对于返回的每件商品，我们需要知道“此”商品在不同时间段内退回的时间：30,60,90,120,180,365天。
项目基于其序列号（Itm_Item_Serial）是唯一的。
使用创建脚本完成示例，并预期结果为here *。
CREATE TABLE ItemReturn
(
[Itm_Id] int,
[Itm_Item_Serial] int,
[Itm_CDate] datetime
);
INSERT INTO ItemReturn ([Itm_Id], [Itm_Item_Serial], [Itm_CDate])
VALUES
(1, 1, '2016-10-02 02:00:00'),
(2, 1, '2016-09-03 02:00:00'),
(3, 1, '2016-11-03 01:00:00')
;
Itm_Id 30d 60d 90d 120d 180d 365d
1 0 0 0 0 0 0
2 1 0 0 0 0 0
3 1 1 0 0 0 0
0或如果此时间范围内没有返回则为null。
[Itm_Id] [Itm_Item_Serial] [Itm_CDate]
1, 1, '2016-10-02 02:00:00'
2, 1, '2016-09-03 02:00:00'
3, 1, '2016-11-03 01:00:00'
*：rextester数据样本和有序数据样本。
答案 0 :(得分：1)
SQL小提琴：http://sqlfiddle.com/#!6/00460/4
不会为您提供您提供的结果，但会提供不同日期失误的计数
; with data_CTE as (
select
m1.Itm_CDate
, DATEDIFF(day, m1.Itm_CDate, m2.Itm_CDate) DayDiff
, m1.Itm_id
, m1.Itm_Item_Serial
from ItemReturn m1
left outer join ItemReturn m2
on m1.Itm_Item_Serial = m2.Itm_Item_Serial
where m1.Itm_CDate < m2.Itm_CDate
and m1.Itm_CDate > DATEADD(day, -30, m2.Itm_CDate)
), aggr_CTE as (
select
Itm_Id, Itm_Item_Serial
, case
when DayDiff < 30 then '30d'
when DayDiff < 60 then '60d'
when DayDiff < 90 then '90d'
when DayDiff < 120 then '120d'
when DayDiff < 180 then '180d'
else '365d'
end DayLapse
from data_CTE
)select
Itm_id, [30d], [60d], [90d], [120d], [180d], [365d]
from (
select Itm_id, Itm_Item_Serial, DayLapse
from aggr_CTE
) src PIVOT (
count(Itm_Item_Serial)
for DayLapse in ([30d], [60d], [90d], [120d], [180d], [365d])
) as PivotTable
data_CTE结果（输出）：
Itm_CDate DayDiff Itm_id Itm_Item_Serial
----------------------- ----------- ----------- ---------------
2016-09-03 02:00:00.000 29 2 1
2016-09-03 02:00:00.000 29 3 1
2016-09-03 02:00:00.000 29 5 2
2016-09-03 02:00:00.000 29 13 6
2016-08-05 02:00:00.000 29 15 6
2016-08-04 02:00:00.000 1 14 6
查询的最终版本
; with data_CTE as (
SELECT
Itm_Id
, Itm_Item_Serial
, ROW_NUMBER() over (Partition By Itm_Item_Serial order by Itm_Item_Serial) Itm_Item_RowNum
, Itm_CDate
FROM [ItemReturn] ir1
), date_CTE as (
select
d2.Itm_Id
, d1.Itm_Item_Serial
, DATEDIFF(day, d1.Itm_CDate, d2.Itm_CDate) DayDiff
from data_CTE d1, data_CTE d2
where d1.Itm_Item_Serial = d2.Itm_Item_Serial
and DATEDIFF(day, d1.Itm_CDate, d2.Itm_CDate) >= 0
), aggr_CTE as (
select
Itm_Id, Itm_Item_Serial
, case
when DayDiff <= 0 then '0d'
when DayDiff <= 30 then '30d'
when DayDiff <= 60 then '60d'
when DayDiff <= 90 then '90d'
when DayDiff <= 120 then '120d'
when DayDiff <= 180 then '180d'
else '365d'
end DayLapse
from date_CTE
)select
Itm_id, [30d], [60d], [90d], [120d], [180d], [365d]
from (
select Itm_id, Itm_Item_Serial, DayLapse
from aggr_CTE
) src PIVOT (
count(Itm_Item_Serial)
for DayLapse in ([0d], [30d], [60d], [90d], [120d], [180d], [365d])
) as PivotTable
答案 1 :(得分：1)
您可以查询如下：
Select * from (
Select Itm_id, Itm_Item_Serial, case when datediff(day, Itm_CDate, getdate()) between 0 and 30 then '30'
when datediff(day, Itm_CDate, getdate()) between 30 and 60 then '60'
when datediff(day, Itm_CDate, getdate()) between 60 and 90 then '90'
when datediff(day, Itm_CDate, getdate()) between 90 and 120 then '120'
when datediff(day, Itm_CDate, getdate()) between 120 and 180 then '180'
when datediff(day, Itm_CDate, getdate()) between 180 and 360 then '365'
Else 0 End as test
from ItemReturn
) a
pivot (count(test) for test in ([30],[60],[90],[120],[180],[365])) p
如果您的差异日期介于下一个日期之间，那么您可能需要使用以下潜在客户：
;With cte as (
Select *, NextDate = lead(itm_cDate) over(partition by Itm_Item_Serial order by Itm_id)
from ItemReturn
), cte2 as (
Select Itm_id, Itm_Item_Serial, case when datediff(day, Itm_CDate, NextDate) between 0 and 30 then '30'
when datediff(day, Itm_CDate, NextDate) between 30 and 60 then '60'
when datediff(day, Itm_CDate, NextDate) between 60 and 90 then '90'
when datediff(day, Itm_CDate, NextDate) between 90 and 120 then '120'
when datediff(day, Itm_CDate, NextDate) between 120 and 180 then '180'
when datediff(day, Itm_CDate, NextDate) between 180 and 360 then '365'
Else 0 End as test
from cte
)
Select * from cte2
pivot (count(test) for test in ([30],[60],[90],[120],[180],[365])) p
Demo此处
答案 2 :(得分：1)
不是最终的答案，而是光滑的方式来获得差异
with cte as
select [Itm_Id], [Itm_Item_Serial], [Itm_CDate]
, row_number() over (partition by [Itm_Item_Serial] order by [Itm_CDate]) as rn
from ItemReturn;
select t1.*
, datediff(dd, t1.[Itm_CDate], t2.[Itm_CDate] desc) as diff
, t2.[Itm_Id], t2.[Itm_CDate]
from cte t1
join cte t2
on t1.[Itm_Item_Serial] = t2.[Itm_Item_Serial]
and t1.rn = 1
and t2.rn <> 1
order by t1.[Itm_Item_Serial], t1.rn