Question

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。

Answer 1

此处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：

结果：

ckeditor.js

Answer 2

你可以试试这个

render [data:['name':'firstname','surname':'secondName'] as JSON]

Answer 3

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

结果：

ID  FromDate    ToDate       Difference IN Months
===================================================    
1   2000-10-08  2011-03-08    125
2   2013-06-08  2015-01-08    19

T-SQL计算不同年份范围之间的持续时间（以月为单位）

3 个答案: