Sql组由最新的重复字段组成
Mc *_*vin 5 sql sql-server sql-server-2016
我甚至不知道这个问题的好标题是什么.
但是我有一张桌子:
create table trans
(
[transid] INT IDENTITY (1, 1) NOT NULL,
[customerid] int not null,
[points] decimal(10,2) not null,
[date] datetime not null
)
和记录:
--cus1
INSERT INTO trans ( customerid , points , date )
VALUES ( 1, 10, '2016-01-01' ) , ( 1, 20, '2017-02-01' ) , ( 1, 22, '2017-03-01' ) ,
( 1, 24, '2018-02-01' ) , ( 1, 50, '2018-02-25' ) , ( 2, 44, '2016-02-01' ) ,
( 2, 20, '2017-02-01' ) , ( 2, 32, '2017-03-01' ) , ( 2, 15, '2018-02-01' ) ,
( 2, 10, '2018-02-25' ) , ( 3, 10, '2018-02-25' ) , ( 4, 44, '2015-02-01' ) ,
( 4, 20, '2015-03-01' ) , ( 4, 32, '2016-04-01' ) , ( 4, 15, '2016-05-01' ) ,
( 4, 10, '2017-02-25' ) , ( 4, 10, '2018-02-27' ) ,( 4, 20, '2018-02-28' ) ,
( 5, 44, '2015-02-01' ) , ( 5, 20, '2015-03-01' ) , ( 5, 32, '2016-04-01' ) ,
( 5, 15, '2016-05-01' ) ,( 5, 10, '2017-02-25' );
-- selecting the data
select * from trans
生产:
transid customerid points date
----------- ----------- --------------------------------------- -----------------------
1 1 10.00 2016-01-01 00:00:00.000
2 1 20.00 2017-02-01 00:00:00.000
3 1 22.00 2017-03-01 00:00:00.000
4 1 24.00 2018-02-01 00:00:00.000
5 1 50.00 2018-02-25 00:00:00.000
6 2 44.00 2016-02-01 00:00:00.000
7 2 20.00 2017-02-01 00:00:00.000
8 2 32.00 2017-03-01 00:00:00.000
9 2 15.00 2018-02-01 00:00:00.000
10 2 10.00 2018-02-25 00:00:00.000
11 3 10.00 2018-02-25 00:00:00.000
12 4 44.00 2015-02-01 00:00:00.000
13 4 20.00 2015-03-01 00:00:00.000
14 4 32.00 2016-04-01 00:00:00.000
15 4 15.00 2016-05-01 00:00:00.000
16 4 10.00 2017-02-25 00:00:00.000
17 4 10.00 2018-02-27 00:00:00.000
18 4 20.00 2018-02-28 00:00:00.000
19 5 44.00 2015-02-01 00:00:00.000
20 5 20.00 2015-03-01 00:00:00.000
21 5 32.00 2016-04-01 00:00:00.000
22 5 15.00 2016-05-01 00:00:00.000
23 5 10.00 2017-02-25 00:00:00.000
我正在尝试将所有customerid分组并总结他们的观点.但是这里有一个问题,如果trans没有活动1年(下一个转换为1年及以上),则积分将过期.
对于这种情况:每个客户的积分应该是:
Customer1 20+22+24+50
Customer2 20+32+15+10
Customer3 10
Customer4 10+20
Customer5 0
这是我到目前为止所拥有的:
select
t1.transid as transid1,
t1.customerid as customerid1,
t1.date as date1,
t1.points as points1,
t1.rank1 as rank1,
t2.transid as transid2,
t2.customerid as customerid2,
t2.points as points2,
isnull(t2.date,getUTCDate()) as date2,
isnull(t2.rank2,t1.rank1+1) as rank2,
cast(case when(t1.date > dateadd(year,-1,isnull(t2.date,getUTCDate()))) Then 0 ELSE 1 END as bit) as ShouldExpire
from
(
select transid,CustomerID,Date,points,
RANK() OVER(PARTITION BY CustomerID ORDER BY date ASC) AS RANK1
from trans
)t1
left join
(
select transid,CustomerID,Date,points,
RANK() OVER(PARTITION BY CustomerID ORDER BY date ASC) AS RANK2
from trans
)t2 on t1.RANK1=t2.RANK2-1
and t1.customerid=t2.customerid
这使
从上表中,我如何检查有MAX(等级-1)为客户ShouldExpire场,如果是1,则totalpoints为0,否则,总结所有的连续的0,直到有没有更多的记录或1被满足？
或者有更好的方法来解决这个问题吗？
以下查询用于LEAD获取同一切片中下一条记录的日期CustomerID：
;WITH CTE AS (
SELECT transid, CustomerID, [Date], points,
LEAD([Date]) OVER (PARTITION BY CustomerID
ORDER BY date ASC) AS nextDate,
CASE
WHEN [date] > DATEADD(YEAR,
-1,
-- same LEAD() here as above
ISNULL(LEAD([Date]) OVER (PARTITION BY CustomerID
ORDER BY date ASC),
getUTCDate()))
THEN 0
ELSE 1
END AS ShouldExpire
FROM trans
)
SELECT transid, CustomerID, [Date], points, nextDate, ShouldExpire
FROM CTE
ORDER BY CustomerID, [Date]
输出：
transid CustomerID Date points nextDate ShouldExpire
-------------------------------------------------------------
1 1 2016-01-01 10.00 2017-02-01 1 <-- last exp. for 1
2 1 2017-02-01 20.00 2017-03-01 0
3 1 2017-03-01 22.00 2018-02-01 0
4 1 2018-02-01 24.00 2018-02-25 0
5 1 2018-02-25 50.00 NULL 0
6 2 2016-02-01 44.00 2017-02-01 1 <-- last exp. for 2
7 2 2017-02-01 20.00 2017-03-01 0
8 2 2017-03-01 32.00 2018-02-01 0
9 2 2018-02-01 15.00 2018-02-25 0
10 2 2018-02-25 10.00 NULL 0
11 3 2018-02-25 10.00 NULL 0 <-- no exp. for 3
12 4 2015-02-01 44.00 2015-03-01 0
13 4 2015-03-01 20.00 2016-04-01 1
14 4 2016-04-01 32.00 2016-05-01 0
15 4 2016-05-01 15.00 2017-02-25 0
16 4 2017-02-25 10.00 2018-02-27 1 <-- last exp. for 4
17 4 2018-02-27 10.00 2018-02-28 0
18 4 2018-02-28 20.00 NULL 0
19 5 2015-02-01 44.00 2015-03-01 0
20 5 2015-03-01 20.00 2016-04-01 1
21 5 2016-04-01 32.00 2016-05-01 0
22 5 2016-05-01 15.00 2017-02-25 0
23 5 2017-02-25 10.00 NULL 1 <-- last exp. for 5
现在，您似乎想计算上次到期后的积分总和。
使用上述CTE作为基础，您可以通过以下方式达到所需的结果：
;WITH CTE AS (
... above query here ...
)
SELECT CustomerID,
SUM(CASE WHEN rnk = 0 THEN points ELSE 0 END) AS sumOfPoints
FROM (
SELECT transid, CustomerID, [Date], points, nextDate, ShouldExpire,
SUM(ShouldExpire) OVER (PARTITION BY CustomerID ORDER BY [Date] DESC) AS rnk
FROM CTE
) AS t
GROUP BY CustomerID
输出：
CustomerID sumOfPoints
-----------------------
1 116.00
2 77.00
3 10.00
4 30.00
5 0.00
| 归档时间： |
|
| 查看次数： |
93 次 |
| 最近记录： |