以下是示例数据:
计算日期PLResult
2014-01-02 100
2014-01-03 200
2014-02-03 300
2014-02-04 400
2014-02-27 500
以下是预期结果(以逻辑格式):
一月 二月
计算日期PLResult计算日期PLResult
2014-01-02 100 2014-02-03 300
2014-01-03 200 2014-02-04 400
2014-02-27 500
以下是预期结果(使用 T-SQL 查询):
一月计算日期一月-PL结果二月-计算日期二月-PL结果
2014-01-02 100 2014-02-03 300
2014-01-03 200 2014-02-04 400
2014-02-27 500
目标:
- 根据月份对结果进行分类。在上面的示例中,一月的结果放置在一月细目中。
- 月数可以是动态的。在上面的示例中,仅显示一月和二月,因为只有 2 个月的结果
- 结果将通过Excel显示。其实我可以查询多个查询表来聚合不同月份的结果,但如果可以通过一次查询返回所有结果,那么维护和调试会更容易。
以下是填充示例数据的脚本:
CREATE TABLE #PLResultPerDay ( CalculationDate DATETIME, PLResult DECIMAL(18,8) )
INSERT INTO #PLResultPerDay ( CalculationDate, PLResult ) VALUES ('2014-01-02' , 100 )
INSERT INTO #PLResultPerDay ( CalculationDate, PLResult ) VALUES ('2014-01-03' , 200 )
INSERT INTO #PLResultPerDay ( CalculationDate, PLResult ) VALUES ('2014-02-03' , 300 )
INSERT INTO #PLResultPerDay ( CalculationDate, PLResult ) VALUES ('2014-02-04' , 400 )
到目前为止,这是我构建查询的尝试:
SELECT
CalculationDate, [January], CalculationDate, [February]
FROM
(
SELECT CalculationDate, PLResult, DATENAME(MONTH, CalculationDate) AS [MTH]
FROM #PLResultPerDay
) x
PIVOT
(
MIN(PLResult)
FOR [MTH] IN ([January], [February])
) p
最佳答案
正如前面所说,这实际上是不可能的,您可以获得的最接近的是:
January2014CalculationDate | January2014PLResult | February2014CalculationDate | February2014PLResult
---------------------------+---------------------+-----------------------------+------------------
2014-01-02 | 100 | 2014-02-03 | 300
2014-01-03 | 200 | 2014-02-04 | 400
NULL | NULL | 2014-02-27 | 500
即使这并不简单,我仍然建议在 sql 之外处理这样的格式。第一步是按月份对数据进行分区,然后对每个月中的日期进行排名:
SELECT CalculationDate,
PLResult,
CalculationMonth,
DenseRank = DENSE_RANK() OVER(PARTITION BY CalculationMonth ORDER BY CalculationDate)
FROM ( SELECT CalculationDate,
PLResult,
CalculationMonth = DATEADD(MONTH, DATEDIFF(MONTH, 0, CalculationDate), 0)
FROM #PLResultPerDay
) pl;
这给出:
CalculationDate PLResult CalculationMonth DenseRank
2014-01-02 100 2014-01-01 1
2014-01-03 200 2014-01-01 2
2014-02-03 300 2014-02-01 1
2014-02-04 400 2014-02-01 2
2014-02-27 500 2014-02-01 3
然后您可以透视此数据:
WITH Data AS
( SELECT CalculationDate,
PLResult,
CalculationMonth,
DenseRank = DENSE_RANK() OVER(PARTITION BY CalculationMonth ORDER BY CalculationDate)
FROM ( SELECT CalculationDate,
PLResult,
CalculationMonth = DATEADD(MONTH, DATEDIFF(MONTH, 0, CalculationDate), 0)
FROM #PLResultPerDay
) pl
)
SELECT Jan2014CalcDate = MIN(CASE WHEN CalculationMonth = '20140101' THEN CalculationDate END),
Jan2014Result = SUM(CASE WHEN CalculationMonth = '20140101' THEN PLResult END),
Feb2014CalcDate = MIN(CASE WHEN CalculationMonth = '20140201' THEN CalculationDate END),
Feb2014Result = SUM(CASE WHEN CalculationMonth = '20140201' THEN PLResult END)
FROM Data
GROUP BY DenseRank
ORDER BY DenseRank;
这给出:
Jan2014CalcDate Jan2014Result Feb2014CalcDate Feb2014Result
2014-01-02 100 2014-02-03 300
2014-01-03 200 2014-02-04 400
NULL NULL 2014-02-27 500
由于您的月份数是动态的,因此您需要动态构建上述语句并使用 SP_EXECUTESQL
来运行它:
DECLARE @SQL NVARCHAR(MAX) = '';
WITH Months AS
( SELECT M,
ColName = DATENAME(MONTH, M) + DATENAME(YEAR, M),
CharFormat = CONVERT(VARCHAR(8), M, 112)
FROM ( SELECT DISTINCT M = DATEADD(MONTH, DATEDIFF(MONTH, 0, CalculationDate), 0)
FROM #PLResultPerDay
) m
)
SELECT @SQL = 'WITH Data AS
( SELECT CalculationDate,
PLResult,
CalculationMonth,
DenseRank = DENSE_RANK() OVER(PARTITION BY CalculationMonth ORDER BY CalculationDate)
FROM ( SELECT CalculationDate,
PLResult,
CalculationMonth = DATEADD(MONTH, DATEDIFF(MONTH, 0, CalculationDate), 0)
FROM #PLResultPerDay
) pl
)
SELECT ' +
STUFF(( SELECT ', ' + ColName + 'CalculationDate = MIN(CASE WHEN CalculationMonth = ''' + CharFormat + ''' THEN CalculationDate END), ' +
ColName + 'PLResult = SUM(CASE WHEN CalculationMonth = ''' + CharFormat + ''' THEN PLResult END)'
FROM Months
ORDER BY M
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)'), 1, 2, '') +
'FROM Data
GROUP BY DenseRank
ORDER BY DenseRank;';
EXECUTE SP_EXECUTESQL @SQL;
请注意,我仍然建议不要使用这种技术,并认为 SQL 应该留给存储/检索数据,以及用于格式化数据的表示层
关于sql-server - 将垂直结果转换为水平模式(T-SQL),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20836576/