Sql 按顺序获取x行数

我有一个表，将“TableAvailable”TableID称为int，并将其称为smallInt（0,1），如下所示

TableID |Available 
1       |1
2       |0
3       |0
4       |1
5       |1
6       |1
7       |0
8       |1

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我需要一个sql，我可以选择一起的前3个表，在我们的示例中，它应该是4,5,6，这是一起可用的前3行

select min(tableid), max(tableid)
from (select ta.*,
             (row_number() over (order by tableid) -
              row_number() over (partition by available order by tableid)
             ) as grp
      from tableavailable ta
     ) ta
where available = 1
group by grp;

然后，添加

having count（*）>=3 order by min（tableid）

将获得第一个

然而，一种更快的方法是只查看下两条记录的可用性。在SQL Server 2012+中，您将使用

lead（）

：

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假设你只需要做3次，假设tableId是连续的，没有间隔。。。两者都可能是错误的假设

SELECT Top 1 A.TableID, B.TableID, C.TableId
FROM TableAvailable A
LEFT JOIN tableAvailable B
 on A.ID = B.ID+1
LEFT JOIn tableAvailable C
 on A.ID = B.ID+2
WHERE A.Available = 1 and B.Availabe=1 and C.Available=1
order by tableID asc

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您没有指定输出的格式，因此我假设您需要表中的行。以下是一个可能的解决方案：

WITH answer AS (
select MIN(TableID) as found
from TableAvailable as T1
where Available = 1
and 1 = (select Available
         from TableAvailable as T2
         where T2.TableID = T1.TableID + 1
         )
and 1 = (select Available
         from TableAvailable as T3
         where T3.TableID = T1.TableID + 2
         )         
  )
select *
from TableAvailable
join answer 
on TableID = found
or TableID = found + 1
or TableID = found + 2

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在这里工作：

什么版本的SQL Server？并且，第一列中的ID是否总是在一行中，没有间隙？

WITH answer AS (
select MIN(TableID) as found
from TableAvailable as T1
where Available = 1
and 1 = (select Available
         from TableAvailable as T2
         where T2.TableID = T1.TableID + 1
         )
and 1 = (select Available
         from TableAvailable as T3
         where T3.TableID = T1.TableID + 2
         )         
  )
select *
from TableAvailable
join answer 
on TableID = found
or TableID = found + 1
or TableID = found + 2