Question

我有一个问题，我解决了，但我写了一个很长的程序，我不能确定它涵盖了所有可能的情况。

问题：

如果我有主间隔时间（From A to B），次要间隔时间（很多或没有）

(`From X to Y AND From X` to Y` AND X`` to  Y`` AND ....`)

我希望在分钟的分钟中 SUM 我的主间隔时间（AB）所有部分的效率和最小数量条件（SQL Server Procedure和C＃方法）？

例如：如果我的主要时间间隔来自02:00 to 10:30 并说一个次要区间来自04:00 to 08:00

现在我想要这个结果：((04:00 - 02:00) + (10:30 -08:00))* 60

图表示例：

在第一种情况下，结果将是：

((X-A) + (B-Y)) * 60

当我有很多次要时期时会更复杂。

注意：

只有在我必须将主要时段[A，B]与的 UNION 进行比较时，才会发生次要时间间隔的重叠，最多两个并行的次要时间间隔。第一组必须只包含一个辅助区间，第二组包含（很多或没有）辅助区间。例如，在比较[A,B]到（2,5组）的图表中第一组(2)由一个辅助区间组成，第二组(5)由三个辅助区间组成。这是最糟糕的情况，我需要处理。

例如：

如果我的主要间隔为[15:00,19:40] 我有两组次要间隔。根据我的规则，这些集合中至少有一个应该由一个次要间隔组成。说第一组是[11:00 ,16:00] 第二组由两个次要区间[10:00,15:00],[16:30,17:45]组成现在我想要结果(16:30 -16:00) +(19:40 -17:45)

根据评论：

我的表是这样的：

第一个表包含次要期间，特定员工在同一日期最多有两组次要期间。第一组在工作日(W) [work_st,work_end]中仅包含一个辅助句点，如果该日期为周末[E]，则此设置为空，在这种情况下，辅助句点之间不会重叠。第二组可能包含同一日期[check_in,check_out]的许多次要期间，因为员工可能会在同一天多次check_in_out。

emp_num  day_date   work_st    work_end   check_in   check_out     day_state

  547    2015-4-1   08:00       16:00     07:45      12:10           W
  547    2015-4-1   08:00       16:00     12:45      17:24           W
  547    2015-4-2   00:00       00:00     07:11      13:11           E

第二个表格包含主要期间[A,B]，它是该员工当天的一个期间（一个记录）

emp_num  day_date   mission_in    mission_out
  547    2015-4-1    15:00          21:30
  547    2015-4-2    8:00           14:00

在前面的例子中，如果我有一个所需的过程或方法，这个过程应该有两个参数：

日期
emp_num

在上一个例子中，它应该像这个('2015-4-1' ,547)

根据我的解释：

第二张表中的主要时期（任务期）[A,B]：该员工在该日期应该只有一个期间

[15:00,21:30]
该员工的通过日期('2015-4-1')的次要期间为2 第一张表中的次要期间（最坏情况）

第一组应仅包含一个辅助句点（或零）句点）[08:00,16:00]第二集可以包含许多次要的期间（或零期）

[07:45,12:10]，[12:45,17:24]

输出应该是[17：24,21：30]转换为分钟

注意

所有day_date,mission_in,mission_out,work_st,work_end,check_in,check_out 是datetime字段，但我只是在示例中加入时间进行简化，我想忽略除day_date之外的日期部分，因为它是我计算的日期除了emp_num。

enter image description here

Answer 1

我必须解决这个问题才能消化一些调度数据。这允许多个在线时间，但假设它们不重叠。

select convert(datetime,'1/1/2015 5:00 AM') StartDateTime,  convert(datetime,'1/1/2015 5:00 PM') EndDateTime, convert(varchar(20),'Online') IntervalType into #CapacityIntervals
insert into #CapacityIntervals select '1/1/2015 4:00 AM' StartDateTime,  '1/1/2015 6:00 AM' EndDateTime, 'Offline' IntervalType
insert into #CapacityIntervals select '1/1/2015 5:00 AM' StartDateTime,  '1/1/2015 6:00 AM' EndDateTime, 'Offline' IntervalType
insert into #CapacityIntervals select '1/1/2015 10:00 AM' StartDateTime,  '1/1/2015 12:00 PM' EndDateTime, 'Offline' IntervalType
insert into #CapacityIntervals select '1/1/2015 11:00 AM' StartDateTime,  '1/1/2015 1:00 PM' EndDateTime, 'Offline' IntervalType
insert into #CapacityIntervals select '1/1/2015 4:00 PM' StartDateTime,  '1/1/2015 6:00 PM' EndDateTime, 'Offline' IntervalType
insert into #CapacityIntervals select '1/1/2015 1:30 PM' StartDateTime,  '1/1/2015 2:00 PM' EndDateTime, 'Offline' IntervalType

