分期付款

Question

我有一个总值，其中包含自他注册以来的用户付款。自注册之日起的每一年，用户都必须支付高级会员资格，但不必在同一年支付。用户可以支付e。例如。三年前。或者提前三年。

年份费率如下：

year | rate
2014 | 100
2015 | 100
2016 | 50
2017 | 150
2018 | 100

这是用户应该支付的价值。但他可以转移一个或多或少的钱，那是需要的。假设用户支付了 438 美元。他什么时候做的并不重要。可能是 2014 年或当年。我们只想知道他该交的有没有，也就是2014年至今年的费用。

我想做的是将他每年应支付的转账总额（计算并存储在一个变量中）除以检查他是否已支付所有款项或有多余的payment/underpayment。

我想得到的结果：

year | rate | payed
2014 | 100  | 100
2015 | 100  | 100
2016 | 50   | 50
2017 | 150  | 150
2018 | 100  | 38

所以用户已经支付了他应该支付的所有费用，并且有 38 美元的超额支付。

使用 loop/cursor 我可以做到，但想不出要使用的 window 功能。有什么提示吗？找不到如何为这种情况查询 google。

示例（解决方案有效但正在寻找 window 等效函数）

DECLARE @total DECIMAL = 438

;WITH data AS 
(
    SELECT *, @total - (SELECT SUM(tt.rate) FROM 
        (VALUES
        (2013, 100),
        (2014, 100),
        (2015, 100),
        (2016, 100),
        (2017, 100),
        (2018, 100)) tt([Year], Rate)
        WHERE tt.[Year] < t.[Year]) AS Payed
    FROM (VALUES
    (2013, 100),
    (2014, 100),
    (2015, 100),
    (2016, 100),
    (2017, 100),
    (2018, 100)) t([Year], Rate)
),
prepared AS
(
    SELECT d.[Year], @total AS Total, d.Rate, 
        CASE 
            WHEN ISNULL(d.Payed, d.Rate) <= 0 THEN NULL 
            WHEN @total < d.Rate THEN @total ELSE ISNULL(d.Payed, d.Rate) 
        END AS Payed
    FROM data d
)
SELECT p.[Year], @total AS Total, p.Rate, 
    CASE 
        WHEN p.Payed >= p.Rate THEN p.Rate 
        ELSE p.Payed 
    END Payed 
FROM prepared p

Answer 1

我不确定我是否 100% 关注，但也许是这样：

工作示例：http://rextester.com/OCHHR61883

DECLARE @total DECIMAL = 438

;WITH data AS 
(
    SELECT *, @total - (SELECT SUM(tt.rate) FROM 
        (VALUES
        (2013, 100),
        (2014, 100),
        (2015, 100),
        (2016, 100),
        (2017, 100),
        (2018, 100)) tt([Year], Rate)
        WHERE tt.[Year] < t.[Year]) AS Payed
    FROM (VALUES
    (2013, 100),
    (2014, 100),
    (2015, 100),
    (2016, 100),
    (2017, 100),
    (2018, 100)) t([Year], Rate)
),
prepared AS
(
    SELECT d.[Year], @total AS Total, d.Rate
          , case when @total-sum(rate) over( Order by year asc)<0  then 0 
                 else @total-sum(rate) over( Order by year asc) end  as RunningTotal

    FROM data d
)
Select * from prepared

这给了我们：

+----+------+-------+------+--------------+
|    | Year | Total | Rate | RunningTotal |
+----+------+-------+------+--------------+
|  1 | 2013 |   438 |  100 |          338 |
|  2 | 2014 |   438 |  100 |          238 |
|  3 | 2015 |   438 |  100 |          138 |
|  4 | 2016 |   438 |  100 |           38 |
|  5 | 2017 |   438 |  100 |            0 |
|  6 | 2018 |   438 |  100 |            0 |
+----+------+-------+------+--------------+

