sql - Sql Server 中的计算

Question

我尝试执行以下计算

示例数据:

CREATE TABLE #Table1
  (
     rno   int identity(1,1),
     ccp   varchar(50),
     [col1] INT,
     [col2] INT,
     [col3] INT,
     col4 as [col2]/100.0
  );

INSERT INTO #Table1
            (ccp,[col1],[col2],[col3])
VALUES      ('ccp1',15,10,1100),
            ('ccp1',20,10,1210),
            ('ccp1',30,10,1331),
            ('ccp2',10,15,900),
            ('ccp2',15,15,1000),
            ('ccp2',20,15,1010)

+-----+------+------+------+------+----------+
| rno | ccp  | col1 | col2 | col3 |   col4   |
+-----+------+------+------+------+----------+
|   1 | ccp1 |   15 |   10 | 1100 | 0.100000 |
|   2 | ccp1 |   20 |   10 | 1210 | 0.100000 |
|   3 | ccp1 |   30 |   10 | 1331 | 0.100000 |
|   4 | ccp2 |   10 |   15 |  900 | 0.150000 |
|   5 | ccp2 |   15 |   15 | 1000 | 0.150000 |
|   6 | ccp2 |   20 |   15 | 1010 | 0.150000 |
+-----+------+------+------+------+----------+

注意:每个 ccp 不只是 3 条记录，可以有 N 条记录

预期结果:

1083.500000 --1100 - (15 * (1+0.100000))
1169.850000 --1210 - ((20 * (1+0.100000)) + (15 * (1+0.100000)* (1+0.100000)) )
1253.835000 --1331 - ((30 * (1+0.100000)) + (20 * (1+0.100000)* (1+0.100000)) + (15 * (1+0.100000)* (1+0.100000) *(1+0.100000)) )
888.500000  --900 - (10 * (1+0.150000))
969.525000  --1000 - ((15 * (1+0.150000)) + (10 * (1+0.150000)* (1+0.150000)) )
951.953750  --1010 - ((20 * (1+0.150000)) + (15 * (1+0.150000)* (1+0.150000)) + (10 * (1+0.150000)* (1+0.150000) *(1+0.150000)) )

我知道我们可以使用递归 CTE 来完成此操作，但效率不高，因为我必须对超过 500 万条记录执行此操作。

我正在寻求实现类似这种基于集合的方法

对于 ccp :ccp1

SELECT col3 - ( col1 * ( 1 + col4 ) )
FROM   #Table1
WHERE  rno = 1

SELECT rno,
       col3 - ( ( col1 * Power(( 1 + col4 ), 1) ) + ( Lag(col1, 1)
                                                        OVER(
                                                          ORDER BY rno ) * Power(( 1 + col4 ), 2) ) )
FROM   #Table1
WHERE  rno IN ( 1, 2 )

SELECT rno,
       col3 - ( ( col1 * Power(( 1 + col4 ), 1) ) + ( Lag(col1, 1)
                                                        OVER(
                                                          ORDER BY rno ) * Power(( 1 + col4 ), 2) ) + ( Lag(col1, 2)
                                                                                                          OVER(
                                                                                                            ORDER BY rno ) * Power(( 1 + col4 ), 3) ) )
FROM   #Table1
WHERE  rno IN ( 1, 2, 3 ) 

有没有办法在单个查询中计算？

更新:

仍然愿意接受建议。我强烈相信应该有一些使用 SUM () Over(Order by) 窗口聚合函数来执行此操作。

Answer 1

最后我使用以下方法实现了结果

SELECT a.*,
       col3 - res AS Result
FROM   #TABLE1 a
       CROSS apply (SELECT Sum(b.col1 * Power(( 1 + b.COL2 / 100.00 ), new_rn)) AS res
                    FROM   (SELECT Row_number()
                                     OVER(
                                       partition BY ccp
                                       ORDER BY rno DESC) new_rn,*
                            FROM   #TABLE1 b
                            WHERE  a.ccp = b.ccp
                                   AND a.rno >= b.rno)b) cs

结果:

+-----+------+------+------+------+----------+-------------+
| rno | ccp  | col1 | col2 | col3 |   col4   |   Result    |
+-----+------+------+------+------+----------+-------------+
|   1 | ccp1 |   15 |   10 | 1100 | 0.100000 | 1083.500000 |
|   2 | ccp1 |   20 |   10 | 1210 | 0.100000 | 1169.850000 |
|   3 | ccp1 |   30 |   10 | 1331 | 0.100000 | 1253.835000 |
|   4 | ccp2 |   10 |   15 |  900 | 0.150000 | 888.500000  |
|   5 | ccp2 |   15 |   15 | 1000 | 0.150000 | 969.525000  |
|   6 | ccp2 |   20 |   15 | 1010 | 0.150000 | 951.953750  |
+-----+------+------+------+------+----------+-------------+