    --Populate your Offline table
    select 
        ROW_NUMBER() over (Order by StartDateTime, EndDateTime) Rownum,
        StartDateTime, 
        EndDateTime
    into #Offline
    from #CapacityIntervals
    where IntervalType in ('Offline','Cleanout')
    group by StartDateTime, EndDateTime

    --Populate your Online table
    select 
        ROW_NUMBER() over (Order by StartDateTime, EndDateTime) Rownum,
        StartDateTime, 
        EndDateTime
    into #Online
    from #CapacityIntervals
    where IntervalType not in ('Offline','Cleanout')


    --If you have overlapping online intervals... check for those here and consolidate.


    -------------------------------
    --find overlaping offline times
    -------------------------------
    declare @Finished as tinyint
    set @Finished = 0

    while @Finished = 0
    Begin
        update #Offline
        set #Offline.EndDateTime = OverlapEndDates.EndDateTime
        from #Offline
        join
            (
            select #Offline.Rownum,
                MAX(Overlap.EndDateTime) EndDateTime
            from #Offline
            join #Offline Overlap
            on Overlap.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
                and #Offline.Rownum <= Overlap.Rownum
            group by #Offline.Rownum
            ) OverlapEndDates
        on #Offline.Rownum = OverlapEndDates.Rownum

        --Remove Online times completely inside of online times
        delete #Offline
        from #Offline
        join #Offline Overlap
        on #Offline.StartDateTime between Overlap.StartDateTime and Overlap.EndDateTime
            and #Offline.EndDateTime between Overlap.StartDateTime and Overlap.EndDateTime
            and #Offline.Rownum > Overlap.Rownum

        --LOOK IF THERE ARE ANY MORE CHAINS LEFT    
        IF NOT EXISTS(
                select #Offline.Rownum,
                    MAX(Overlap.EndDateTime) EndDateTime
                from #Offline
                join #Offline Overlap
                on  Overlap.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
                    and #Offline.Rownum < Overlap.Rownum
                group by #Offline.Rownum
                )
            SET @Finished = 1
    END

    -------------------------------
    --Modify Online times with offline ranges
    -------------------------------

    --delete any Online times completely inside offline range
    delete #Online 
    from #Online
    join #Offline
    on #Online.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
        and #Online.EndDateTime between #Offline.StartDateTime and #Offline.EndDateTime

    --Find Online Times with offline range at the beginning
    update #Online
    set #Online.StartDateTime = #Offline.EndDateTime
    from #Online
    join #Offline
    on #Online.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
        and #Online.EndDateTime >= #Offline.EndDateTime

    --Find Online Times with offline range at the end
    update #Online
    set #Online.EndDateTime = #Offline.StartDateTime
    from #Online
    join #Offline
    on #Online.StartDateTime <= #Offline.StartDateTime
        and #Online.EndDateTime between #Offline.StartDateTime and #Offline.EndDateTime

    --Find Online Times with offline range punched in the middle
    select #Online.Rownum, 
        #Offline.Rownum OfflineRow,
        #Offline.StartDateTime,
        #Offline.EndDateTime,
        ROW_NUMBER() over (Partition by #Online.Rownum order by #Offline.Rownum Desc) OfflineHoleNumber
    into #OfflineHoles
    from #Online
    join #Offline
    on #Offline.StartDateTime between #Online.StartDateTime and #Online.EndDateTime
        and #Offline.EndDateTime between #Online.StartDateTime and #Online.EndDateTime

    declare @HoleNumber as integer
    select @HoleNumber = isnull(MAX(OfflineHoleNumber),0) from #OfflineHoles

