我正在尝试为以下内容编写查询：
计算过去90天内的交易次数。注意：这需要对新成员进行平均= =（交易数/天为成员）x 90
新成员是DateCreated从今天起0到90天之间的成员。
表格结构：
Column Name Datatype
---------------------------------
Member_No nvarchar(255)
Order_No int
Transaction_Date datetime
Net money
Date_Created datetime
此外，最后一步是在计算交易后，我需要给出一个分数。因此，如果一名成员的数量超过8，则得分为5。
以下是范围：`
Transaction count Score
>8 5
6-8 4
4-6 3
2-4 2
0-2 1
示例数据：
Member No Sales Order Number Date Created Transaction Date Net
M1 2332 01-10-15 10-07-16 354
M2 2311 12-12-16 14-12-16 53
M3 5422 04-10-14 07-10-14 35
M5 4535 10-10-16 12-11-16 54
M9 5522 03-05-15 07-10-15 55
M3 5422 04-10-14 02-12-16 83
M5 4534 10-10-16 13-12-16 73
M3 5432 04-10-14 09-10-14 35
M3 5484 04-10-14 11-10-16 34
M3 5453 04-10-14 07-11-16 67
M3 5474 04-10-14 09-11-16 56
M3 5493 04-10-14 07-12-16 52
M3 5452 04-10-14 10-12-16 75
M3 5496 04-10-14 11-12-16 34
M3 5442 04-10-14 13-12-16 90
M3 5494 04-10-14 14-12-16 757
M3 5464 04-10-14 16-12-16 72
M5 4542 10-10-16 15-12-16 76
M5 4502 10-10-16 17-12-16 72
M5 4535 10-10-16 18-12-16 43
输出：
Member No Order Count (In last 90 days) Score
M1 0 1
M2 10 5
M3 9 5
M5 5 3
M9 0 1
[注意：
M1为0，因为DateCreated从今天起不到90天（不是新成员），并且该成员在过去90天内没有成交 天
M2订单数量为10，因为DaetCreated是从今天起90天内（新会员），交易日期也在90以内 天所以应用公式（No of 交易/天是会员）x 90）=（1/10）* 90 = 10这是订单数> 8+因此得分为5.
M3订单数量为9，因为该会员在过去90天内已成交9次。得分为5。
M5订单数量为5，因为DaetCreated是从今天起90天内（新会员），他的交易日期也在 90天所以应用公式（交易数量/天数为a 成员）x 90）=（4/72）* 90 = 5因此是4-6范围内的ordercount 得分为3。
如果有任何疑问，请告诉我。
答案 0 :(得分：0)
以下内容完全未经测试，但至少应该让您了解可以使其运行的过程。
我建议你让每一步都执行，并尝试了解他们做了什么以及如何做。
第一步：计算过去90天内的交易。
select Member_No, count(Order_No) as TransactionCount
from Sales
where datediff(day, Transaction_Date, getdate()) <= 90
group by Member_No
第二步：获取每位成员的开始日期和年龄
select
Member_No,
min(Date_Created) as Date_Created,
datediff(day, min(Date_Created), getdate()) as Days_Old
from Sales
group by Member_No
第三步：结合以上
select *
from (
select
Member_No,
min(Date_Created) as Date_Created,
datediff(day, min(Date_Created), getdate()) as Days_Old
from Sales
group by Member_No
) dc
join (
select Member_No, count(Order_No) as TransactionCount
from Sales
where datediff(day, Transaction_Date, getdate()) <= 90
) tc on tc.Member_No=dc.Member_No
第四步：新成员的按比例交易计数
select
Member_No,
case when dc.Days_Old > 90
then tc.Transaction_Count
else tc.Transaction_Count * 90 / dc.Days_Old
end as Transaction_Count
from (
select
Member_No,
min(Date_Created) as Date_Created,
datediff(day, min(Date_Created), getdate()) as Days_Old
from Sales
group by Member_No
) dc
join (
select Member_No, count(Order_No) as TransactionCount
from Sales
where datediff(day, Transaction_Date, getdate()) <= 90
) tc on tc.Member_No=dc.Member_No
第五步，根据交易次数添加分数（请注意，我更改了分数表，使其不重叠）：
select
*,
case
when x.Transaction_Count > 8 then 5
when x.Transaction_Count between 6 and 7 then 4
when x.Transaction_Count between 4 and 5 then 3
when x.Transaction_Count between 2 and 3 then 2
else 1
end as Score
from (
select
Member_No,
case when dc.Days_Old > 90
then tc.Transaction_Count
else tc.Transaction_Count * 90 / dc.Days_Old
end as Transaction_Count
from (
select
Member_No,
min(Date_Created) as Date_Created,
datediff(day, min(Date_Created), getdate()) as Days_Old
from Sales
group by Member_No
) dc
join (
select Member_No, count(Order_No) as TransactionCount
from Sales
where datediff(day, Transaction_Date, getdate()) <= 90
) tc on tc.Member_No=dc.Member_No
) x
答案 1 :(得分：0)
我相信这是一个基本的聚合查询，还有一些额外的逻辑：
select memberno,
order_count_90_days,
(case when days_ago_start >= 90
then (case when order_count_90_days > 8 then 5
when order_count_90_days > 6 then 4
when order_count_90_days > 4 then 3
when order_count_90_days > 2 then 2
else 1
end)
else (case when (order_count_90_days * 90.0 / days_ago_start) > 8 then 5
when (order_count_90_days * 90.0 / days_ago_start) > 6 then 4
when (order_count_90_days * 90.0 / days_ago_start) > 4 then 3
when (order_count_90_days * 90.0 / days_ago_start) > 2 then 2
else 1
end)
end) as score
from (select memberno,
datediff(day, min(transactiondate), getdate()) as days_ago_start,
sum(case when transactiondate >= getdate() - 90 then 1 else 0 end) as order_count_90_days
from sample
group by memberno
) m;