SQL Server查询日期范围的语句

Question

我有一个包含数千个条目的数据库，有两个日期/时间字段，我们将它们称为In和Out:

In                                      Out
2011-06-16 13:45:11.000    2011-06-16 13:49:12.000
2011-06-16 13:51:31.000    2011-06-16 14:23:17.000
2011-06-16 14:51:54.000    2011-06-16 14:58:43.000

等等…

在输入上我有一个时间框架(timeIn和timeOut)。在输出中，我需要知道属于这个时间框架的每个记录的"In"one_answers"Out"时间之间的累积秒数。用这个图表更容易说明:

http://i150.photobucket.com/albums/s99/dc2000_bucket/sql_graph.jpg

从图中取名字，我需要知道"时间间隔1(以秒为单位)"+"时间间隔2(秒)"+"时间间隔3(秒)"。如何编写这样的SQL语句?

Answer 1

select sum(datediff(second, [In], case when @TimeOut < [Out] 
                                    then @TimeOut 
                                    else [Out] 
                                  end))
from YourTable
where [In] between @TimeIn and @TimeOut

或者像这样，如果你想包括在@TimeIn - @TimeOut结束的间隔。

select sum(datediff(second, case when @TimeIn < [In] 
                              then [In]
                              else @TimeIn
                            end, 
                            case when @TimeOut < [Out] 
                              then @TimeOut 
                              else [Out] 
                            end))
from YourTable
where [In] <= @TimeOut and 
      [Out] >= @TimeIn

Answer 2

SQL-Server中没有GREATEST()或LEAST()函数，否则会像这样:

SELECT LEAST(t.timeOut, @TimeOut) - GREATEST(t.timeIn,@TimeIn) 
FROM tableX AS t
WHERE t.timeIn <= @TimeOut
  AND @TimeIn <= t.timeOut

使用CASE:

SELECT DATEDIFF( second
               , CASE WHEN t.timeIn <= @TimeIn
                        THEN @TimeIn
                        ELSE t.timeIn
                 END
               , CASE WHEN t.timeOut <= @TimeOut
                        THEN t.timeOut
                        ELSE @TimeOut
                 END 
               )
       AS TimeInterval 
FROM tableX AS t
WHERE t.timeIn <= @TimeOut
  AND @TimeIn <= t.timeOut
ORDER BY t.timeIn