    --Punch the holes out of the online times
    While @HoleNumber > 0
    Begin
        insert into #Online 
        select
            -1 Rownum,
            #OfflineHoles.EndDateTime StartDateTime,
            #Online.EndDateTime EndDateTime
        from #Online
        join #OfflineHoles
        on #Online.Rownum = #OfflineHoles.Rownum
        where OfflineHoleNumber = @HoleNumber

        update #Online
        set #Online.EndDateTime = #OfflineHoles.StartDateTime
        from #Online
        join #OfflineHoles
        on #Online.Rownum = #OfflineHoles.Rownum
        where OfflineHoleNumber = @HoleNumber

        set @HoleNumber=@HoleNumber-1
    end

--Output total hours
select SUM(datediff(second,StartDateTime, EndDateTime)) / 3600.0 TotalHr
from #Online

--see how it split up the online intervals
select * 
from #Online
order by StartDateTime, EndDateTime

Answer 2

我已经使用您的数据示例更新了我的答案，并且我正在为使用图表中的案例2和5的员工248添加另一个示例。

--load example data for emply 547
select CONVERT(int, 547) emp_num, 
    Convert(datetime, '2015-4-1') day_date, 
    Convert(datetime, '2015-4-1 08:00') work_st,
    Convert(datetime, '2015-4-1 16:00') work_end, 
    Convert(datetime, '2015-4-1 07:45') check_in, 
    Convert(datetime, '2015-4-1 12:10') check_out, 
    'W' day_state
into #SecondaryIntervals
insert into #SecondaryIntervals select 547, '2015-4-1', '2015-4-1 08:00', '2015-4-1 16:00', '2015-4-1 12:45', '2015-4-1 17:24', 'W'
insert into #SecondaryIntervals select 547, '2015-4-2', '2015-4-2 00:00', '2015-4-2 00:00', '2015-4-2 07:11', '2015-4-2 13:11', 'E'

select CONVERT(int, 547) emp_num, 
    Convert(datetime, '2015-4-1') day_date, 
    Convert(datetime, '2015-4-1 15:00') mission_in,
    Convert(datetime, '2015-4-1 21:30') mission_out
into #MainIntervals
insert into #MainIntervals select 547, '2015-4-2', '2015-4-2 8:00', '2015-4-2 14:00'

--load more example data for an employee 548 with overlapping secondary intervals
insert into #SecondaryIntervals select 548, '2015-4-1', '2015-4-1 06:00', '2015-4-1 11:00', '2015-4-1 9:00', '2015-4-1 10:00', 'W'
insert into #SecondaryIntervals select 548, '2015-4-1', '2015-4-1 06:00', '2015-4-1 11:00', '2015-4-1 10:30', '2015-4-1 12:30', 'W'
insert into #SecondaryIntervals select 548, '2015-4-1', '2015-4-1 06:00', '2015-4-1 11:00', '2015-4-1 13:15', '2015-4-1 16:00', 'W'

insert into #MainIntervals select 548, '2015-4-1', '2015-4-1 8:00', '2015-4-1 14:00'

--Populate your Offline table with the intervals in #SecondaryIntervals
select 
    ROW_NUMBER() over (Order by emp_num, day_date, StartDateTime, EndDateTime) Rownum,
    emp_num,
    day_date,
    StartDateTime, 
    EndDateTime
into #Offline
from 
    (select emp_num,
        day_date,
        work_st StartDateTime, 
        work_end EndDateTime
    from #SecondaryIntervals
    where day_state = 'W'
    Group by emp_num,
        day_date,
        work_st, 
        work_end
    union
    select 
        emp_num,
        day_date,
        check_in StartDateTime, 
        check_out EndDateTime
    from #SecondaryIntervals
    Group by emp_num,
        day_date,
        check_in, 
        check_out
    ) SecondaryIntervals


--Populate your Online table
select 
    ROW_NUMBER() over (Order by emp_num, day_date, mission_in, mission_out) Rownum,
    emp_num,
    day_date,
    mission_in StartDateTime, 
    mission_out EndDateTime
into #Online
from #MainIntervals
group by emp_num,
    day_date,
    mission_in,
    mission_out


-------------------------------
--find overlaping offline times
-------------------------------
declare @Finished as tinyint
set @Finished = 0

while @Finished = 0
Begin
    update #Offline
    set #Offline.EndDateTime = OverlapEndDates.EndDateTime
    from #Offline
    join
        (
        select #Offline.Rownum,
            MAX(Overlap.EndDateTime) EndDateTime
        from #Offline
        join #Offline Overlap
        on #Offline.emp_num = Overlap.emp_num
            and #Offline.day_date = Overlap.day_date
            and Overlap.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
            and #Offline.Rownum <= Overlap.Rownum
        group by #Offline.Rownum
        ) OverlapEndDates
    on #Offline.Rownum = OverlapEndDates.Rownum

