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社区首页 >问答首页 >获取不在数据库中的范围

获取不在数据库中的范围
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Stack Overflow用户
提问于 2014-10-23 23:52:46
回答 2查看 58关注 0票数 1

我想要获得每个房间都没有发生事件的所有时间。一天的开始是9:00:00,结束是22:00:00

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Event       EventStart  EventEnd    Days                Rooms   DayStarts
CISC 3660   09:00:00    12:30:00    Monday              7-3     9/19/2014   
MATH 2501   15:00:00    17:00:00    Monday:Wednesday    7-2     10/13/2014  
CISC 1110   14:00:00    16:00:00    Monday              7-3     9/19/2014   

我想获取不在数据库中的时间。

例如。对于SelectedDate (2014年9月19日),该表应返回:

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Room  FreeTimeStart  FreeTimeEnd
7-3   12:30:00       14:00:00
7-3   16:00:00       22:00:00

ex2。SelectedDate (10/13/2014):

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Room  FreeTimeStart  FreeTimeEnd
7-2    9:00:00       15:00:00
7-2   17:00:00       22:00:00  

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select * from Events where ________ NOT BETWEEN eventstart AND eventend;

但我不知道该把什么放在空间的位置上。

EN

回答 2

Stack Overflow用户

发布于 2014-10-24 01:27:28

这是一个相当复杂的请求。SQL最适合使用集合,而不是逐行查看。这是我想出来的。为了更容易弄清楚,我把它写成一系列的CTE,这样我就可以一步一步地解决这个问题。我并不是说这是最好的方法,但它不需要使用任何游标。您需要Events表和一个包含房间名称的表(否则,您看不到没有任何预订的房间)。

这里是查询,我将解释方法。

代码语言:javascript
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DECLARE @Events TABLE (Event varchar(20), EventStart Time, EventEnd Time, Days varchar(50), Rooms varchar(10), DayStarts date)

INSERT INTO @Events
SELECT 'CISC 3660',   '09:00:00',    '12:30:00',    'Monday',              '7-3',     '9/19/2014' UNION
SELECT 'MATH 2501',   '15:00:00',    '17:00:00',    'Monday:Wednesday',    '7-2',     '10/13/2014' UNION
SELECT 'CISC 1110',   '14:00:00',    '16:00:00',    'Monday',              '7-3',     '9/19/2014' 

DECLARE @Rooms TABLE (RoomName varchar(10))
INSERT INTO @Rooms
SELECT '7-2' UNION 
SELECT '7-3'

DECLARE @SelectedDate date = '9/19/2014'
DECLARE @MinTimeInterval int = 30 --smallest time unit room can be reserved for
;WITH
  D1(N) AS (
            SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
            SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
            SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
           ),
  D2(N) AS (SELECT 1 FROM D1 a, D1 b),
  D4(N) AS (SELECT 1 FROM D2 a, D2 b),
  Numbers AS (SELECT TOP 3600 ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 AS Number FROM D4),
  AllTimes AS 
    (SELECT CAST(DATEADD(n,Numbers.Number*@MinTimeInterval,'09:00:00') as time) AS m FROM Numbers
    WHERE DATEADD(n,Numbers.Number*@MinTimeInterval,'09:00:00') <= '22:00:00'),
  OccupiedTimes AS (
    SELECT e.Rooms, ValidTimes.m
    FROM @Events E
    CROSS APPLY (SELECT m FROM AllTimes WHERE m BETWEEN CASE WHEN e.EventStart = '09:00:00' THEN e.EventStart ELSE DATEADD(n,1,e.EventStart) END and CASE WHEN e.EventEnd = '22:00:00' THEN e.EventEnd ELSE DATEADD(n,-1,e.EventEnd) END) ValidTimes
    WHERE e.DayStarts = @SelectedDate
    ),
    AllRoomsAllTimes AS (
        SELECT * FROM @Rooms R CROSS JOIN AllTimes
    ), AllOpenTimes AS (
    SELECT a.*, ROW_NUMBER() OVER( PARTITION BY (a.RoomName) ORDER BY a.m) AS pos
    FROM AllRoomsAllTimes A
    LEFT OUTER JOIN OccupiedTimes o ON a.RoomName = o.Rooms AND a.m = o.m
    WHERE o.m IS NULL
    ), Finalize AS (
    SELECT a1.RoomName,
        CASE WHEN a3.m IS NULL OR  DATEDIFF(n,a3.m, a1.m) > @MinTimeInterval THEN a1.m else NULL END AS FreeTimeStart,
        CASE WHEN a2.m IS NULL OR DATEDIFF(n,a1.m,a2.m) > @MinTimeInterval THEN A1.m ELSE NULL END AS FreeTimeEnd,
        ROW_NUMBER() OVER( ORDER BY a1.RoomName )  AS Pos
    FROM AllOpenTimes A1
    LEFT OUTER JOIN AllOpenTimes A2 ON a1.RoomName = a2.RoomName and a1.pos = a2.pos-1
    LEFT OUTER JOIN AllOpenTimes A3 ON a1.RoomName = a3.RoomName and a1.pos = a3.pos+1
    WHERE A2.m IS NULL OR DATEDIFF(n,a1.m,a2.m) > @MinTimeInterval
    OR
    A3.m IS NULL OR DATEDIFF(n,a3.m, a1.m) > @MinTimeInterval
    )
    SELECT F1.RoomName, f1.FreeTimeStart, f2.FreeTimeEnd FROM Finalize F1
    LEFT OUTER JOIN Finalize F2 ON F1.Pos = F2.pos-1 AND f1.RoomName = f2.RoomName
    WHERE f1.pos % 2 = 1

在前几行中,我创建了temp变量来模拟您的桌子、事件和房间。变量@MinTimeInterval确定房间计划的时间间隔(每30分钟、15分钟等-这个数字需要均匀地除以60)。

接下来,我们将获得所有占用时间(OccupiedTimes)的列表,然后将此表向左外部连接到AllTimes表,这将为我们提供所有可用时间。因为我们只需要每个空闲时间的开始和结束,所以创建Finalize表,它自己将每条记录连接到表中的前一条记录和下一条记录。如果这些行中的时间大于@MinTimeInterval,则我们知道这是空闲时间的开始或结束。

票数 1
EN

Stack Overflow用户

发布于 2014-10-24 01:58:30

这是一个解决方案,它将返回“完整的图片”,包括在有问题的一天根本没有预订的房间:

代码语言:javascript
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Declare @Date char(8) = '20141013'
;
WITH cte as
(
    SELECT *
    FROM -- use your table name instead of the VALUES construct
    (VALUES
    ('09:00:00','12:30:00' ,'7-3', '20140919'),
    ('15:00:00','17:00:00' ,'7-2', '20141013'),
    ('14:00:00','16:00:00' ,'7-3', '20140919')) x(EventStart , EventEnd,Rooms, DayStarts)
), cte_Days_Rooms AS
-- get a cartesian product for the day specified and all rooms as well as the start and end time to compare against
(
    SELECT y.EventStart,y.EventEnd, x.rooms,a.DayStarts FROM 
    (SELECT @Date DayStarts) a
    CROSS JOIN
    (SELECT DISTINCT Rooms FROM cte)x
    CROSS JOIN
    (SELECT '09:00:00' EventStart,'09:00:00' EventEnd UNION ALL
     SELECT '22:00:00' EventStart,'22:00:00' EventEnd) y        
), cte_1 AS
-- Merge the original data an the "base data"
(
    SELECT * FROM cte WHERE DayStarts=@Date
    UNION ALL
    SELECT * FROM cte_Days_Rooms
), cte_2 as
-- use the ROW_NUMBER() approach to sort the data
(
    SELECT *, ROW_NUMBER() OVER(PARTITION BY DayStarts, Rooms ORDER BY EventStart) as pos
    FROM cte_1
)
-- final query: self join with an offest of one row, eliminating duplicate rows if a room is booked starting 9:00 or ending 22:00
SELECT c2a.DayStarts, c2a.Rooms , c2a.EventEnd, c2b.EventStart 
FROM cte_2 c2a
INNER JOIN cte_2 c2b on c2a.DayStarts = c2b.DayStarts AND c2a.Rooms =c2b.Rooms AND c2a.pos = c2b.pos -1
WHERE c2a.EventEnd <> c2b.EventStart
ORDER BY c2a.DayStarts, c2a.Rooms
票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/26532158

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