Question

我有一张名为Theater(Sn, SeatVacant)

的表格

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

我想预订3个座位（应该是连续的）。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

像这样，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'
    )

选择具有相同列值超过3次的行

3 个答案: