我正在建立一个网站，您可以在线获取任命。我不会详细解释所有内容，但我有一张桌子，上面有我可以预约的时间。分布到5分钟的间隔。这是一个例子：
ID StartDate EndDate
492548 2016-12-16 08:00:00.000 2016-12-16 08:05:00.000
492549 2016-12-16 08:05:00.000 2016-12-16 08:10:00.000
492550 2016-12-16 08:10:00.000 2016-12-16 08:15:00.000
492551 2016-12-16 08:15:00.000 2016-12-16 08:20:00.000
492552 2016-12-16 08:20:00.000 2016-12-16 08:25:00.000
492553 2016-12-16 08:25:00.000 2016-12-16 08:30:00.000
492554 2016-12-16 08:30:00.000 2016-12-16 08:35:00.000
492555 2016-12-16 08:35:00.000 2016-12-16 08:40:00.000
492556 2016-12-16 08:40:00.000 2016-12-16 08:45:00.000
492557 2016-12-16 08:45:00.000 2016-12-16 08:50:00.000
492558 2016-12-16 08:50:00.000 2016-12-16 08:55:00.000
492559 2016-12-16 08:55:00.000 2016-12-16 09:00:00.000
492560 2016-12-16 09:00:00.000 2016-12-16 09:05:00.000
492561 2016-12-16 09:05:00.000 2016-12-16 09:10:00.000
492562 2016-12-16 09:10:00.000 2016-12-16 09:15:00.000
492563 2016-12-16 09:15:00.000 2016-12-16 09:20:00.000
492564 2016-12-16 09:20:00.000 2016-12-16 09:25:00.000
492565 2016-12-16 09:25:00.000 2016-12-16 09:30:00.000
492566 2016-12-16 09:30:00.000 2016-12-16 09:35:00.000
根据咨询时间，根据咨询的原因，我必须将这些行分组为一个并知道min（IDSchedulingInterval）和max（IDSchedulingInterval）。
如果我的持续时间为15分钟，这是我想要的结果示例：
Min(ID) Max(ID) StartDate EndDate
492548 492550 2016-12-16 08:00:00.000 2016-12-16 08:15:00.000
492551 492553 2016-12-16 08:15:00.000 2016-12-16 08:30:00.000
492554 492556 2016-12-16 08:30:00.000 2016-12-16 08:45:00.000
492557 492559 2016-12-16 08:45:00.000 2016-12-16 09:00:00.000
持续时间可以改变。我不知道如何继续进行此查询..
修改 以下是您必须检查的一些例外情况。这是我的表
ID StartDate EndDate Isreserved
492548 2016-12-16 08:00:00.000 2016-12-16 08:05:00.000 0
492549 2016-12-16 08:05:00.000 2016-12-16 08:10:00.000 0
492550 2016-12-16 08:10:00.000 2016-12-16 08:15:00.000 0
492551 2016-12-16 08:15:00.000 2016-12-16 08:20:00.000 0
492552 2016-12-16 08:20:00.000 2016-12-16 08:25:00.000 0
492555 2016-12-16 08:35:00.000 2016-12-16 08:40:00.000 0
492556 2016-12-16 08:40:00.000 2016-12-16 08:45:00.000 0
492557 2016-12-16 08:45:00.000 2016-12-16 08:50:00.000 1
492558 2016-12-16 08:50:00.000 2016-12-16 08:55:00.000 1
492559 2016-12-16 08:55:00.000 2016-12-16 09:00:00.000 1
492560 2016-12-16 09:00:00.000 2016-12-16 09:05:00.000 0
492561 2016-12-16 09:05:00.000 2016-12-16 09:10:00.000 0
492562 2016-12-16 09:10:00.000 2016-12-16 09:15:00.000 0
492563 2016-12-16 09:15:00.000 2016-12-16 09:20:00.000 0
492564 2016-12-16 09:20:00.000 2016-12-16 09:25:00.000 0
492565 2016-12-16 09:25:00.000 2016-12-16 09:30:00.000 0
492566 2016-12-16 09:30:00.000 2016-12-16 09:35:00.000 0
这里保留8:45到9:00之间的时间，所以你不能接受它。 你也没有时间在8:25和8:35之间，所以你也不能保留它。举个例子，如果我想预约30分钟，那么我应该得到像这样的结果：
Min(ID) Max(ID) StartDate EndDate
492560 492565 2016-12-16 09:00:00.000 2016-12-16 09:30:00.000
由于您没有足够的时间间隔
，因此只会返回1行编辑2
感谢DVT，我已经修改了查询，我几乎有我的查询工作，这里唯一的hic是重叠时间。这是我的疑问：
DECLARE @newinterval INT = 60;
;with cte as (
SELECT
t1.IdSchedulingByInterval AS IdSchedulingByIntervalMin
, t2.IdSchedulingByInterval AS IdSchedulingByIntervalMax
, t1.SchedulingByIntervalStartDate
, t2.SchedulingByIntervalEndDate
FROM
RDV_tbSchedulingByInterval t1
JOIN RDV_tbSchedulingByInterval t2 ON t2.SchedulingByIntervalStartDate = DATEADD(minute, @newinterval - 5, t1.SchedulingByIntervalStartDate)
) select * from cte where (select SUM(5) from RDV_tbSchedulingByInterval where IdSchedulingByInterval
between cte.IdSchedulingByIntervalMin and cte.IdSchedulingByIntervalMax ) = @newinterval
order by cte.SchedulingByIntervalStartDate
这是我的结果：
492551 492562 2016-12-16 08:15:00.000 2016-12-16 09:15:00.000
492552 492563 2016-12-16 08:20:00.000 2016-12-16 09:20:00.000
492553 492564 2016-12-16 08:25:00.000 2016-12-16 09:25:00.000
492554 492565 2016-12-16 08:30:00.000 2016-12-16 09:30:00.000
492555 492566 2016-12-16 08:35:00.000 2016-12-16 09:35:00.000
492556 492567 2016-12-16 08:40:00.000 2016-12-16 09:40:00.000
492557 492568 2016-12-16 08:45:00.000 2016-12-16 09:45:00.000
492558 492569 2016-12-16 08:50:00.000 2016-12-16 09:50:00.000
492559 492570 2016-12-16 08:55:00.000 2016-12-16 09:55:00.000
492560 492571 2016-12-16 09:00:00.000 2016-12-16 10:00:00.000
492561 492572 2016-12-16 09:05:00.000 2016-12-16 10:05:00.000
492562 492573 2016-12-16 09:10:00.000 2016-12-16 10:10:00.000
492563 492574 2016-12-16 09:15:00.000 2016-12-16 10:15:00.000
492564 492575 2016-12-16 09:20:00.000 2016-12-16 10:20:00.000
492565 492576 2016-12-16 09:25:00.000 2016-12-16 10:25:00.000
492566 492577 2016-12-16 09:30:00.000 2016-12-16 10:30:00.000
492567 492578 2016-12-16 09:35:00.000 2016-12-16 10:35:00.000
492568 492579 2016-12-16 09:40:00.000 2016-12-16 10:40:00.000
492569 492580 2016-12-16 09:45:00.000 2016-12-16 10:45:00.000
预期结果：
492551 492562 2016-12-16 08:15:00.000 2016-12-16 09:15:00.000
492563 492574 2016-12-16 09:15:00.000 2016-12-16 10:15:00.000
我不希望时间重叠其他
答案 0 :(得分：1)
由于您的会话持续时间可能会发生变化，因此这是一种稍微灵活的方法：
declare @t table (IDSchedulingByInterval int identity(1,1)
,SchedulingByIntervalStartDate datetime
,SchedulingByIntervalEndDate datetime
);
insert into @t(SchedulingByIntervalStartDate, SchedulingByIntervalEndDate)
values('2016-12-16 08:00:00.000','2016-12-16 08:05:00.000'),('2016-12-16 08:05:00.000','2016-12-16 08:10:00.000'),('2016-12-16 08:10:00.000','2016-12-16 08:15:00.000'),('2016-12-16 08:15:00.000','2016-12-16 08:20:00.000'),('2016-12-16 08:20:00.000','2016-12-16 08:25:00.000'),('2016-12-16 08:25:00.000','2016-12-16 08:30:00.000'),('2016-12-16 08:30:00.000','2016-12-16 08:35:00.000'),('2016-12-16 08:35:00.000','2016-12-16 08:40:00.000'),('2016-12-16 