Question

我在银行工作。考虑到保持平衡，我们计划根据以下规则为客户提供一些要点：

Balance    ----  Point(per 1000)
1000-5000        10
5000-10000       20
10000-20000      30

例如，某人的帐户中有20000，我们根据以下几点给他/她450点：

(5*10) + (5*20)+ (10*30)=450

我们大约有300万客户，我们应该计算他们的积分。

我们应该如何使用T-SQL编写此查询？

Answer 1

有多种方法。这是一种方法：

select t.*,
       ((case when balance > 1000 and balance < 5000
              then (balance - 1000) * 10.0 / 1000
              else (5000 - 1000) * 10.0 / 1000
        ) +
        (case when balance > 5000 and balance < 10000
              then (balance - 5000) * 20.0 / 1000
              else (10000 - 5000) * 20.0 / 1000
        ) +
        (case when balance > 10000 and balance < 20000
              then (balance - 10000) * 30.0 / 1000
              else (20000 - 10000) * 30.0 / 1000
        )
       ) as points
from t

Answer 2

一种解决方法如下。关键假设是范围没有差距。 Working demo link

create table temp (Balance   nvarchar(1000),Points int);
insert into temp values
('1000-5000'  , 10)
,('5000-10000' ,20)
,('10000-20000',30);

create table Points(customerid int, balance int);
insert into Points values
(1,2637),(2,7888),(3,12782),(4,20000);

; with rangeTable as (
 select 
startrange=isnull
    (lag(cast(right(Balance, len(Balance)-charindex('-',Balance)) as int)) over ( order by Points asc),0) ,
endrange=cast(right(Balance, len(Balance)-charindex('-',Balance)) as int) ,
Points
from temp
    )


 select sum(case when p.Balance >r.endrange then (r.endrange-r.startrange) else (p.Balance-r.startrange) end  *r.Points)/1000,p.customerid,p.Balance from Points p cross join rangeTable r
 where p.Balance >r.startrange
 group by p.customerid,p.Balance

Answer 3

我喜欢将层保持在表中。这消除了代码中的逻辑，并允许您在一个结构中支持多层。

示例

Declare @Tier table (TierGrp varchar(50),TierTitle varchar(50),TierR1 money,TierR2 money,Pnts int)
Insert Into @Tier values
 ('Sample','0 - 5'   ,0     ,5000   ,10)
,('Sample','5 - 10'  ,5000  ,10000  ,20)
,('Sample','10 - 20' ,10000 ,20000  ,30)
,('Sample','20 - 50' ,20000 ,50000  ,40)
,('Sample','50 - 100',50000 ,100000 ,50)
,('Sample','10+'     ,100000,9999999,60)

Declare @YourTable table (ID int,RemainBalance money)
Insert Into @YourTable values
 (1,20000)
,(2,15500)
,(3,30000)

Select A.ID
      ,A.RemainBalance
      ,Points = sum(floor(((IIF(RemainBalance>TierR2,TierR2,RemainBalance)-TierR1)/1000)) * Pnts )
 From  @YourTable A
 Join  @Tier      B on TierGrp='Sample' and RemainBalance >=TierR1
 Group By A.ID,A.RemainBalance
 Order By A.ID

返回

ID  RemainBalance   Points
1   20000.00        450.00
2   15500.00        300.00
3   30000.00        850.00

按不同的值对bigint列进行分组

3 个答案: