我有一个名为Bookings的简化表，其中有两列BookDate和BookSlot。 BookDate列仅包含日期（没有时间），BookSlot列将包含一天中的时间，间隔为30分钟，从0到1410（含）。 （即600 = 10:00 am）
如何找到未来可用的第一个插槽（未预订）没有在循环中运行？
这是表定义和测试数据：
Create Table Bookings(
BookDate DateTime Not Null,
BookSlot Int Not Null
)
Go
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-01',0);
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-01',30);
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-01',60);
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-01',630);
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-02',60);
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-02',90);
Insert Into Bookings(BookDate,BookSlot) Values('2014-07-02',120);
我想要一种方法来返回表中没有的第一个可用插槽，以及将来（基于服务器时间）。
基于以上测试数据：
如果将来没有预订，该功能将简单地返回将来最接近的半小时。
-
SQL小提琴就在这里：
答案 0 :(得分：1)
这有点复杂，但试试这个：
WITH DATA
AS (SELECT *,
Row_number()
OVER (
ORDER BY BOOKDATE, BOOKSLOT) RN
FROM BOOKINGS)
SELECT CASE
WHEN T.BOOKSLOT = 1410 THEN Dateadd(DAY, 1, BOOKDATE)
ELSE BOOKDATE
END Book_Date,
CASE
WHEN T.BOOKSLOT = 1410 THEN 0
ELSE BOOKSLOT + 30
END Book_Slot
FROM (SELECT TOP 1 T1.*
FROM DATA T1
LEFT JOIN DATA t2
ON t1.RN = T2.RN - 1
WHERE t2.BOOKSLOT - t1.BOOKSLOT > 30
OR ( t1.BOOKDATE != T2.BOOKDATE
AND ( t2.BOOKSLOT != 0
OR t1.BOOKSLOT != 630 ) )
OR t2.BOOKSLOT IS NULL)T
以下是SQL fiddle示例。
<强>解释
此解决方案包含两部分：
修改
在查询中添加了TOP 1，以便只按请求返回第一个广告位。
<强>更新
这是更新版本，包括2个新元素（获取当前日期+时间和处理空表）：
DECLARE @Date DATETIME = '2014-07-01',
@Slot INT = 630
DECLARE @time AS TIME = Cast(Getdate() AS TIME)
SELECT @Slot = Datepart(HOUR, @time) * 60 + Round(Datepart(MINUTE, @time) / 30,
0) * 30
+ 30
SET @Date = Cast(Getdate() AS DATE)
;WITH DATA
AS (SELECT *,
Row_number()
OVER (
ORDER BY BOOKDATE, BOOKSLOT) RN
FROM BOOKINGS
WHERE BOOKDATE > @Date
OR ( BOOKDATE = @Date
AND BOOKSLOT >= @Slot ))
SELECT TOP 1 BOOK_DATE,
BOOK_SLOT
FROM (SELECT CASE
WHEN RN = 1
AND NOT (@slot = BOOKSLOT
AND @Date = BOOKDATE) THEN @Date
WHEN T.BOOKSLOT = 1410 THEN Dateadd(DAY, 1, BOOKDATE)
ELSE BOOKDATE
END Book_Date,
CASE
WHEN RN = 1
AND NOT (@slot = BOOKSLOT
AND @Date = BOOKDATE) THEN @Slot
WHEN T.BOOKSLOT = 1410 THEN 0
ELSE BOOKSLOT + 30
END Book_Slot,
1 AS ID
FROM (SELECT TOP 1 T1.*
FROM DATA T1
LEFT JOIN DATA t2
ON t1.RN = T2.RN - 1
WHERE t2.BOOKSLOT - t1.BOOKSLOT > 30
OR ( t1.BOOKDATE != T2.BOOKDATE
AND ( t2.BOOKSLOT != 0
OR t1.BOOKSLOT != 1410 ) )
OR t2.BOOKSLOT IS NULL)T
UNION
SELECT @date AS bookDate,
@slot AS BookSlot,
2 ID)X
ORDER BY X.ID
使用SQL fiddle，让我知道你的想法。
答案 1 :(得分：1)
以下是一种方法，允许将来预订最多256天，并允许空的预订表。我假设您使用的是SQL Server 2005，因为您的BookDate是dateTime而不是date 在任何情况下，您可以考虑将插槽存储为完整的日期时间而不是单独的列。这将有助于查询和提高性能。
DECLARE @now DATETIME = '2014-07-01 00:10:00';
WITH T4
AS (SELECT N
FROM (VALUES(0),
(0),
(0),
(0),
(0),
(0),
(0),
(0)) AS t(N)),
T256
AS (SELECT Row_number()
OVER(
ORDER BY (SELECT 0)) - 1 AS n
FROM T4 AS a
CROSS JOIN T4 AS b
CROSS JOIN T4 AS c),
START_DATE
AS (SELECT Dateadd(DAY, Datediff(DAY, '', @now), '') AS start_date),
START_TIME
AS (SELECT Dateadd(MINUTE, Datediff(MINUTE, '', @now) / 30 * 30, '') AS
start_time),
DAILY_INTERVALS
AS (SELECT N * 30 AS interval
FROM T256
WHERE N < 48)
SELECT TOP (1) Dateadd(DAY, future_days.N, START_DATE) AS BookDate,
DAILY_INTERVALS.INTERVAL AS BookSlot
FROM START_DATE
CROSS APPLY START_TIME
CROSS APPLY DAILY_INTERVALS
CROSS APPLY T256 AS future_days
WHERE Dateadd(MINUTE, DAILY_INTERVALS.INTERVAL,
Dateadd(DAY, future_days.N, START_DATE)) > START_TIME
AND NOT EXISTS(SELECT *
FROM DBO.BOOKINGS
WHERE BOOKDATE = START_DATE
AND BOOKSLOT = DAILY_INTERVALS.INTERVAL)
ORDER BY BOOKDATE,
BOOKSLOT;
请参阅此SQL Fiddle
答案 2 :(得分：0)
在SQL Server 2012及更高版本中，您可以使用lead()功能。由于所有的边界条件，逻辑有点复杂。我认为这抓住了它：
select top 1
(case when BookSlot = 1410 then BookDate else BookDate + 1 end) as BookDate,
(case when BookSlot = 1410 then 0 else BookSlot + 30 end) as BookSlot
from (select b.*,
lead(BookDate) over (order by BookDate) as next_dt,
lead(BookSlot) over (partition by BookDate order by BookSlot) as next_bs
from bookings b
) b
where (next_bs is null and BookSlot < 1410 or
next_bs - BookSlot > 30 or
BookSlot = 1410 and (next_dt <> BookDate + 1 or next_dt = BookDate and next_bs <> 0)
)
order by BookDate, BookSlot;
答案 3 :(得分：0)
使用计数表生成6周内原始可用预订广告位列表（可在下面调整）：
declare @Date as date = getdate();
declare @slot as int = 30 * (datediff(n,@Date,getdate()) /30);
with
slots as (
select (ROW_NUMBER() over (order by s)-1) * 30 as BookSlot
from(
values (1),(1),(1),(1),(1),(1),(1),(1) -- 4 hour block
)slots(s)
cross join (
values (1),(1),(1),(1),(1),(1) -- 6 blocks of 4 hours each day
)QuadHours(t)
)
,days as (
select (ROW_NUMBER() over (order by s)-1) + getdate() as BookDate
from (
values (1),(1),(1),(1),(1),(1),(1) -- 7 days in a week
)dayList(s)
cross join (
-- set this to number of weeks out to allow bookings to be made
values (1),(1),(1),(1),(1),(1) -- allow 6 weeks of bookings at a time
)weeks(t)
)
,tally as (
select
cast(days.BookDate as date) as BookDate
,slots.BookSlot as BookSLot
from slots
cross join days
)
select top 1
tally.BookDate
,tally.BookSlot
from tally
left join #Bookings book
on tally.BookDate = book.BookDate
and tally.BookSlot = book.BookSlot
where book.BookSlot is null
and ( tally.BookDate > @Date or tally.BookSlot > @slot )
order by tally.BookDate,tally.BookSlot;
go
答案 4 :(得分：-1)
试试这个：
SELECT a.bookdate, ((a.bookslot/60.)+.5) * 60
FROM bookings a LEFT JOIN bookings b
ON a.bookdate=b.bookdate AND (a.bookslot/60.)+.50=b.bookslot/60.
WHERE b.bookslot IS null