08:40:00.000','2016-12-16 08:45:00.000'),('2016-12-16 08:45:00.000','2016-12-16 08:50:00.000'),('2016-12-16 08:50:00.000','2016-12-16 08:55:00.000'),('2016-12-16 08:55:00.000','2016-12-16 09:00:00.000'),('2016-12-16 09:00:00.000','2016-12-16 09:05:00.000'),('2016-12-16 09:05:00.000','2016-12-16 09:10:00.000'),('2016-12-16 09:10:00.000','2016-12-16 09:15:00.000'),('2016-12-16 09:15:00.000','2016-12-16 09:20:00.000'),('2016-12-16 09:20:00.000','2016-12-16 09:25:00.000'),('2016-12-16 09:25:00.000','2016-12-16 09:30:00.000'),('2016-12-16 09:30:00.000','2016-12-16 09:35:00.000'),('2016-12-16 09:35:00.000','2016-12-16 09:40:00.000'),('2016-12-16 09:40:00.000','2016-12-16 09:45:00.000'),('2016-12-16 09:45:00.000','2016-12-16 09:50:00.000'),('2016-12-16 09:50:00.000','2016-12-16 09:55:00.000'),('2016-12-16 09:55:00.000','2016-12-16 10:00:00.000');
declare @Interval int = 15; -- This is the number of minutes for each session. Must be divisible by 5 as base data is at a 5 minute granualarity.
select s.IDSchedulingByInterval as MinIDSchedulingByInterval
,e.IDSchedulingByInterval as MaxIDSchedulingByInterval
,s.SchedulingByIntervalStartDate
,e.SchedulingByIntervalEndDate
from @t s
left join @t e -- Find the corresponding end time for the session's start time
on(dateadd(minute,@Interval,s.SchedulingByIntervalStartDate) = e.SchedulingByIntervalEndDate)
where datediff(minute
,(select min(SchedulingByIntervalStartDate) from @t)
,s.SchedulingByIntervalStartDate
) % @Interval = 0; -- This is the check that start time is at the start of one of your defined intervals.
declare @t table (IDSchedulingByInterval int identity(1,1)
,SchedulingByIntervalStartDate datetime
,SchedulingByIntervalEndDate datetime
,Reserved bit
);
insert into @t(SchedulingByIntervalStartDate, SchedulingByIntervalEndDate,Reserved)
values('2016-12-16 08:00:00.000','2016-12-16 08:05:00.000',0),('2016-12-16 08:05:00.000','2016-12-16 08:10:00.000',0),('2016-12-16 08:10:00.000','2016-12-16 08:15:00.000',0),('2016-12-16 08:15:00.000','2016-12-16 08:20:00.000',0),('2016-12-16 08:20:00.000','2016-12-16 08:25:00.000',0),('2016-12-16 08:25:00.000','2016-12-16 08:30:00.000',0),('2016-12-16 08:30:00.000','2016-12-16 08:35:00.000',0),('2016-12-16 08:35:00.000','2016-12-16 08:40:00.000',0),('2016-12-16 08:40:00.000','2016-12-16 08:45:00.000',0),('2016-12-16 08:45:00.000','2016-12-16 08:50:00.000',1),('2016-12-16 08:50:00.000','2016-12-16 08:55:00.000',1),('2016-12-16 08:55:00.000','2016-12-16 09:00:00.000',1),('2016-12-16 09:00:00.000','2016-12-16 09:05:00.000',0),('2016-12-16 09:05:00.000','2016-12-16 09:10:00.000',0),('2016-12-16 09:10:00.000','2016-12-16 09:15:00.000',0),('2016-12-16 09:15:00.000','2016-12-16 09:20:00.000',0),('2016-12-16 09:20:00.000','2016-12-16 09:25:00.000',0),('2016-12-16 09:25:00.000','2016-12-16 09:30:00.000',0),('2016-12-16 09:30:00.000','2016-12-16 09:35:00.000',0),('2016-12-16 09:35:00.000','2016-12-16 09:40:00.000',0),('2016-12-16 09:40:00.000','2016-12-16 09:45:00.000',0),('2016-12-16 09:45:00.000','2016-12-16 09:50:00.000',0),('2016-12-16 09:50:00.000','2016-12-16 09:55:00.000',0),('2016-12-16 09:55:00.000','2016-12-16 10:00:00.000',0);