    --Remove Online times completely inside of online times
    delete #Offline
    from #Offline
    join #Offline Overlap
    on #Offline.emp_num = Overlap.emp_num
        and #Offline.day_date = Overlap.day_date
        and #Offline.StartDateTime between Overlap.StartDateTime and Overlap.EndDateTime
        and #Offline.EndDateTime between Overlap.StartDateTime and Overlap.EndDateTime
        and #Offline.Rownum > Overlap.Rownum

    --LOOK IF THERE ARE ANY MORE CHAINS LEFT    
    IF NOT EXISTS(
            select #Offline.Rownum,
                MAX(Overlap.EndDateTime) EndDateTime
            from #Offline
            join #Offline Overlap
            on #Offline.emp_num = Overlap.emp_num
                and #Offline.day_date = Overlap.day_date
                and Overlap.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
                and #Offline.Rownum < Overlap.Rownum
            group by #Offline.Rownum
            )
        SET @Finished = 1
END

-------------------------------
--Modify Online times with offline ranges
-------------------------------

--delete any Online times completely inside offline range
delete #Online 
from #Online
join #Offline
on #Online.emp_num = #Offline.emp_num
    and #Online.day_date = #Offline.day_date
    and #Online.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
    and #Online.EndDateTime between #Offline.StartDateTime and #Offline.EndDateTime

--Find Online Times with offline range at the beginning
update #Online
set #Online.StartDateTime = #Offline.EndDateTime
from #Online
join #Offline
on #Online.emp_num = #Offline.emp_num
    and #Online.day_date = #Offline.day_date
    and #Online.StartDateTime between #Offline.StartDateTime and #Offline.EndDateTime
    and #Online.EndDateTime >= #Offline.EndDateTime

--Find Online Times with offline range at the end
update #Online
set #Online.EndDateTime = #Offline.StartDateTime
from #Online
join #Offline
on #Online.emp_num = #Offline.emp_num
    and #Online.day_date = #Offline.day_date
    and #Online.StartDateTime <= #Offline.StartDateTime
    and #Online.EndDateTime between #Offline.StartDateTime and #Offline.EndDateTime

--Find Online Times with offline range punched in the middle
select #Online.Rownum, 
    #Offline.Rownum OfflineRow,
    #Offline.StartDateTime,
    #Offline.EndDateTime,
    ROW_NUMBER() over (Partition by #Online.Rownum order by #Offline.Rownum Desc) OfflineHoleNumber
into #OfflineHoles
from #Online
join #Offline
on #Online.emp_num = #Offline.emp_num
    and #Online.day_date = #Offline.day_date
    and #Offline.StartDateTime between #Online.StartDateTime and #Online.EndDateTime
    and #Offline.EndDateTime between #Online.StartDateTime and #Online.EndDateTime

declare @HoleNumber as integer
select @HoleNumber = isnull(MAX(OfflineHoleNumber),0) from #OfflineHoles

--Punch the holes out of the online times
While @HoleNumber > 0
Begin
    insert into #Online 
    select
        -1 Rownum,
        #Online.emp_num,
        #Online.day_date,
        #OfflineHoles.EndDateTime StartDateTime,
        #Online.EndDateTime EndDateTime
    from #Online
    join #OfflineHoles
    on #Online.Rownum = #OfflineHoles.Rownum
    where OfflineHoleNumber = @HoleNumber

    update #Online
    set #Online.EndDateTime = #OfflineHoles.StartDateTime
    from #Online
    join #OfflineHoles
    on #Online.Rownum = #OfflineHoles.Rownum
    where OfflineHoleNumber = @HoleNumber

    set @HoleNumber=@HoleNumber-1
end

--Output total hours
select emp_num, day_date, 
    SUM(datediff(second,StartDateTime, EndDateTime)) / 3600.0 TotalHr, 
    SUM(datediff(second,StartDateTime, EndDateTime)) / 60.0 TotalMin
from #Online
group by emp_num, day_date
order by 1, 2

--see how it split up the online intervals
select emp_num, day_date, StartDateTime, EndDateTime
from #Online
order by 1, 2, 3, 4

输出是：

emp_num     day_date                TotalHr                                 TotalMin
----------- ----------------------- --------------------------------------- ---------------------------------------
547         2015-04-01 00:00:00.000 4.100000                                246.000000
547         2015-04-02 00:00:00.000 0.816666                                49.000000
548         2015-04-01 00:00:00.000 0.750000                                45.000000

