T-SQL计算不同年份范围之间的持续时间(以月为单位)

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几个月.请帮我找一个解决方案.

Answer 1

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