Sql 需要帮助执行以下计算吗
我有一个特定的值，比如说10，作为我计算的基础。现在，与值10对应的第一个期间的增长为5。我想要的结果是10*1+5/100，基本上是增长的基数*1+%。第一个期间的结果值将成为下一个期间的新基数。假设下一个增长是6，下一个时期的结果是101+5/100*1+6/100。这基本上是运行乘法，可以通过多种方式实现。现在请有人提出实现这一计算的最终最佳方法 10 , 5 -> 10 1 + 5/100 = 10.50 10.50 , 6 -> 10.50 1 + 6/100 = 11.1300 11.13、任何值等 我尝试过使用其他数据样本的方法，但基本上是运行乘法Sql 需要帮助执行以下计算吗,sql,sql-server,oracle,Sql,Sql Server,Oracle,我有一个特定的值，比如说10，作为我计算的基础。现在，与值10对应的第一个期间的增长为5。我想要的结果是10*1+5/100，基本上是增长的基数*1+%。第一个期间的结果值将成为下一个期间的新基数。假设下一个增长是6，下一个时期的结果是101+5/100*1+6/100。这基本上是运行乘法，可以通过多种方式实现。现在请有人提出实现这一计算的最终最佳方法 10 , 5 -> 10 1 + 5/100 = 10.50 10.50 , 6 -> 10.50 1 + 6/100 = 11
CREATE TABLE #t1
(
projection_details_sid INT,
period_sid INT,
growth NUMERIC(22, 6)
)
INSERT INTO #t1
(projection_details_sid,
period_sid,
growth)
VALUES ( 1,601,2 ),
( 1,602,2 ),
( 1,603,2 ),
( 1,604,1 ),
( 1,605,6 ),
( 1,606,3 )
SELECT *,
Exp(Sum(Log(growth))
OVER (
PARTITION BY projection_details_sid
ORDER BY projection_details_sid ROWS UNBOUNDED PRECEDING ))
FROM #t1
您可以从下面的查询中获得想法：-
DECLARE @Growth INT = 5
,@Base DECIMAL(18,2) = 10.0;
;WITH Test(Base, Growth)
AS
(
SELECT @Base, @Growth
UNION ALL
SELECT CAST(t.Base * (1 + t.Growth/100.0) AS DECIMAL(18,2)) , t.Growth + 1
FROM Test t
WHERE t.Base < 600000000
)
SELECT *
FROM Test
option (maxrecursion 0)
您可以从下面的查询中获得想法：-
DECLARE @Growth INT = 5
,@Base DECIMAL(18,2) = 10.0;
;WITH Test(Base, Growth)
AS
(
SELECT @Base, @Growth
UNION ALL
SELECT CAST(t.Base * (1 + t.Growth/100.0) AS DECIMAL(18,2)) , t.Growth + 1
FROM Test t
WHERE t.Base < 600000000
)
SELECT *
FROM Test
option (maxrecursion 0)
WITH our_recursive_query(projection_details_sid, period_sid, growth, base, our_result)
AS (
select projection_details_sid, period_sid, growth,
10.0 as base,
10 * ( 1 + growth/100) As our_result
from t1 where period_sid = 601
UNION ALL
SELECT a.projection_details_sid, a.period_sid, a.growth,
b.our_result as base,
b.our_result * ( 1 + a.growth/100) As our_result
FROM t1 a
JOIN our_recursive_query b
ON a.period_sid = b.period_sid + 1
)
SELECT * FROM our_recursive_query
PROJECTION_DETAILS_SID PERIOD_SID GROWTH BASE OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 10.00000000 10.20000000
1 602 2 10.20000000 10.40400000
1 603 2 10.40400000 10.61208000
1 604 1 10.61208000 10.71820080
1 605 6 10.71820080 11.36129285
1 606 3 11.36129285 11.70213163
PROJECTION_DETAILS_SID PERIOD_SID GROWTH MULTIPLIER OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 1.02000000 10.20000000
1 602 2 1.04040000 10.40400000
1 603 2 1.06120800 10.61208000
1 604 1 1.07182008 10.71820080
1 605 6 1.13612928 11.36129285
1 606 3 1.17021316 11.70213163
@科迪科，谢谢。有没有其他可能达到这个结果 而不是递归cte，因为递归cte具有类似的性能 当然是while循环 对。在Oracle中，您可以创建自己的聚合函数来执行数字链的乘法运算，其方式与内置求和函数的方式类似==>X1+X2+…+Xn，但实际上是X1*X2*…*Xn。 例如：
create or replace TYPE MyCumulativeMultiply_type
AS OBJECT (
cumulativeMultiplyResult NUMBER,
STATIC FUNCTION odciaggregateinitialize(ctx IN OUT MyCumulativeMultiply_type) RETURN NUMBER,
MEMBER FUNCTION odciaggregateiterate(self IN OUT MyCumulativeMultiply_type, your_parameter_to_aggregate IN NUMBER) RETURN NUMBER,
MEMBER FUNCTION odciaggregatemerge(self IN OUT MyCumulativeMultiply_type, ctx2 IN MyCumulativeMultiply_type) RETURN NUMBER,
MEMBER FUNCTION odciaggregateterminate(self IN MyCumulativeMultiply_type, returnvalue OUT NUMBER, flags IN NUMBER) RETURN NUMBER
);
/
CREATE OR REPLACE TYPE BODY MyCumulativeMultiply_type
IS
STATIC FUNCTION odciaggregateinitialize(ctx IN OUT MyCumulativeMultiply_type)
RETURN NUMBER
IS
BEGIN