(3 row(s) affected)

emp_num     day_date                StartDateTime           EndDateTime
----------- ----------------------- ----------------------- -----------------------
547         2015-04-01 00:00:00.000 2015-04-01 17:24:00.000 2015-04-01 21:30:00.000
547         2015-04-02 00:00:00.000 2015-04-02 13:11:00.000 2015-04-02 14:00:00.000
548         2015-04-01 00:00:00.000 2015-04-01 12:30:00.000 2015-04-01 13:15:00.000

(3 row(s) affected)

我留下了我的另一个答案，因为它更通用，以防其他人想要抓住它。我看到你为这个问题增加了一笔赏金。让我知道，如果我的答案有一些特定的东西不能满足你，我会尽力帮助你。我使用这种方法处理数千个间隔，并在几秒钟后返回。

Answer 3

我的解决方案与弗拉基米尔·巴拉诺夫非常相似。

链接到.NetFiddle

一般想法

我的算法基于interval tree的修改。它假设最小的时间单位是1分钟（易于修改）。

每个树节点处于3状态中的1个状态：未访问，访问和使用。该算法基于递归搜索功能，可通过以下步骤进行描述：

如果节点已使用或搜索间隔为空，则返回空间隔。
如果节点未访问且节点间隔等于搜索间隔，则将当前节点标记为 used 并返回节点间隔。
将节点标记为已访问，分割搜索间隔并为左右儿童返回搜索的总和。

步骤解决方案

计算最大间隔。
添加到树“次要间隔”。
添加到树“主要间隔”。
计算间隔总和。

请注意我认为间隔是[开始;结束]，即两个区间都是包容性的，很容易改变。

<强>要求

假设

n - “次要间隔”的数量

m - 基本单位的最长时间

构造需要O（2n）存储空间并在O（n log n + m）时间内工作。

这是我的代码

  public class Interval
    {
        public int Start { get; set; }

        public int End { get; set; }
    };
    enum Node
    {
        Unvisited = 0,
        Visited = 1,
        Used = 2
    };
    Node[] tree;

    public void Calculate()
    {
        var secondryIntervalsAsDates = new List<Tuple<DateTime,DateTime>> { new Tuple<DateTime, DateTime>( new DateTime(2015, 03, 15, 4, 0, 0), new DateTime(2015, 03, 15, 5, 0, 0))};
        var mainInvtervalAsDate = new Tuple<DateTime, DateTime>(new DateTime(2015, 03, 15, 3, 0, 0), new DateTime(2015, 03, 15, 7, 0, 0));
        // calculate biggest interval
        var startDate = secondryIntervalsAsDates.Union( new List<Tuple<DateTime,DateTime>>{mainInvtervalAsDate}).Min(s => s.Item1).AddMinutes(-1);
        var endDate = secondryIntervalsAsDates.Union(new List<Tuple<DateTime, DateTime>> { mainInvtervalAsDate }).Max(s => s.Item2);
        var mainInvterval = new Interval { Start = (int)(mainInvtervalAsDate.Item1 - startDate).TotalMinutes, End = (int)(mainInvtervalAsDate.Item2 - startDate).TotalMinutes };
        var wholeInterval = new Interval { Start = 1, End = (int)(endDate - startDate).TotalMinutes};
        //convert intervals to minutes
        var secondaryIntervals = secondryIntervalsAsDates.Select(s => new Interval { Start = (int)(s.Item1 - startDate).TotalMinutes, End = (int)(s.Item2 - startDate).TotalMinutes}).ToList();
        tree = new Node[wholeInterval.End * 2 + 1];
        //insert secondary intervals
        secondaryIntervals.ForEach(s => Search(wholeInterval, s, 1));
        //insert main interval
        var result = Search(wholeInterval, mainInvterval, 1);
        //calculate result
        var minutes = result.Sum(r => r.End - r.Start) + result.Count();
    }

    public IEnumerable<Interval> Search(Interval current, Interval searching, int index)
    {
        if (tree[index] == Node.Used || searching.End < searching.Start)
        {
            return new List<Interval>();
        }
        if (tree[index] == Node.Unvisited && current.Start == searching.Start && current.End == searching.End)
        {
            tree[index] = Node.Used;
            return new List<Interval> { current };
        }
        tree[index] = Node.Visited;
        return Search(new Interval { Start = current.Start, End = current.Start + (current.End - current.Start) / 2 },
                  new Interval { Start = searching.Start, End = Math.Min(searching.End, current.Start + (current.End - current.Start) / 2)  }, index * 2).Union(
            Search(new Interval { Start = current.Start + (current.End - current.Start) / 2 + 1 , End = current.End},
              new Interval { Start = Math.Max(searching.Start, current.Start + (current.End - current.Start) / 2 + 1), End = searching.End }, index * 2 + 1));
    }

