查找忽略重叠的总分钟数(将基于光标的答案转换为CTE)

Question

有一个问题,询问如何查找多个日期范围中有多少分钟,忽略重叠.

给出的示例数据是(userID不是特别相关)

--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'

我可以使用游标解决问题,但我认为它应该适应CTE,但我无法弄清楚如何做到这一点.

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,直到我们找到一个不与我们的合并范围重叠的范围.然后我们计算合并范围内的分钟数,并记住这个我们继续下一个范围,再次合并任何重叠.我们每次得到一个非重叠的起始点时累积分钟.最后,我们将累积的分钟数加到最后一个范围的长度上

很容易看出,由于订单的原因,一旦范围与之前的范围不同,那么没有其他范围可能与以前的范围重叠,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 coalesced 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;

让CTE工作,在递归中只是一个愚蠢的错误.查询计划爆炸令人印象深刻:

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 

这是否适用于CTE,是否有一种通用方式累积列表的一般模式？

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

Answer 1

以下查询根据您的定义查找数据中的句点.它首先使用相关子查询来确定记录是否是句点的开始(即,与早期时间段没有重叠).然后它将"periodStart"指定为最近的开始,即非重叠时段的开始.

以下(未经测试的)查询采用以下方法:

with TimeWithOverlap as (
     select t.*,
            (case when exists (select * from dbo.Available tbefore where t.availStart > tbefore.availStart and tbefore.availEnd >= t.availStart)
                  then 0
                  else 1
             end) as IsPeriodStart
     from dbo.Available t 
    ),
    TimeWithPeriodStart as (
     select two.*,
            (select MAX(two1.AvailStart) from TimeWithOverlap two1 where IsPeriodStart = 1 and two1.AvailStart <= two.AvailStart
            ) as periodStart
     from TimeWithOverlap two
    )
select periodStart, MAX(AvailEnd) as periodEnd
from TimeWithPeriodStart twps
group by periodStart;

http://sqlfiddle.com/#!6/3483c/20(第二次查询)

如果两个时段都同时开始,那么它仍然有效,因为AvailStart值是相同的.由于相关的子查询,即使是中等大小的数据集,这也可能表现不佳.

还有其他方法可以解决这个问题.例如,如果您有SQL Server 2012,则可以使用累积和函数,它提供了一种更简单的方法.