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[英]Calculate length of service of an employee in Years, Months and Days using T-SQL when there is a gap in service?
[英]T-SQL Calculate duration in months between different years of ranges
我在SQL Server中有一个表,其中包含用于不同作业的用户的持续时间。 我需要计算用户的总体验数。
Declare @temp table(Id int, FromDate DATETIME, ToDate DATETIME)
INSERT INTO @temp ( Id ,FromDate ,ToDate )
VALUES ( 1 , '2003-1-08 06:55:56' , '2005-5-08 06:55:56'),
( 2 , '2000-10-08 06:55:56' , '2008-7-08 06:55:56'),
( 3 , '2013-6-08 06:55:56' , '2015-1-08 06:55:56'),
( 4 , '2006-4-08 06:55:56' , '2011-3-08 06:55:56' )
SELECT * FROM @temp
我想计算经验总数;
Id FromDate ToDate Difference IN Months
===================================================
1 2003-01-08 2005-05-08 28
2 2000-10-08 2008-07-08 93
3 2013-06-08 2015-01-08 19
4 2006-04-08 2011-03-08 59
去除了像2003-2005这样的年份重叠在2000 - 2008年; 我有这样的事情:
Id FromDate ToDate Difference IN Months
===================================================
1 2000-10-08 2011-03-08 125
2 2013-06-08 2015-01-08 19
所以答案是125+19 = 144
个月。 请帮我找一个解决方案。
这里的语法是查找没有重叠FromDate和ToDate间隔的所有FromDate以及没有重叠FromDate和ToDate间隔的所有ToDate。 根据日期值给他们一个rownumber并在rownumber上匹配它们:
;WITH CTE as
(
SELECT min(Id) Id ,FromDate, row_number() over (ORDER BY FromDate) rn
FROM @temp x
WHERE
not exists
(SELECT * FROM @temp WHERE x.FromDate > FromDate and x.FromDate <= Todate)
GROUP BY FromDate
), CTE2 as
(
SELECT Max(Id) Id ,ToDate, row_number() over (ORDER BY ToDate) rn
FROM @temp x
WHERE
not exists
(SELECT * FROM @temp WHERE x.ToDate >= FromDate and x.ToDate < Todate)
GROUP BY ToDate
)
SELECT SUM(DateDiff(month, CTE.FromDate, CTE2.ToDate))
FROM CTE
JOIN CTE2
ON CTE.rn = CTE2.rn
结果:
144
你可以试试这个
SELECT Set1.FromDate,MIN(List1.ToDate) AS ToDate, DATEDIFF(MONTH,Set1.FromDate,MIN(List1.ToDate))
FROM @temp Set1
INNER JOIN @temp List1 ON Set1.FromDate <= List1.ToDate
AND NOT EXISTS(SELECT * FROM @temp List2
WHERE List1.ToDate >= List2.FromDate AND List1.ToDate < List2.ToDate)
WHERE NOT EXISTS(SELECT * FROM @temp Set2
WHERE Set1.FromDate > Set2.FromDate AND Set1.FromDate <= Set2.ToDate)
GROUP BY Set1.FromDate
ORDER BY Set1.FromDate
你可以试试这段代码:
DECLARE @temp TABLE (ID INT, FromDate DATETIME, ToDate DATETIME)
INSERT INTO @temp (ID, FromDate, ToDate)
VALUES ( 1 , '2003-1-08 06:55:56' , '2005-5-08 06:55:56'),
( 2 , '2000-10-08 06:55:56' , '2008-7-08 06:55:56'),
( 3 , '2013-6-08 06:55:56' , '2015-1-08 06:55:56'),
( 4 , '2006-4-08 06:55:56' , '2011-3-08 06:55:56' )
SELECT
ID,
CONVERT(DATE, FromDate) AS FromDate,
CONVERT(DATE, ToDate) AS ToDate,
DATEDIFF(MONTH, FromDate, ToDate) AS [Difference IN Months]
INTO #tmp
FROM @temp
SELECT T1.ID AS ID1, T2.ID AS ID2, T2.ToDate, T1.[Difference IN Months] + T2.[Difference IN Months] AS [Difference IN Months]
INTO #tmp2
FROM #tmp T1
INNER JOIN #tmp T2
ON CAST(T1.ToDate AS DATE) = CAST(T2.FromDate AS DATE)
OR (YEAR(T1.ToDate) = YEAR(T2.FromDate) AND CAST(T1.ToDate AS DATE) < CAST(T2.FromDate AS DATE))
OR YEAR(T1.ToDate) = YEAR(T2.FromDate) - 1
DELETE #tmp WHERE ID IN (SELECT ID2 FROM #tmp2)
UPDATE #tmp
SET ToDate = (SELECT ToDate FROM #tmp2 WHERE #tmp.ID = ID1),
[Difference IN Months] = (SELECT [Difference IN Months] FROM #tmp2 WHERE #tmp.ID = ID1)
WHERE ID IN (SELECT ID1 FROM #tmp2)
SELECT
*,
ROW_NUMBER() OVER(ORDER BY FromDate) AS RF
INTO #tmp3
FROM #tmp
SELECT T1.ID AS ID1, T2.ID AS ID2, T1.ToDate
INTO #tmp4
FROM #tmp3 T1
INNER JOIN #tmp3 T2 ON T1.RF = T2.RF + 1
WHERE CAST(T1.FromDate AS DATE) < CAST(T2.ToDate AS DATE)
UPDATE #tmp
SET ToDate = (SELECT ToDate FROM #tmp4 WHERE #tmp.ID = ID2)
WHERE ID IN (SELECT ID2 FROM #tmp4)
DELETE #tmp WHERE ID IN (SELECT ID1 FROM #tmp4)
UPDATE #tmp
SET[Difference IN Months] = DATEDIFF(MONTH, FromDate, ToDate)
SELECT
ROW_NUMBER() OVER(ORDER BY FromDate) AS ID,
FromDate, ToDate, [Difference IN Months]
FROM #tmp
DROP TABLE #tmp
DROP TABLE #tmp2
DROP TABLE #tmp3
DROP TABLE #tmp4
结果:
ID FromDate ToDate Difference IN Months
===================================================
1 2000-10-08 2011-03-08 125
2 2013-06-08 2015-01-08 19
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