declare @Interval int = 60; -- This is the number of minutes for each session. Must be divisible by 5 as base data is at a 5 minute granualarity.
with cte
as
(
select s.IDSchedulingByInterval as MinIDSchedulingByInterval
,e.IDSchedulingByInterval as MaxIDSchedulingByInterval
,s.SchedulingByIntervalStartDate
,e.SchedulingByIntervalEndDate
from @t s
left join @t e -- Find the corresponding end time for the session's start time
on(dateadd(minute,@Interval,s.SchedulingByIntervalStartDate) = e.SchedulingByIntervalEndDate)
where datediff(minute
,(select min(SchedulingByIntervalStartDate) from @t)
,s.SchedulingByIntervalStartDate
) % @Interval = 0 -- This is the check that start time is at the start of one of your defined intervals.
)
select c.MinIDSchedulingByInterval
,c.MaxIDSchedulingByInterval
,c.SchedulingByIntervalStartDate
,c.SchedulingByIntervalEndDate
from cte c
left join @t t
on(t.SchedulingByIntervalStartDate <= c.SchedulingByIntervalEndDate
and t.SchedulingByIntervalEndDate > c.SchedulingByIntervalStartDate
)
group by c.MinIDSchedulingByInterval
,c.MaxIDSchedulingByInterval
,c.SchedulingByIntervalStartDate
,c.SchedulingByIntervalEndDate
having sum(cast(t.Reserved as int)) = 0
答案 1 :(得分：1)
-- This converts the period to date-time format
SELECT
-- note the 15, the "minute", and the starting point to convert the
-- period back to original time
DATEADD(minute, AP.FifteenMinutePeriod * 15, '2010-01-01T00:00:00') AS Period,
AP.AvgValue
FROM
-- this groups by the period and gets the average
(SELECT
P.FifteenMinutePeriod,
AVG(P.Value) AS AvgValue
FROM
-- This calculates the period (fifteen minutes in this instance)
(SELECT
-- note the division by 15 and the "minute" to build the 15 minute periods
-- the '2010-01-01T00:00:00' is the starting point for the periods
datediff(minute, '2010-01-01T00:00:00', T.Time)/15 AS FifteenMinutePeriod,
T.Value
FROM Test T) AS P
GROUP BY P.FifteenMinutePeriod) AP
答案 2 :(得分：1)
这处理可变的分钟间隔。请注意，这尚未经过性能测试：
var onlineState,
origOnLineProp;
beforeEach(function() {
// Remember original config
origOnLineProp = Object.getOwnPropertyDescriptor(Navigator.prototype, "onLine");
onlineState = true;
// New behavior
Object.defineProperty(Navigator.prototype, "onLine", {
enumerable: origOnLineProp.enumerable,
configurable: origOnLineProp.configurable,
get: function() { return onlineState }
});
});
