T-SQL计算不同年份范围之间的持续时间(以月为单位)
我在SQL Server中有一个表,其中包含用户为不同作业工作的持续时间。我需要计算用户体验的总数T-SQL计算不同年份范围之间的持续时间(以月为单位),sql,sql-server,tsql,sql-server-2012,sqldatetime,Sql,Sql Server,Tsql,Sql Server 2012,Sqldatetime,我在SQL Server中有一个表,其中包含用户为不同作业工作的持续时间。我需要计算用户体验的总数 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
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
我想计算经验的总数
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
在2000-2008年消除了2003-2005年等重叠年份的掩盖之后;
我得到了这样的东西:
Id FromDate ToDate Difference IN Months
===================================================
1 2000-10-08 2011-03-08 125
2 2013-06-08 2015-01-08 19
因此答案是125+19=144
monds。
请帮助我找到解决方案。这里的语法是查找所有FromDate和ToDate间隔不重叠的FromDate,以及所有FromDate和ToDate间隔不重叠的ToDate。根据日期值为他们指定一个行号,并在该行号上匹配他们:
;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
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
选择Set1.FromDate,MIN(List1.ToDate)作为ToDate,DATEDIFF(MONTH,Set1.FromDate,MIN(List1.ToDate))
从@temp Set1开始
Set1.FromDate=List2.FromDate和List1.ToDateSet2.FromDate和Set1.FromDate您可以尝试以下代码:
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
每天的部分总是8吗?所以你从来没有需要考虑的部分月份吗?你尝试过DATEDIFF函数吗?SQL Server的版本是什么?SQL Server 2012,但这并不重要。这正是我想要的,谢谢
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
ID FromDate ToDate Difference IN Months
===================================================
1 2000-10-08 2011-03-08 125
2 2013-06-08 2015-01-08 19