我有下表(#temp1),我需要替换列中的字符串' Formula'使用匹配的输入' VALUE'基于小组'年纪'。
的专栏'公式'列可以是任何数学表达式以便更好地理解我在下面提到了一个简单的例子。
IDNUM formula INPUTNAME VALUE YEARMONTH
---------------------------------------------------------------------
1 imports(398)+imports(399) imports(398) 17.000 2003:1
2 imports(398)+imports(399) imports(398) 56.000 2003:2
3 imports(398)+imports(399) imports(399) 15.000 2003:1
4 imports(398)+imports(399) imports(399) 126.000 2003:2
例如:从上表中我需要输出为
Idnum Formula Yearmonth
1. 17.00 +15.00 2003:1
2. 56.00 +126.00 2003:2
我尝试了以下不同的查询来自各种建议,但是coludnt实现了它。有人可以帮帮我这个吗?
Type1:
SELECT
REPLACE(FORMULA, INPUTName, AttributeValue) AS realvalues,
yearmonth
FROM #temp1
GROUP BY yearmonth
TYPE2: 使用XML PATH ...在这种情况下它已经工作但我只需要用值替换字符串而不是根据数学运算填充字符串。(因为公式可能是任何类型)。
SELECT
IDNUM = MIN(IDNUM),
FORMULA =
(SELECT STUFF(
(SELECT ' +' + CONVERT(VARCHAR(10), Value)
FROM #temp1
WHERE YEARMONTH = t1.YEARMONTH
FOR XML PATH(''))
,1, 2, '')),
YEARMONTH
FROM #TEMP1 t1
GROUP BY YEARMONTH
TYPE3:使用递归...这只返回空值......
;with t as (
select t.*,
row_number() over (partition by yearmonth order by idnum) as seqnum,
count(*) over (partition by yearmonth) as cnt
from #temp1 t
)
,cte as (
select t.seqnum, t.yearmonth, t.cnt,
replace(formula, inputname, AttributeValue) as formula1
from t
where seqnum = 1
union all
select cte.seqnum, cte.yearmonth, cte.cnt,
replace(CTE.formula1, T.inputname, T.AttributeValue) as formula2
from cte join
t
on cte.yearmonth = t.yearmonth
AND cte.seqnum = t.seqnum + 1
)
select row_number() over (order by (select null)) as id,formula1
from cte
where seqnum = cnt
答案 0 :(得分:2)
这是使用递归CTE的完整工作示例:
DECLARE @DataSource TABLE
(
[IDNUM] TINYINT
,[formula] VARCHAR(MAX)
,[INPUTNAME] VARCHAR(128)
,[VALUE] DECIMAL(9,3)
,[YEARMONTH] VARCHAR(8)
);
INSERT INTO @DataSource ([IDNUM], [formula], [INPUTNAME], [VALUE], [YEARMONTH])
VALUES ('1', 'imports(398)+imports(399)', 'imports(398)', '17.000', '2003:1')
,('2', 'imports(398)+imports(399)', 'imports(398)', '56.000', '2003:2')
,('3', 'imports(398)+imports(399)', 'imports(399)', '15.000', '2003:1')
,('4', 'imports(398)+imports(399)', 'imports(399)', '126.000', '2003:2')
,('5', '(imports(391)+imports(392)-imports(393))/imports(394)', 'imports(391)', '5.000', '2003:3')
,('6', '(imports(391)+imports(392)-imports(393))/imports(394)', 'imports(392)', '10.000', '2003:3')
,('7', '(imports(391)+imports(392)-imports(393))/imports(394)', 'imports(393)', '3.000', '2003:3')
,('8', '(imports(391)+imports(392)-imports(393))/imports(394)', 'imports(394)', '-5.000', '2003:3');
WITH DataSource AS
(
SELECT ROW_NUMBER() OVER(PARTITION BY [YEARMONTH] ORDER BY [IDNUM]) AS [ReplacementOrderID]
,[YEARMONTH]
,[formula]
,[INPUTNAME] AS [ReplacementString]
,[VALUE] AS [ReplacementValue]
FROM @DataSource
),
RecursiveDataSource AS
(
SELECT [ReplacementOrderID]
,[YEARMONTH]
,REPLACE([formula], [ReplacementString], [ReplacementValue]) AS [formula]
FROM DataSource
WHERE [ReplacementOrderID] = 1
UNION ALL
SELECT DS.[ReplacementOrderID]
,DS.[YEARMONTH]
,REPLACE(RDS.[formula], DS.[ReplacementString], DS.[ReplacementValue]) AS [formula]
FROM RecursiveDataSource RDS
INNER JOIN DataSource DS
ON RDS.[ReplacementOrderID] + 1 = DS.[ReplacementOrderID]
AND RDS.[YEARMONTH] = DS.[YEARMONTH]
)
SELECT RDS.[YEARMONTH]
,RDS.[formula]
FROM RecursiveDataSource RDS
INNER JOIN
(
SELECT [YEARMONTH]
,MAX([ReplacementOrderID]) AS [ReplacementOrderID]
FROM DataSource
GROUP BY [YEARMONTH]
) DS
ON RDS.[YEARMONTH] = DS.[YEARMONTH]
AND RDS.[ReplacementOrderID] = DS.[ReplacementOrderID]
ORDER BY RDS.[YEARMONTH]
通常,您只想在一个语句中对字符串执行多次替换。您可以使用MAXRECURSION
选项获得许多替换值。
答案 1 :(得分:0)
--Create sample data
DROP TABLE #temp1
CREATE TABLE #temp1 (IDNUM int, formula varchar(max), INPUTNAME varchar(max), VALUE decimal, YEARMONTH varchar(max))
INSERT INTO #temp1 VALUES
(1, 'imports(398)+imports(399)', 'imports(398)', 17.000, '2003:1'),
(2, 'imports(398)+imports(399)', 'imports(398)', 56.000, '2003:2'),
(3, 'imports(398)+imports(399)', 'imports(399)', 15.000, '2003:1'),
(4, 'imports(398)+imports(399)', 'imports(399)', 126.000, '2003:2')
--Query
;WITH t as (
SELECT formula, YEARMONTH, IDNUM
FROM #temp1
UNION ALL
SELECT REPLACE(a.formula, b.INPUTNAME, CAST(b.VALUE AS varchar(100))) AS formula, a.YEARMONTH, a.IDNUM
FROM t a
JOIN #temp1 b ON a.YEARMONTH = b.YEARMONTH AND a.formula LIKE '%' + b.INPUTNAME + '%'
)
SELECT MIN(IDNUM) AS IDNUM, formula, YEARMONTH
FROM t
WHERE formula not LIKE '%imports(%'
GROUP BY formula, YEARMONTH