Question

如果您有开始时间：时间列

例10:10:01到10:10:31

需要按10个间隔进行分组：

示例

     10:10:01  to   10:10:04 ,
     10:10:04  to   10:10:07 ,
     10:10:07  to   10:10:10 ,
     10:10:10  to   10:10:13 ,
     10:10:13  to   10:10:16 ,
     10:10:16  to   10:10:19 ,
     10:10:19  to   10:10:22 ,
     10:10:22  to   10:10:25 ,
     10:10:25  to   10:10:28 ,
     10:10:28  to   10:10:31

使用SQL Server

Answer 1

这是一个解决方案（SQL Fiddle demo）：

declare @from time = '10:10:01', @to time = '10:10:31'

declare @interval int = datediff(millisecond, @from, @to) / 10

select dateadd(millisecond, @interval * (i-1), @from) as [From], 
  dateadd(millisecond, @interval * i, @from) as [To]
from (values (1), (2), (3), (4), (5), (6), (7), (8), (9), (10)) x(i)

它是如何工作的：需要差异 - 以毫秒为单位进行良好测量，因为时间可能并不总是很好地划分 - 在指定的@from和@to之间，并除以10以获得间隔。然后它使用一个值表作为一个穷人的循环（因为我不想写出10个查询的UNION ALL）来生成每个句点的时间和时间

Answer 2

这是另一个解决方案，即使您更改了所需的间隔数量也能正常工作：

create function "dbo"."splitIntervals"
(
    @starttime time
    , @endtime time
    , @intervals int
)
returns table
as
return
(
    with cte as
    (
        select 
            cast( @starttime as time ) as "starttime"
            , cast( dateadd( ms, datediff(ms, @starttime, @endtime) / @intervals, @starttime) as time ) as "endtime" 
        union all
        select 
            cast(dateadd( ms, datediff(ms, @starttime, @endtime) / @intervals, "starttime") as time ) as "starttime" 
            , cast( dateadd( ms, datediff(ms, @starttime, @endtime) / @intervals, "endtime") as time ) as "entime" 
        from cte where "starttime" < cast( @endtime as time )
    )
    select 
        * from "cte"
    where
        endtime <= @endtime
);

;select * from "dbo"."splitIntervals" ( '10:10:01.000', '10:10:31.000', 15)

See SQL-Fiddle for working version

将开始时间除以结束时间间隔10

2 个答案: