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SQL Server中的计算

sql

我试图执行以下计算

样本数据：

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： 不仅3每个ccp记录都可以有Nno.of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进行此操作，但效率不高，因为我必须对500万条以上的记录进行此操作。

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 )

有没有一种方法可以在单个查询中进行计算？

更新 ：

仍然愿意接受建议。我坚信应该使用SUM () Over(Order by)窗口聚合函数来执行此操作。


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2021-04-15

共1个答案

小编典典

用的方法self join。不知道这是否会比您使用的版本更有效cross apply。

WITH T AS
  (SELECT *,
          ROW_NUMBER() OVER(PARTITION BY CCP
                            ORDER BY RNO) AS RN
   FROM #TABLE1)
SELECT T1.RNO,
       T1.CCP,
       T1.COL1,
       T1.COL2,
       T1.COL3,
       T1.COL3-SUM(T2.COL1*POWER(1+T1.COL2/100.0,T1.RN-T2.RN+1)) AS RES
FROM T T1
JOIN T T2 ON T1.CCP=T2.CCP
AND T1.RN>=T2.RN
GROUP BY T1.RNO,
         T1.CCP,
         T1.COL1,
         T1.COL2,
         T1.COL3

Sample Demo

2021-04-15