Answer 4

以下是SQLFiddle完整查询。

我将展示如何构建一个返回每个emp_num, day_date的分钟数的查询。如果事实证明特定emp_num, day_date没有剩余时间，那么不的结果会有一行0，根本就没有这样的行。

一般想法

我将使用table of numbers。我们只需要24*60=1440个数字，但最好在数据库中为其他报告提供此类表格。我个人拥有10万行。这是very good article比较生成此类表的不同方法。

对于每个间隔，我将使用数字表生成一组行 - 间隔中每分钟一行。我假设间隔是[start; end)，即开始分钟是包含的，结束分钟是独占的。例如，从07:00到08:00的时间间隔为60分钟，而不是61。

生成数字表

DECLARE @Numbers TABLE (N int);
INSERT INTO @Numbers(N)
SELECT TOP(24*60)
    ROW_NUMBER() OVER(ORDER BY S.object_id) - 1 AS N
FROM
    sys.all_objects AS S
ORDER BY N
;

对于此任务，最好使用从0开始的数字。通常，您可以将其作为N上具有主键的永久表。

示例数据

DECLARE @Missions TABLE (emp_num int, day_date datetime, mission_in datetime, mission_out datetime);
DECLARE @Periods TABLE (emp_num int, day_date datetime, work_st datetime, work_end datetime, check_in datetime, check_out datetime, day_state char(1));

INSERT INTO @Missions (emp_num, day_date, mission_in, mission_out) VALUES
(547, '2015-04-01', '2015-04-01 15:00:00', '2015-04-01 21:30:00'),
(547, '2015-04-02', '2015-04-02 08:00:00', '2015-04-02 14:00:00');

INSERT INTO @Periods (emp_num, day_date, work_st, work_end, check_in, check_out, day_state) VALUES
(547, '2015-04-01', '2015-04-01 08:00:00', '2015-04-01 16:00:00', '2015-04-01 07:45:00', '2015-04-01 12:10:00', 'W'),
(547, '2015-04-01', '2015-04-01 08:00:00', '2015-04-01 16:00:00', '2015-04-01 12:45:00', '2015-04-01 17:24:00', 'W'),
(547, '2015-04-02', '2015-04-02 00:00:00', '2015-04-02 00:00:00', '2015-04-02 07:11:00', '2015-04-02 13:11:00', 'E');

我的解决方案不会使用day_state列。我希望00:00:00和work_st都有work_end。解决方案期望同一行中的日期组件相同，并且day_date没有时间组件。

如果我为此任务设计了架构，我将有三个表而不是两个：Missions，WorkPeriods和CheckPeriods。我会将您的表格Periods拆分为两个，以避免在多行中重复work_st和work_end。但是这个解决方案将处理您当前的架构，它将基本上生成第三个表。在实践中，这意味着可以提高绩效。

任务分钟

WITH
CTE_MissionMinutes
AS
(
    SELECT emp_num, day_date, N.N
    FROM
        @Missions AS M
        CROSS JOIN @Numbers AS N
    WHERE
        N.N >= DATEDIFF(minute, M.day_date, M.mission_in) AND
        N.N < DATEDIFF(minute, M.day_date, M.mission_out)
)

@Missions中的每个原始行都会变成一组行，每隔一分钟就有一行(mission_in, mission_out)。

工作期

,CTE_WorkPeriods
AS
(
    SELECT P.emp_num, P.day_date, P.work_st, P.work_end
    FROM @Periods AS P
    GROUP BY P.emp_num, P.day_date, P.work_st, P.work_end
)

