具有相同的列值的3倍以上选择行

Question

我有一个名为Theater(Sn, SeatVacant)

e.g SN SEATVACANT 
    1 Y 
    2 Y 
    3 N 
    . . 
    . . 
    100 Y 

我要预订3个席位（应该是连续的）表。我怎样才能获得连续空缺的座位。

Answer 1

select f1.sn, f2.sn, f3.sn from Theater f1 
inner join Theater f2 on f1.sn=f2.sn + 1 
inner join Theater f3 on f1.sn=f3.sn + 2 
where f1.SEATVACANT='Y' and f2.SEATVACANT='Y' and f3.SEATVACANT='Y' 

Answer 2

with tablenewkey as(
select ROW_NUMBER() over(order by f1.sn) newkey, f1.* from theater f1 
), 
nbplacevacant as (
select 3 as NbrowByGroup 
), 
calculdiff as (
select f1.*, isnull(f3.newkey, 0) newkeylastN, f1.newkey - isnull(f3.newkey, 0) DiffYWithLasN 
from tablenewkey f1 
outer apply 
(
select top 1 * from tablenewkey f2 
where f2.newkey<f1.newkey and f2.SEATVACANT='N' 
order by f2.newkey desc 
) f3 
where f1.SEATVACANT='Y' and (f1.newkey - isnull(f3.newkey, 0))>=(select NbrowByGroup from nbplacevacant) 
), 
possibilite as (
select f0.*, f1.newkey Groupement, f1.DiffYWithLasN 
from tablenewkey f0 inner join calculdiff f1 
on f0.newkey between (f1.newkey - DiffYWithLasN +1) and f1.newkey 
where f0.SEATVACANT='Y' 
) 
select newkey, sn, Groupement, DENSE_RANK() over(order by Groupement) PossiblilityRang from possibilite 
order by groupement, sn 

Answer 3

这样，如果你想要一个动态的解决方案：

--Theater(Sn, SeatVacant) 

DECLARE @ContiguougsSeats AS INT 
SET @ContiguougsSeats = 4 

SELECT Sn, ' to ', Sn+@ContiguougsSeats-1 
FROM Theater As t1 
WHERE t1.Sn+@ContiguougsSeats-1 <= (Select MAX(Sn) From Theater) 
    AND NOT EXISTS(
    Select * 
    From Theater As t2 
    Where t2.Sn Between t1.Sn AND t1.Sn+@ContiguougsSeats-1 
     And t2.SeatVacant = 'N' 
    )