SQL 查找组中的连续数字

Question

我有一张类似于所示的表。它包含一个用户 ID 列表、一天中每个小时的小时值和一个 Avail 标志，以确定该用户在那个小时是否有空。

我需要列出在定义为 @n 的连续几个小时内可用的所有用户 ID

#####################
# UID # Avail # Hour#
#####################
# 123 #   1   #  0  #
# 123 #   1   #  1  #
# 123 #   0   #  2  #
# 123 #   0   #  3  #
# 123 #   0   #  4  #
# 123 #   1   #  5  #
# 123 #   1   #  6  #
# 123 #   1   #  7  #
# 123 #   1   #  8  #
# 341 #   1   #  0  #
# 341 #   1   #  1  #
# 341 #   0   #  2  #
# 341 #   1   #  3  #
# 341 #   1   #  4  #
# 341 #   0   #  5  #
# 341 #   1   #  6  # 
# 341 #   1   #  7  #
# 341 #   0   #  8  #
######################

这应该会导致@n=3 的以下输出

#######
# UID #
#######
# 123 #
#######

我曾尝试使用
ROW_NUMBER() over (partition by UID,Avail ORDER BY UID,Hour)
为由 UID 分区的每一行分配一个数字，以及它们是否被标记为可用。但是，这不起作用，因为可用性时间段一天可能会更改多次，并且 ROW_NUMBER() 函数仅根据可用性标志为每个用户保留两个计数。

Answer 1

如果您使用的是 SQL Server 2012+，您可以使用窗口 SUM，但您必须提前指定窗口框架中的行数，因为它不接受变量，因此它不是那么灵活:

;with cte as 
(
    select distinct 
       UID, 
       SUM(avail) over (partition by uid 
                        order by hour 
                        rows between current row and 2 following
                       ) count 
    from table1
)
select uid from cte where count = 3;

如果您想要灵活性，您可以将其设为存储过程并使用动态 SQL 来构建和执行语句，如下所示:

create procedure testproc (@n int) as
declare @sql nvarchar(max)
set @sql = concat('
    ;with cte as 
    (
       select distinct 
          UID, 
          SUM(avail) over (partition by uid 
                        order by hour 
                        rows between current row and ', @n - 1 , ' following
                        ) count 
       from table1
    )
    select uid from cte where count = ' , @n , ';')
exec sp_executesql @sql

并使用 execute testproc 3 执行它

一个更不灵活的解决方案是使用相关子查询，但是您必须为每个添加的计数添加另一个子查询:

select distinct uid 
from Table1 t1
where Avail = 1
  and exists (select 1 from Table1 where Avail = 1 and UID = t1.UID and Hour = t1.Hour + 1)
  and exists (select 1 from Table1 where Avail = 1 and UID = t1.UID and Hour = t1.Hour + 2);

还有另一种方法，使用 row_number 查找岛屿，然后按每个岛屿的总和进行过滤:

;with c as (
    select 
       uid, avail, 
       row_number() over (partition by uid order by hour) 
       - row_number() over (partition by uid, avail order by hour) grp
from table1
)

select uid from c
group by uid, grp
having sum(avail) >= 3