或者现在我仔细观察可能是这样的：

http://rextester.com/AFR44863

DECLARE @total DECIMAL = 438

;WITH data AS 
(
    SELECT *, @total - (SELECT SUM(tt.rate) FROM 
        (VALUES
        (2013, 100),
        (2014, 100),
        (2015, 100),
        (2016, 100),
        (2017, 100),
        (2018, 100)) tt([Year], Rate)
        WHERE tt.[Year] < t.[Year]) AS Payed
    FROM (VALUES
    (2013, 100),
    (2014, 100),
    (2015, 100),
    (2016, 100),
    (2017, 100),
    (2018, 100)) t([Year], Rate)
),
prepared AS
(
    SELECT d.[Year], @total AS Total, d.Rate
          , case when @total-sum(rate) over( Order by year asc) > 0  then d.rate
                 when @total-sum(rate) over( Order by year asc) < 0 and sum(rate) over( Order by year asc) > 0 and d.rate+@total-sum(rate) over( Order by year asc) < 0 then 0              
                 when @total-sum(rate) over( Order by year asc) < 0 and sum(rate) over( Order by year asc) > 0 then d.rate+@total-sum(rate) over( Order by year asc)
                 end  as Paid

    FROM data d
)
Select * from prepared


+----+------+-------+------+------+
|    | Year | Total | Rate | Paid |
+----+------+-------+------+------+
|  1 | 2013 |   438 |  100 |  100 |
|  2 | 2014 |   438 |  100 |  100 |
|  3 | 2015 |   438 |  100 |  100 |
|  4 | 2016 |   438 |  100 |  100 |
|  5 | 2017 |   438 |  100 |   38 |
|  6 | 2018 |   438 |  100 |    0 |
+----+------+-------+------+------+

或基于第一个查询的不同方法： http://rextester.com/BVSDS50299

DECLARE @total DECIMAL = 438

;WITH data AS 
(
    SELECT *, @total - (SELECT SUM(tt.rate) FROM 
        (VALUES
        (2013, 100),
        (2014, 100),
        (2015, 100),
        (2016, 100),
        (2017, 100),
        (2018, 100)) tt([Year], Rate)
        WHERE tt.[Year] < t.[Year]) AS Payed
    FROM (VALUES
    (2013, 100),
    (2014, 100),
    (2015, 100),
    (2016, 100),
    (2017, 100),
    (2018, 100)) t([Year], Rate)
),
prepared AS
(
    SELECT d.[Year], @total AS Total, d.Rate
          , case when @total-sum(rate) over( Order by year asc) < 0  then 0 
                 else case when @total-sum(rate) over( Order by year asc)> d.rate then d.rate else @total-sum(rate) over( Order by year asc) end  end
    as RunningTotal

    FROM data d
)
Select * from prepared

这给了我们与第二个查询相同的结果。

Answer 2

如果您使用的是 SQL Server 2012+，使用 SUM() OVER (ORDER BY...).

可以很容易地计算未付（或多付）总额

示例数据

DECLARE @TotalPaid money = 438;

DECLARE @Rates TABLE (Y int, Rate money);
INSERT INTO @Rates (Y, Rate) VALUES
(2014, 100),
(2015, 100),
(2016,  50),
(2017, 150),
(2018, 100),
(2019, 100);

查询

WITH
CTE
AS
(
    SELECT
        Y
        ,Rate
        ,SUM(Rate) OVER (ORDER BY Y 
        ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS SumRate
    FROM @Rates
)
,CTE2
AS
(
    SELECT
        Y
        ,Rate
        ,SumRate
        ,@TotalPaid - SumRate AS PaidExtra
    FROM CTE
)
SELECT
    Y
    ,Rate
    ,SumRate
    ,PaidExtra
    ,CASE WHEN PaidExtra >= 0
        THEN Rate
        ELSE Rate + PaidExtra
    END AS Paid
FROM CTE2
ORDER BY Y
;

结果

+------+--------+---------+-----------+--------+
|  Y   |  Rate  | SumRate | PaidExtra |  Paid  |
+------+--------+---------+-----------+--------+
| 2014 | 100.00 | 100.00  | 338.00    | 100.00 |
| 2015 | 100.00 | 200.00  | 238.00    | 100.00 |
| 2016 | 50.00  | 250.00  | 188.00    | 50.00  |
| 2017 | 150.00 | 400.00  | 38.00     | 150.00 |
| 2018 | 100.00 | 500.00  | -62.00    | 38.00  |
| 2019 | 100.00 | 600.00  | -162.00   | -62.00 |
+------+--------+---------+-----------+--------+

分期付款

divide payments for periods

sql

tsql

sql-server

window-functions