生成第三个帮助表 - 每个emp_num, day_date, work_st, work_end一行 - (work_st, work_end)的所有间隔。

工作和检查分钟

,CTE_WorkMinutes
AS
(
    SELECT emp_num, day_date, N.N
    FROM
        CTE_WorkPeriods
        CROSS JOIN @Numbers AS N
    WHERE
        N.N >= DATEDIFF(minute, CTE_WorkPeriods.day_date, CTE_WorkPeriods.work_st) AND
        N.N < DATEDIFF(minute, CTE_WorkPeriods.day_date, CTE_WorkPeriods.work_end)
)
,CTE_CheckMinutes
AS
(
    SELECT emp_num, day_date, N.N
    FROM
        @Periods AS P
        CROSS JOIN @Numbers AS N
    WHERE
        N.N >= DATEDIFF(minute, P.day_date, P.check_in) AND
        N.N < DATEDIFF(minute, P.day_date, P.check_out)
)

与Missions完全相同。

联盟＆＃34;次要时间间隔＆＃34;

,CTE_UnionPeriodMinutes
AS
(
    SELECT emp_num, day_date, N
    FROM CTE_WorkMinutes

    UNION ALL -- can be not ALL here, but ALL is usually faster

    SELECT emp_num, day_date, N
    FROM CTE_CheckMinutes
)

从主要

中减去辅助间隔

,CTE_FinalMinutes
AS
(
    SELECT emp_num, day_date, N
    FROM CTE_MissionMinutes

    EXCEPT

    SELECT emp_num, day_date, N
    FROM CTE_UnionPeriodMinutes
)

总结分钟数

SELECT
    emp_num
    ,day_date
    ,COUNT(*) AS FinalMinutes
FROM CTE_FinalMinutes
GROUP BY emp_num, day_date
ORDER BY emp_num, day_date;

要使最终查询只是将所有CTE放在一起。

结果集

emp_num day_date                FinalMinutes
547     2015-04-01 00:00:00.000 246
547     2015-04-02 00:00:00.000 49

There are 246 minutes between 17:24 and 21:30.
There are  49 minutes between 13:11 and 14:00.

以下是SQLFiddle完整查询。

显示导致此SUM分钟的实际间隔非常容易，但您说您只需要SUM。

Answer 5

我发现可能是最简单的解决方案。

.netFiddle

排序＆＃34;次要间隔＆＃34;开始日期。
寻找＆＃34;次要间隔＆＃34; （简单迭代）

将差距与＆＃34;主要时间间隔＆＃34;。

进行比较

    //declare intervals
var secondryIntervals = new List<Tuple<DateTime, DateTime>> {
        new Tuple<DateTime, DateTime>( new DateTime(2015, 03, 15, 4, 0, 0), new DateTime(2015, 03, 15, 5, 0, 0)),
        new Tuple<DateTime, DateTime>( new DateTime(2015, 03, 15, 4, 10, 0), new DateTime(2015, 03, 15, 4, 40, 0)),
        new Tuple<DateTime, DateTime>( new DateTime(2015, 03, 15, 4, 40, 0), new DateTime(2015, 03, 15, 5, 20, 0))};
var mainInterval = new Tuple<DateTime, DateTime>(new DateTime(2015, 03, 15, 3, 0, 0), new DateTime(2015, 03, 15, 7, 0, 0));
// add two empty intervals before and after main interval
secondryIntervals.Add(new Tuple<DateTime, DateTime>(mainInterval.Item1.AddMinutes(-1), mainInterval.Item1.AddMinutes(-1)));
secondryIntervals.Add(new Tuple<DateTime, DateTime>(mainInterval.Item2.AddMinutes(1), mainInterval.Item2.AddMinutes(1)));
secondryIntervals = secondryIntervals.OrderBy(s => s.Item1).ToList();
// endDate will rember 'biggest' end date
var endDate = secondryIntervals.First().Item1;
var result = secondryIntervals.Select(s =>
{
    var temp = endDate;
    endDate = endDate < s.Item2 ? s.Item2 : endDate;
    if (s.Item1 > temp)
    {
        return new Tuple<DateTime, DateTime>(temp < mainInterval.Item1 ? mainInterval.Item1 : temp,
                                             mainInterval.Item2 < s.Item1 ? mainInterval.Item2 : s.Item1);
    }
    return null;
})
    // remove empty records
                .Where(s => s != null && s.Item2 > s.Item1).ToList();
var minutes = result.Sum(s => (s.Item2 - s.Item1).TotalMinutes);

该算法需要O（n log n）时间（用于排序），无需额外的存储和假设。

如何计算时间间隔？

注意：

注意

5 个答案: