7

场景是用户指定他们何时可用，这些指定的时间可以相互重叠。我正在尝试获取他们可用的总时间。SQL Fiddle 示例：

--Available--
ID  userID  availStart          availEnd
1   456     '2012-11-19 16:00'  '2012-11-19 17:00'
2   456     '2012-11-19 16:00'  '2012-11-19 16:50'
3   456     '2012-11-19 18:00'  '2012-11-19 18:30'
4   456     '2012-11-19 17:30'  '2012-11-19 18:10'
5   456     '2012-11-19 16:00'  '2012-11-19 17:10'
6   456     '2012-11-19 16:00'  '2012-11-19 16:50'

输出应该是 130 分钟：

1: 60
2: 0 as falls inside 1
3: 30
4: 30 as the last 10 mins is covered by 3
5: 10 as first 60 mins is covered by 1
6: 0 as falls inside 1

我可以获得总重叠分钟数，但这比可用分钟数的总和还要多：

SQL小提琴

有什么想法可以实现这一目标吗？

12 月 21 日编辑：到目前为止，感谢大家的解决方案——在某种程度上，我很高兴看到这不是一个“简单”的查询。

12 年 11 月 23 日编辑：这都是很棒的工作。在内部，我们认为最好确保用户不能输入重叠时间（例如，强制他们修改现有条目）！

4

5 回答 5

4

Gordon Linoff有一个基于 CTE 的答案

我已经对所有工作算法进行了一些性能分析空白值意味着它花费了太长时间。这是在单个 Core i7 X920 @2GHz 芯片上测试的，由几个 SSD 支持。创建的唯一索引是 UserID 上的集群 AvailStart。如果您认为可以提高任何性能，请告诉我。

这个 CTE 版本比线性差，SQL Server 不能以有效的方式进行 RN = RN + 1 连接。我用下面的混合方法纠正了这个问题，我将第一个 CTE 保存并索引到一个表变量中。这仍然需要十倍于基于游标的方法的 IO。

With OrderedRanges as (
  Select
    Row_Number() Over (Partition By UserID Order By AvailStart) AS RN,
    AvailStart,
    AvailEnd
  From
    dbo.Available
  Where
    UserID = 456
),
AccumulateMinutes (RN, Accum, CurStart, CurEnd) as (
  Select
    RN, 0, AvailStart, AvailEnd
  From
    OrderedRanges
  Where 
    RN = 1
  Union All
  Select
    o.RN, 
    a.Accum + Case When o.AvailStart <= a.CurEnd Then
        0
      Else 
        DateDiff(Minute, a.CurStart, a.CurEnd)
      End,
    Case When o.AvailStart <= a.CurEnd Then 
        a.CurStart
      Else
        o.AvailStart
      End,
    Case When o.AvailStart <= a.CurEnd Then
        Case When a.CurEnd > o.AvailEnd Then a.CurEnd Else o.AvailEnd End
      Else
        o.AvailEnd
      End
  From
    AccumulateMinutes a
        Inner Join 
    OrderedRanges o On 
        a.RN = o.RN - 1
)

Select Max(Accum + datediff(Minute, CurStart, CurEnd)) From AccumulateMinutes

http://sqlfiddle.com/#!6/ac021/2

在做了一些性能分析之后，这是一个混合 CTE/表变量版本，它的性能比基于游标的方法更好

Create Function dbo.AvailMinutesHybrid(@UserID int) Returns Int As
Begin

Declare @UserRanges Table (
  RN int not null primary key, 
  AvailStart datetime, 
  AvailEnd datetime
)
Declare @Ret int = Null

;With OrderedRanges as (
  Select
    Row_Number() Over (Partition By UserID Order By AvailStart) AS RN,
    AvailStart,
    AvailEnd
  From
    dbo.Available
  Where
    UserID = @UserID
)
Insert Into @UserRanges Select * From OrderedRanges


;With AccumulateMinutes (RN,Accum, CurStart, CurEnd) as (
  Select
    RN, 0, AvailStart, AvailEnd
  From
    @UserRanges
  Where 
    RN = 1
  Union All
  Select
    o.RN, 
    a.Accum + Case When o.AvailStart <= a.CurEnd Then
        0
      Else 
        DateDiff(Minute, a.CurStart, a.CurEnd)
      End,
    Case When o.AvailStart <= a.CurEnd Then 
        a.CurStart
      Else
        o.AvailStart
      End,
    Case When o.AvailStart <= a.CurEnd Then
        Case When a.CurEnd > o.AvailEnd Then a.CurEnd Else o.AvailEnd End
      Else
        o.AvailEnd
      End
  From
    AccumulateMinutes a
        Inner Join 
    @UserRanges o On 
        a.RN + 1 = o.RN
)

Select 
  @Ret = Max(Accum + datediff(Minute, CurStart, CurEnd)) 
From 
  AccumulateMinutes 
Option
  (MaxRecursion 0)

Return @Ret

End

http://sqlfiddle.com/#!6/bfd94

于 2012-11-21T16:28:29.197 回答
2

主要问题是您可以拥有重叠条目链，因此您需要组合无限次以删除所有重叠 - 这比 SQL 更适合程序方法。但是，如果您不想使用临时表，这里有一个 CTE 方法 - 请记住，CTE 只能递归给定次数，因此如果您有任何特别长的链，它将失败。

WITH MergedAvailable
AS
(
  SELECT Available.UserID, Available.AvailStart, MAX(Available.AvailEnd) AS AvailEnd
    FROM Available
   WHERE (
           SELECT COUNT(*)
             FROM Available AS InnerAvailable
            WHERE InnerAvailable.AvailStart < Available.AvailStart
                  AND
                  InnerAvailable.AvailEnd >= Available.AvailStart
         ) = 0
   GROUP BY Available.UserID, Available.AvailStart
  UNION ALL
  SELECT MergedAvailable.UserID, MergedAvailable.AvailStart,
         LongestExtensionToAvailableInterval.NewIntervalEnd
    FROM MergedAvailable
   CROSS APPLY GetLongestExtensionToAvailableInterval(MergedAvailable.UserID,
               MergedAvailable.AvailStart,
               MergedAvailable.AvailEnd) AS LongestExtensionToAvailableInterval
   WHERE LongestExtensionToAvailableInterval.NewIntervalEnd IS NOT NULL
)

SELECT SUM(DATEDIFF(MINUTE,
                    FinalAvailable.AvailStart,
                    FinalAvailable.AvailEnd)) AS MinsAvailable
  FROM (
         SELECT MergedAvailable.UserID, MergedAvailable.AvailStart,
                MAX(MergedAvailable.AvailEnd) AS AvailEnd
           FROM MergedAvailable
          GROUP BY MergedAvailable.UserID, MergedAvailable.AvailStart
       ) AS FinalAvailable

此表函数是必需的：

CREATE FUNCTION GetLongestExtensionToAvailableInterval
(
  @UserID int,
  @CurrentIntervalStart datetime,
  @CurrentIntervalEnd datetime
)
RETURNS TABLE
AS
RETURN 
  SELECT MAX(Available.AvailEnd) AS NewIntervalEnd
    FROM Available
   WHERE Available.UserID = @UserID
         AND
         Available.AvailStart > @CurrentIntervalStart
         AND
         Available.AvailStart <= @CurrentIntervalEnd
         AND
         Available.AvailEnd > @CurrentIntervalEnd

一般的想法是，它从范围开始不重叠任何东西的所有范围开始，然后随着每次递归，它将当前范围扩展到当前重叠范围的最远范围。需要 table 函数来确定最远范围，因为 CTE 的递归部分不允许包含普通聚合。

使用您提供的数据，起始行是：

456 2012-11-19 16:00 2012-11-19 17:10
456 2012-11-19 17:30 2012-11-19 18:10

最终通过递归添加的唯一行是：

456 2012-11-19 17:30 2012-11-19 18:30

举例来说，假设您有一行 ID 为 7，从 18:20 到 19:20。然后会有第二次递归，它带回了行：

456 2012-11-19 17:30 2012-11-19 19:20

因此，虽然查询将到达每个重叠范围的开始和结束，但它也将带回所有中间阶段。这就是为什么我们需要为 CTE 之后的每个开始日期取最大结束日期的总和，以删除它们。

于 2012-11-20T03:28:40.260 回答
2

这是另一种使用光标的方法。我觉得这种技术应该适用于 CTE，但我不知道该怎么做

方法是按开始时间排列每个范围然后我们构建一个按顺序合并范围的范围，直到找到一个不与我们合并的范围重叠的范围。然后我们计算在合并范围内有多少分钟，并记住这一点 我们继续下一个范围，再次合并任何重叠的范围。每次我们得到一个不重叠的起点时，我们都会累积分钟最后我们将累积的分钟添加到最后一个范围的长度上

很容易看出，由于顺序的原因，一旦一个范围与之前的范围不同，那么没有更多的范围可以与之前的范围重叠，因为它们的开始日期都更大。

Declare
  @UserID int = 456,
  @CurStart datetime, -- our current coalesced range start
  @CurEnd datetime, -- our current coalsced range end
  @AvailStart datetime, -- start or range for our next row of data
  @AvailEnd datetime, -- end of range for our next row of data
  @AccumMinutes int = 0 -- how many minutes so far accumulated by distinct ranges

Declare MinCursor Cursor Fast_Forward For
Select
  AvailStart, AvailEnd
From
  dbo.Available
Where
  UserID = @UserID
Order By
  AvailStart

Open MinCursor

Fetch Next From MinCursor Into @AvailStart, @AvailEnd
Set @CurStart = @AvailStart
Set @CurEnd = @AvailEnd

While @@Fetch_Status = 0
Begin
  If @AvailStart <= @CurEnd -- Ranges Overlap, so coalesce and continue
    Begin
    If @AvailEnd > @CurEnd 
      Set @CurEnd = @AvailEnd
    End
  Else -- Distinct range, coalesce minutes from previous range
  Begin
    Set @AccumMinutes = @AccumMinutes + DateDiff(Minute, @CurStart, @CurEnd)
    Set @CurStart = @AvailStart -- Start coalescing a new range
    Set @CurEnd = @AvailEnd
  End
  Fetch Next From MinCursor Into @AvailStart, @AvailEnd
End

Select @AccumMinutes + DateDiff(Minute, @CurStart, @CurEnd) As TotalMinutes

Close MinCursor
Deallocate MinCursor;

http://sqlfiddle.com/#!6/3483c/15

于 2012-11-20T21:44:09.353 回答
2

条件t1.availStart > t2.availEnd 或 t1.availEnd < t2.availStart永远不会跨越的检查周期。如果超过，则最小availStart 或最大availEnd esleavailStart 或availEnd。

Probably more than one crossing period.
In your case it 
16:00:00 - 17:10:00 includes the ranges:16:00:00 - 16:50:00,
                                        16:00:00 - 16:50:00,
                                        16:00:00 - 17:00:00,
                                        16:00:00 - 17:10:00
17:30:00 - 18:30:00 includes the ranges:17:30:00 - 18:10:00,
                                        18:00:00 - 18:30:00

2012 年 11 月 21 日更新；2012 年 11 月 30 日；04.01.2013

CREATE FUNCTION dbo.Overlap
 (
  @availStart datetime,
  @availEnd datetime,
  @availStart2 datetime,
  @availEnd2 datetime
  )
RETURNS TABLE
RETURN
  SELECT CASE WHEN @availStart >= @availEnd2 OR @availEnd <= @availStart2
              THEN @availStart ELSE
                               CASE WHEN @availStart > @availStart2 THEN @availStart2 ELSE @availStart END
                               END AS availStart,
         CASE WHEN @availStart >= @availEnd2 OR @availEnd <= @availStart2
              THEN @availEnd ELSE
                             CASE WHEN @availEnd > @availEnd2 THEN @availEnd ELSE @availEnd2 END
                             END AS availEnd

;WITH cte AS
 (
  SELECT userID, availStart, availEnd, ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY AvailStart) AS Id
  FROM dbo.test53
  ), cte2 AS
 (
  SELECT Id, availStart, availEnd
  FROM cte
  WHERE Id = 1
  UNION ALL
  SELECT c.Id, o.availStart, o.availEnd
  FROM cte c JOIN cte2 ct ON c.Id = ct.Id + 1
             CROSS APPLY dbo.Overlap(c.availStart, c.availEnd, ct.availStart, ct.availEnd) AS o
  )
  SELECT TOP 1 SUM(DATEDIFF(minute, availStart, MAX(availEnd))) OVER()
  FROM cte2
  GROUP BY availStart

SQLFiddle上的演示

于 2012-11-21T09:47:51.057 回答
1 
Create Table #Available (
  ID int not null primary key,
  UserID int not null,
  AvailStart datetime not null,
  AvailEnd datetime not null
)


Insert Into #Available (ID,UserID, AvailStart, AvailEnd) Values
  (1,456, '2012-11-19 16:00', '2012-11-19 17:00'),
  (2,456, '2012-11-19 16:00', '2012-11-19 16:50'),
  (3,456, '2012-11-19 18:00', '2012-11-19 18:30'),
  (4,456, '2012-11-19 17:30', '2012-11-19 18:10'),
  (5,456, '2012-11-19 16:00', '2012-11-19 17:10'),
  (6,456, '2012-11-19 16:00', '2012-11-19 16:50'),
  (7,457, '2012-11-19 16:00', '2012-11-19 17:10'),
  (8,457, '2012-11-19 16:00', '2012-11-19 16:50');  
Select Distinct UserID 
into #users
from #Available


Create Table #mins(UserID int,atime datetime,aset tinyint )
Declare @start Datetime
Declare @end Datetime

Select @start=min(AvailStart),@end=max(AvailEnd) from #Available 
While @start<@end
    begin
     insert into #mins(UserID,atime) 
     Select UserID ,@Start from #users
     Select @start=DateAdd(mi,1,@start)
    end

update #mins set aset=1
from #Available
where atime>=AvailStart and atime<Availend and #mins.UserID = #Available.UserID


select UserID,SUM(aset) as [Minutes] 
from #mins
Group by UserID 
Drop table #Available
Drop table #mins
Drop table #users
于 2012-11-20T00:02:08.327 回答