是否可以在SQL Server 2008 R2中找到不同记录的天数差异?
SELECT OrderDate FROM OrdersTbl WHERE SKU='AA0000' ORDER BY ORDERDATE DESC
OrderDate
-----------------------
2009-12-03 00:00:00.000
2009-04-03 00:00:00.000
2008-02-22 00:00:00.000
2008-02-21 00:00:00.000
2007-02-18 00:00:00.000
2007-01-27 00:00:00.000
2006-10-13 00:00:00.000
我想要一种方法来获取每个订单日期之间的天数,以便找到平均频率。提前谢谢。
答案 0 :(得分:5)
您可以使用公用表表达式和ROW_NUMBER:
来完成WITH OrderDates AS (
SELECT
ROW_NUMBER() OVER (ORDER BY OrderDate DESC) AS RowNumber,
OrderDate
FROM OrdersTable
WHERE SKU = 'AA0000'
)
SELECT
AVG(DATEDIFF(DD, O2.OrderDate, O1.OrderDate)) AS AverageFrequency
FROM OrderDates O1
LEFT JOIN OrderDates O2
ON O2.RowNumber = O1.RowNumber + 1
答案 1 :(得分:1)
;With cteDifference as (
Select SKU, OrderDate, Row_Number() OVER (Partition by SKU Order by OrderDate) as RowNumber
from OrdersTbl
)
select cur.SKU,
cur.OrderDate as CurrentDate,
prev.OrderDate as PreviousDate,
DATEDIFF(DD,prev.OrderDate, cur.OrderDate) as DaysDifference
from cteDifference cur
left join cteDifference prev
on cur.SKU = prev.SKU
and cur.RowNumber = prev.RowNumber + 1
where cur.SKU = 'AA0000'
order by cur.OrderDate desc
答案 2 :(得分:1)
在SQL Server中没有LEAD / LAG支持:
SELECT z.orderdate,
z.prev_date,
DATEDIFF(dd, z.prev_date, z.orderdate)
FROM (SELECT OrderDate,
(SELECT MAX(y.orderdate)
FROM ORDERSTBL y
WHERE y.orderdate < x.orderdate
AND y.sku = x.sku) AS prev_date
FROM OrdersTbl x
WHERE x.sku ='AA0000') z
ORDER BY z.orderdate DESC