-- instantiate our type, NULL the dummy attribute
ctx := MyCumulativeMultiply_type( 1 );
RETURN odciconst.success;
END odciaggregateinitialize;
MEMBER FUNCTION odciaggregateiterate(self IN OUT MyCumulativeMultiply_type, your_parameter_to_aggregate IN NUMBER)
RETURN NUMBER
IS
BEGIN
self.cumulativeMultiplyResult := self.cumulativeMultiplyResult * your_parameter_to_aggregate;
RETURN odciconst.success;
END odciaggregateiterate;
MEMBER FUNCTION odciaggregatemerge(self IN OUT MyCumulativeMultiply_type, ctx2 IN MyCumulativeMultiply_type)
RETURN NUMBER
IS
BEGIN
self.cumulativeMultiplyResult := self.cumulativeMultiplyResult * ctx2.cumulativeMultiplyResult;
RETURN odciconst.success;
END odciaggregatemerge;
MEMBER FUNCTION odciaggregateterminate(self IN MyCumulativeMultiply_type,
returnvalue OUT NUMBER,
flags IN NUMBER
)
RETURN NUMBER
IS
BEGIN
returnvalue := self.cumulativeMultiplyResult;
RETURN odciconst.success;
END odciaggregateterminate;
END;
/
CREATE OR REPLACE FUNCTION cumulative_multiply(arg NUMBER)
RETURN NUMBER
PARALLEL_ENABLE
AGGREGATE USING MyCumulativeMultiply_type;
/
select t1.*
, cumulative_multiply( 1 + growth/100 ) OVER (order by period_sid ) as multiplier
, 10 * cumulative_multiply( 1 + growth/100 ) OVER (order by period_sid ) as our_result
from t1;
PROJECTION_DETAILS_SID PERIOD_SID GROWTH BASE OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 10.00000000 10.20000000
1 602 2 10.20000000 10.40400000
1 603 2 10.40400000 10.61208000
1 604 1 10.61208000 10.71820080
1 605 6 10.71820080 11.36129285
1 606 3 11.36129285 11.70213163
PROJECTION_DETAILS_SID PERIOD_SID GROWTH MULTIPLIER OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 1.02000000 10.20000000
1 602 2 1.04040000 10.40400000
1 603 2 1.06120800 10.61208000
1 604 1 1.07182008 10.71820080
1 605 6 1.13612928 11.36129285
1 606 3 1.17021316 11.70213163
WITH our_recursive_query(projection_details_sid, period_sid, growth, base, our_result)
AS (
select projection_details_sid, period_sid, growth,
10.0 as base,
10 * ( 1 + growth/100) As our_result
from t1 where period_sid = 601
UNION ALL
SELECT a.projection_details_sid, a.period_sid, a.growth,
b.our_result as base,
b.our_result * ( 1 + a.growth/100) As our_result
FROM t1 a
JOIN our_recursive_query b
ON a.period_sid = b.period_sid + 1
)
SELECT * FROM our_recursive_query
WITH our_recursive_query(projection_details_sid, period_sid, growth, base, our_result)
AS (
select projection_details_sid, period_sid, growth,
10.0 as base,
10 * ( 1 + growth/100) As our_result
from t1 where period_sid = 601
UNION ALL
SELECT a.projection_details_sid, a.period_sid, a.growth,
b.our_result as base,
b.our_result * ( 1 + a.growth/100) As our_result
FROM t1 a
JOIN our_recursive_query b
ON a.period_sid = b.period_sid + 1
)
SELECT * FROM our_recursive_query
PROJECTION_DETAILS_SID PERIOD_SID GROWTH BASE OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 10.00000000 10.20000000
1 602 2 10.20000000 10.40400000
1 603 2 10.40400000 10.61208000
1 604 1 10.61208000 10.71820080
1 605 6 10.71820080 11.36129285
1 606 3 11.36129285 11.70213163
PROJECTION_DETAILS_SID PERIOD_SID GROWTH MULTIPLIER OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 1.02000000 10.20000000
1 602 2 1.04040000 10.40400000
1 603 2 1.06120800 10.61208000
1 604 1 1.07182008 10.71820080
1 605 6 1.13612928 11.36129285
1 606 3 1.17021316 11.70213163
@科迪科，谢谢。有没有其他可能达到这个结果 而不是递归cte，因为递归cte具有类似的性能 当然是while循环 对。在Oracle中，您可以创建自己的聚合函数来执行数字链的乘法运算，其方式与内置求和函数的方式类似==>X1+X2+…+Xn，但实际上是X1*X2*…*Xn。 例如：
create or replace TYPE MyCumulativeMultiply_type
AS OBJECT (
cumulativeMultiplyResult NUMBER,
STATIC FUNCTION odciaggregateinitialize(ctx IN OUT MyCumulativeMultiply_type) RETURN NUMBER,
MEMBER FUNCTION odciaggregateiterate(self IN OUT MyCumulativeMultiply_type, your_parameter_to_aggregate IN NUMBER) RETURN NUMBER,
MEMBER FUNCTION odciaggregatemerge(self IN OUT MyCumulativeMultiply_type, ctx2 IN MyCumulativeMultiply_type) RETURN NUMBER,
MEMBER FUNCTION odciaggregateterminate(self IN MyCumulativeMultiply_type, returnvalue OUT NUMBER, flags IN NUMBER) RETURN NUMBER
);
/
CREATE OR REPLACE TYPE BODY MyCumulativeMultiply_type
IS
STATIC FUNCTION odciaggregateinitialize(ctx IN OUT MyCumulativeMultiply_type)
RETURN NUMBER
IS
BEGIN
-- instantiate our type, NULL the dummy attribute
ctx := MyCumulativeMultiply_type( 1 );
RETURN odciconst.success;
END odciaggregateinitialize;
MEMBER FUNCTION odciaggregateiterate(self IN OUT MyCumulativeMultiply_type, your_parameter_to_aggregate IN NUMBER)
RETURN NUMBER
IS
BEGIN
self.cumulativeMultiplyResult := self.cumulativeMultiplyResult * your_parameter_to_aggregate;
RETURN odciconst.success;
END odciaggregateiterate;
MEMBER FUNCTION odciaggregatemerge(self IN OUT MyCumulativeMultiply_type, ctx2 IN MyCumulativeMultiply_type)
RETURN NUMBER
IS
BEGIN
self.cumulativeMultiplyResult := self.cumulativeMultiplyResult * ctx2.cumulativeMultiplyResult;
RETURN odciconst.success;
END odciaggregatemerge;
MEMBER FUNCTION odciaggregateterminate(self IN MyCumulativeMultiply_type,
returnvalue OUT NUMBER,
flags IN NUMBER
)
RETURN NUMBER
IS
BEGIN
returnvalue := self.cumulativeMultiplyResult;
RETURN odciconst.success;
END odciaggregateterminate;
END;
/
CREATE OR REPLACE FUNCTION cumulative_multiply(arg NUMBER)
RETURN NUMBER
PARALLEL_ENABLE
AGGREGATE USING MyCumulativeMultiply_type;
/
select t1.*
, cumulative_multiply( 1 + growth/100 ) OVER (order by period_sid ) as multiplier
, 10 * cumulative_multiply( 1 + growth/100 ) OVER (order by period_sid ) as our_result
from t1;
PROJECTION_DETAILS_SID PERIOD_SID GROWTH BASE OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 10.00000000 10.20000000
1 602 2 10.20000000 10.40400000
1 603 2 10.40400000 10.61208000
1 604 1 10.61208000 10.71820080
1 605 6 10.71820080 11.36129285
1 606 3 11.36129285 11.70213163
PROJECTION_DETAILS_SID PERIOD_SID GROWTH MULTIPLIER OUR_RESULT
--------------------------------------- ---------- ---------- ------------ ------------
1 601 2 1.02000000 10.20000000
1 602 2 1.04040000 10.40400000
1 603 2 1.06120800 10.61208000
1 604 1 1.07182008 10.71820080
1 605 6 1.13612928 11.36129285
1 606 3 1.17021316 11.70213163
不幸的是，我不知道上述方法在SQL Server中是可行的。@reto感谢您的指导。我试过用计算机运行乘法运算。但任何使用超前、滞后或任何其他技术的方法都对我有好处。@reto谢谢你的指导。我试过用计算机运行乘法运算。但任何使用超前、滞后或其他技术的方法都对我有好处。@kordiko谢谢。由于递归cte与while循环具有类似的性能，因此除了递归cte之外，还有其他方法可以实现该结果吗？@kordiko，谢谢。由于递归cte的性能与while loop.Thank相似，因此除了递归cte之外，还有其他可能实现此结果的方法吗。由于递归cte的性能与while loop.Thank相似，因此除了递归cte之外，还有其他可能实现此结果的方法吗。由于递归cte具有类似while循环的性能，因此除了递归cte之外，还有其他可能实现该结果的方法吗。
WITH our_recursive_query(projection_details_sid, period_sid, growth, base, our_result)
AS (
select projection_details_sid, period_sid, growth,
10.0 as base,
10 * ( 1 + growth/100) As our_result
from t1 where period_sid = 601
UNION ALL
SELECT a.projection_details_sid, a.period_sid, a.growth,
b.our_result as base,
b.our_result * ( 1 + a.growth/100) As our_result
FROM t1 a
JOIN our_recursive_query b
ON a.period_sid = b.period_sid + 1
)
SELECT * FROM our_recursive_query