获取每个小时的时间范围

Question

所以我想要的是生成特定时间范围内的所有小时.

所以考虑到上午11点到下午2点,我会得到:

 11:00 AM
 12:00 PM
 1:00 PM
 2:00 PM

我试图避免必须存储商店可能打开的每个特定小时,只是存储范围(我需要比较其他时间的小时)

谢谢

Answer 1

无需循环,递归CTE或数字表.

DECLARE 
  @start TIME(0) = '11:00 AM', 
  @end   TIME(0) =  '2:00 PM';

WITH x(n) AS 
(
  SELECT TOP (DATEDIFF(HOUR, @start, @end) + 1) 
  rn = ROW_NUMBER() OVER (ORDER BY [object_id]) 
  FROM sys.all_columns ORDER BY [object_id]
)
SELECT t = DATEADD(HOUR, n-1, @start) FROM x ORDER BY t;

Answer 2

您可以使用递归CTE.这将产生11到14之间的小时:

;with   Hours as
        (
        select  11 as hr
        union all
        select  hr + 1
        from    Hours
        where   hr < 14
        )
select  *
from    Hours

SQL Fiddle的实例.

Answer 3

如果你有一个数字表(如果你没有,点击链接创建一个)...

create table test(
    startTime time
,   endTime time
)

insert into test
select '11:00', '14:00'

select
    dateadd(hh, n.n, t.startTime) as times
from test t
  inner join Numbers n
    -- assuming your numbers start at 1 rather than 0
    on n.n-1 <= datediff(hh, t.startTime, t.endTime)

如果这是专门的,您可以创建一个只有24个值的小时表.

create table HoursInADay(
    [hours] time not null
,   constraint PK_HoursInADay primary key ([hours])
)

-- insert
insert into HoursInADay select '1:00'
insert into HoursInADay select '2:00'
insert into HoursInADay select '3:00'
insert into HoursInADay select '4:00'
insert into HoursInADay select '5:00'
insert into HoursInADay select '6:00'
insert into HoursInADay select '7:00'
...

select
    h.[hours]
from test t
  inner join HoursInADay h
    on h.[hours] between t.startTime and t.endTime