_AverageOnHand]
from Erp.Part as Part
right outer join Erp.PartTran as PartTran on
Part.PartNum = PartTran.PartNum
inner join Erp.PartWhse as PartWhse on
Part.PartNum = PartWhse.PartNum
group by [Part].[PartNum]
答案 0 :(得分：2)
撤消交易以计算每日数量。添加缺失日期并向后查看最近填写每日数量的日期。我想我会尝试一个比这个更好的解决方案。
with trn as (
select PartNum, TranDate, TranQty from PartTran
union all
select PartNum, cast('20160601' as date), 0 from PartWhse
union all
select PartNum, cast('20160630' as date), 0 from PartWhse
), qty as (
select
t.PartNum, t.TranDate,
-- assumes that end date corresponds with OnHandQty
min(w.OnHandQty) + sum(t.TranQty)
- sum(sum(t.TranQty))
over (partition by t.PartNum order by t.TranDate desc) as DailyOnHand,
coalesce(
lead(t.TranDate) over (partition by t.PartNum order by t.TranDate),
dateadd(day, 1, t.TranDate)
) as NextTranDate
-- if lead() isn't available...
-- coalesce(
-- (
-- select min(t2.TranDate) from trn as t2
-- where t2.PartNum = t.PartNum and t2.TranDate > t.TranDate
-- ),
-- dateadd(day, 1, t.TranDate)
-- ) as NextTranDate
from PartWhse as w inner join trn as t on t.PartNum = w.PartNum
where t.TranDate between '20160601' and '20160630'
group by t.PartNum, t.TranDate
)
select
PartNum,
sum(datediff(day, TranDate, NextTranDate) * DailyOnHand) * 1.00
/ sum(datediff(day, TranDate, NextTranDate)) as DailyAvg
from qty
group by PartNum;
答案 1 :(得分：2)
这是一个有趣的sql-server 2012 +方法。
;WITH cte AS (
SELECT
p.PartNum
,CAST(t.TranDate AS DATE) AS TranDate
,i.OnHandQty
--,SUM(SUM(t.TranQty)) OVER (PARTITION BY p.PartNum ORDER BY CAST(t.TranDate AS DATE) DESC) AS InventoryChange
,i.OnHandQty - SUM(SUM(t.TranQty)) OVER (PARTITION BY p.PartNum ORDER BY CAST(t.TranDate AS DATE) DESC) AS InventoryOnDate
,DATEDIFF(day,
CAST(ISNULL(LAG(MAX(TranDate)) OVER (PARTITION BY p.PartNum ORDER BY CAST(t.TranDate AS DATE) ASC),@StartDate) AS DATE)
,CAST(t.TranDate AS DATE)
) AS DaysAtInventory
FROM
#Parts p
LEFT JOIN #Transact t
ON p.PartNum = t.PartNum
LEFT JOIN #Inventory i
ON p.PartNum = i.PartNum
GROUP BY
p.PartNum
,CAST(t.TranDate AS DATE)
,i.OnHandQty
)
SELECT
PartNum
,(SUM(ISNULL(DaysAtInventory,0) * ISNULL(InventoryOnDate,0))
+ ((DATEDIFF(day,MAX(TranDate),CAST(GETDATE() AS DATE)) + 1) * ISNULL(MAX(OnHandQty),0)))
/((DATEDIFF(day,CAST(@StartDate AS DATE),CAST(GETDATE() AS DATE)) + 1) * 1.00) AS AvgDailyInventory
FROM
cte
GROUP BY
PartNum
这个实际上给了我22.9但是引入了3.33333 333，因为1天必须放在某处，所以我把它当作当前的库存。
这是我之前回答的一个方法，这个方法更容易概念化数据......我会对两种方法之间的性能差异感到好奇。
这些步骤中的一些可以结合起来更简洁但是这有效（虽然我得到了22.6而不是.1或.9 ....）我把所有内容都整理成了一个完整的日期，这样你就不会＃ 39;不得不担心一天的开始和结束。
DECLARE @StartDate DATETIME = '6/1/2016'
;WITH cteDates AS (
SELECT @StartDate AS d
UNION ALL
SELECT
d + 1 AS d
FROM
cteDates c
WHERE c.d + 1 <= CAST(CAST(GETDATE() AS DATE) AS DATETIME)
--get dates to today beginning of day
)
, ctePartsDaysCross AS (
SELECT
d.d
,p.PartNum
,ISNULL(i.OnHandQty,0) AS OnHandQty
FROM
cteDates d
CROSS JOIN #Parts p
LEFT JOIN #Inventory i
ON p.PartNum = i.PartNum
)
, cteTransactsQuantityByDate AS (
SELECT
CAST(t.TranDate AS DATE) as d
,t.PartNum
,TranQty = SUM(t.TranQty)
FROM
#Transact t
GROUP BY
CAST(t.TranDate AS DATE)
,t.PartNum
)
,cteDailyInventory AS (
SELECT
c.d
,c.PartNum
,c.OnHandQty - SUM(ISNULL(t.TranQty,0)) OVER (PARTITION BY c.PartNum ORDER BY c.d DESC) AS DailyOnHand
FROM
ctePartsDaysCross c
LEFT JOIN cteTransactsQuantityByDate t
ON c.d = t.d
AND c.PartNum = t.PartNum
)
SELECT
PartNum
,AVG(CAST(DailyOnHand AS DECIMAL(6,3)))
FROM
cteDailyInventory
GROUP BY
PartNum
以下是测试数据：
IF OBJECT_ID('tempdb..#Parts') IS NOT NULL
BEGIN
DROP TABLE #Parts
END
IF OBJECT_ID('tempdb..#Transact') IS NOT NULL
BEGIN
DROP TABLE #Transact
END
IF OBJECT_ID('tempdb..#Inventory') IS NOT NULL
BEGIN
DROP TABLE #Inventory
END
CREATE TABLE #Parts (
PartNum CHAR(2)
)
CREATE TABLE #Transact (
AutoId INT IDENTITY(1,1) NOT NULL
,PartNum CHAR(2)
,TranDate DATETIME
,TranQty INT
)
CREATE TABLE #Inventory (
PartNum CHAR(2)
,OnHandQty INT
)
INSERT INTO #Parts (PartNum) VALUES ('P1'),('P2'),('P3')
INSERT INTO #Transact (PartNum, TranDate, TranQty)
VALUES ('P1','6/28/2016',5),('P1','6/26/2016',3),('P1','6/26/2016',-1)
,('P1','6/15/2016',2) ,('P2','6/15/2016',1)
INSERT INTO #Inventory (PartNum, OnHandQty) VALUES ('P1',30),('P2',2)
我在想1递归cte可能更简单，可能会将其作为更新发布。
答案 2 :(得分：0)
我能用一笔钱来解决这个问题。首先，我将手头的最终数量乘以该范围内的天数。接下来，我将每个库存变化乘以从@StartDate到TransDate的时间。
select
[Part].[PartNum] as [Part_PartNum],
(max(PartWhse.OnHandQty)*datediff(day,@StartDate,Constants.Today)-
sum(PartTran.TranQty*datediff(day,@StartDate,PartTran.TranDate))) as [Calculated_WeightedSum],
(WeightedSum/DATEDIFF(day, @StartDate, Constants.Today)) as [Calculated_AverageOnHand]
from Erp.Part as Part
right outer join Erp.PartTran as PartTran on
Part.PartNum = PartTran.PartNum
inner join Erp.PartWhse as PartWhse on
Part.PartNum = PartWhse.PartNum
group by [Part].[PartNum]
感谢大家的帮助！你真的帮我思考过了。