Question

我有这个我需要做的事情，一些建议将不胜感激。我有一个带有一些电话的SQL服务器表。对于每个电话，我有开始和结束时间。

我需要完成的事情：一段时间内的存储过程，假设x间隔为5小时，让2分钟返回已连接的呼叫数。

Something like:

Interval                                    Nr of Calls Connected
01-01-2010 12:00:00 - 01-01-2010 12:05:00   30
01-01-2010 12:05:01 - 01-01-2010 12:10:00   10
.............

哪种方法最快？谢谢你的帮助

Answer 1

这适用于有来电的时间间隔 ...

Declare @datetimestart datetime
Declare @interval int
Set @datetimestart = '2009-01-01 12:00:00'
Set @interval = 5 --in minutes

Select
  [start_interval], [end_interval] , count([start_interval]) as [calls]
From
  (
    Select 
     DateAdd( Minute,Floor(DateDiff(Minute,@datetimestart,[date])/@interval)*@interval
,@datetimestart) , 
    DateAdd( Minute,@interval + Floor(DateDiff(Minute,@datetimestart,[date])/@interval)*@interval
,@datetimestart) 
    From yourTable
  ) As W([start_interval],[end_interval])
group by [start_interval], [end_interval]

无论通话次数多少，这都适用于所有时间间隔 ..

Declare @datetimestart datetime, @datetimeend datetime, @datetimecurrent datetime
Declare @interval int
Set @datetimestart = '2009-01-01 12:00:00'
Set @interval = 10
Set @datetimeend = (Select max([date]) from yourtable)

SET @datetimecurrent = @datetimestart

declare @temp as table ([start_interval] datetime, [end_interval] datetime)

while  @datetimecurrent < @datetimeend
BEGIN
  insert into @temp select (@datetimecurrent), dateAdd( minute, @interval, @datetimecurrent)
  set @datetimecurrent = dateAdd( minute, @interval, @datetimecurrent)
END

Select
  *
From
  (
    Select 
      [start_interval],[end_interval], count(d.[start_time]) 
    From @temp t left join yourtable d on d.[start_time] between t.[start_interval] and t.[end_interval]
  ) As W([start_interval],[end_interval], [calls])

Answer 2

我改变了Gaby的例子，做了一点你想做的事情

Declare @datetimeend datetime
,@datetimecurrent datetime
,@interval int
Set @interval = 10
Set @datetimeend = (Select max([end_time]) from Calls)
SET @datetimecurrent = '2010-04-17 14:20:00'

declare @temp as table ([start_interval] datetime, [end_interval] datetime)

while  @datetimecurrent < @datetimeend
BEGIN
  insert into @temp select (@datetimecurrent), dateAdd( minute, @interval, @datetimecurrent)
  set @datetimecurrent = dateAdd( minute, @interval, @datetimecurrent)
END

Select 
  [start_interval],[end_interval], count(d.id) [COUNT]
From @temp t 
left join Calls d on 
    d.end_time >= t.start_interval
AND d.start_time <= t.end_interval
GROUP BY [start_interval],[end_interval]

用它来创建表并填充它

CREATE TABLE dbo.Calls
    (
    id int NOT NULL IDENTITY (1, 1),
    start_time datetime NOT NULL,
    end_time datetime NULL,
    caller nvarchar(50) NULL,
    receiver nvarchar(50) NULL
    )  ON [PRIMARY]
GO
ALTER TABLE dbo.Calls ADD CONSTRAINT
    PK_Calls PRIMARY KEY CLUSTERED 
    (
    id
    )  ON [PRIMARY]

GO

DECLARE @I INT
SET @I = 0
WHILE @I < 100
BEGIN
INSERT INTO Calls
(start_time, end_time)
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-10,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-9,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-9,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-8,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-8,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-7,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-7,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-6,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-6,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-5,GETDATE()))
UNION
SELECT
DATEADD(HOUR,-@I,DATEADD(MINUTE,-5,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-4,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-4,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-3,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-3,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-2,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-2,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-1,GETDATE()))
UNION
select 
DATEADD(HOUR,-@I,DATEADD(MINUTE,-1,GETDATE()))
,DATEADD(HOUR,-@I,DATEADD(MINUTE,-0,GETDATE()));
SET @I = @I + 1
END

在SQL Server 2008中完成但该脚本可以在其他版本中使用

Answer 3

我会使用Numbers数据透视表来获取时间间隔，然后计算所有与间隔重叠的调用：

SELECT Intervals.IntervalStart
    ,Intervals.IntervalEnd
    ,COUNT(*)
FROM (
    SELECT DATEADD(MINUTE, Numbers * 2, @StartTime) AS IntervalStart
        ,DATEADD(MINUTE, (Numbers + 1) * 2, @StartTime) AS IntervalEnd
    FROM Numbers
    WHERE Numbers BETWEEN 0 AND (5 * 60 / 2)
) AS Intervals
LEFT JOIN Calls
    ON Calls.CallEnd >= Intervals.IntervalStart
        AND Calls.CallStart < Intervals.IntervalEnd
GROUP BY Intervals.IntervalStart
    ,Intervals.IntervalEnd

要获得空间隔，您需要从另一个“间隔”派生表中LEFT JOIN到此。

Answer 4

这种方法怎么样：

select Year(StartTime) as Year, Month(StartTime) as Month, Day(StartTime) as Day, datepart(hh, StartTime) as Hour, datepart(mm, StartTime) / 2 as TwoMinuteSegment, count(*)
from MyTable
where StartDate between '01-01-2010 12:00:00' and '01-01-2010 17:00:00'
group by Year(StartTime), Month(StartTime), Day(StartTime), datepart(hh, StartTime), datepart(mm, StartTime) / 2

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