it("...", function() {
onlineState = false;
expect("code").toBe("correctly functioning in offline mode");
});
afterEach(function() {
// Restore original behavior
Object.defineProperty(Navigator.prototype, "onLine", {
enumerable: origOnLineProp.enumerable,
configurable: origOnLineProp.configurable,
get: origOnLineProp.get
});
});
答案 3 :(得分：1)
这是我能提出的最简单的方法
;WITH Tally AS
(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) AS N FROM master.sys.objects A, master.sys.objects B
)
,Intervals AS
(
SELECT
N AS Id,
DATEADD(MINUTE, (N-1)*5, '20160101') AS StartDate,
DATEADD(MINUTE, (N)*5, '20160101') AS EndDate
FROM Tally
)
SELECT MIN(Id) AS MinId, MAX(Id) AS MaxId, MIN(StartDate), MAX(EndDate) FROM Intervals
GROUP BY CAST(StartDate AS Date), DATEPART(HOUR,StartDate), DATEPART(MINUTE, StartDate) / 15
ORDER BY MinId
修改
只需用您的表替换名称，如
SELECT
MIN(IdSchedulingByInterval) AS MinId,
MAX(IdSchedulingByInterval) AS MaxId,
MIN(SchedulingByIntervalStartDate), ¨
MAX(SchedulingByIntervalEndDate)
FROM RDV_tbSchedulingByInterval
GROUP BY CAST(SchedulingByIntervalStartDate AS Date), DATEPART(HOUR,SchedulingByIntervalStartDate), DATEPART(MINUTE, SchedulingByIntervalStartDate) / 15
ORDER BY MinId
答案 4 :(得分：1)
我认为您可以使用如下查询：
SELECT MIN(ID) AS minID, MAX(ID) AS maxID,
MIN(StartDate) AS StartDate, MAX(EndDate) AS EndDate
FROM (
SELECT ID, StartDate, EndDate,
ROW_NUMBER() OVER (ORDER BY StartDate) -
ROW_NUMBER() OVER (PARTITION BY x.v
ORDER BY StartDate) AS grp
FROM mytable
CROSS APPLY (SELECT CAST(CONVERT(DATE, EndDate) AS VARCHAR(10)) +
CAST(DATEPART(HOUR, StartDate) AS VARCHAR(10)) +
CAST(DATEPART(MINUTE, StartDate) / 15 AS VARCHAR(10))) AS x(v)) AS t
GROUP BY t.grp
ORDER BY EndDate
答案 5 :(得分：0)
DECLARE @newinterval INT = 15;
SELECT
t1.IdSchedulingByInterval AS IdSchedulingByIntervalMin
, t2.IdSchedulingByInterval AS IdSchedulingByIntervalMax
, t1.SchedulingByIntervalStartDate
, t2.SchedulingByIntervalEndDate
FROM
<table> t1
JOIN <table> t2 ON t2.SchedulingByIntervalStartDate = DATEADD(minute, @newinterval - 5, t1.SchedulingByIntervalStartDate)
WHERE
DATEPART(minute,t1.SchedulingByIntervalStartDate) % @newinterval = 0;
答案 6 :(得分：0)
我能想到的最简单的查询..
select MIN(ID), MAX(ID), MIN(StartDate), MAX(EndDate)
from
(
select
ID,
case
when substring(CONVERT(varchar, StartDate),16,2)in ('00','05','10') then 1
when substring(CONVERT(varchar, StartDate),16,2)in ('15','20','25') then 2
when substring(CONVERT(varchar, StartDate),16,2)in ('30','35','40') then 3
when substring(CONVERT(varchar, StartDate),16,2)in ('45','50','55') then 4
end as MinOfDate,
substring(CONVERT(varchar, StartDate),13,2)as HourOfDate,
substring(CONVERT(varchar, GETDATE()),1,6) as DayOfDate
StartDate,
EndDate
from SourceTable
where IsReserved = 0
) t
group by DayOfDate,HourOfDate,MinOfDate