sql - 获取一系列日期之间的日期

Question

我需要使用 SQL Server 2005 获取日期范围内的所有日期

Answer 1

干得好：

DECLARE @DateFrom smalldatetime, @DateTo smalldatetime;
SET @DateFrom='20000101';
SET @DateTo='20081231';
-------------------------------
WITH T(date)
AS
( 
SELECT @DateFrom 
UNION ALL
SELECT DateAdd(day,1,T.date) FROM T WHERE T.date < @DateTo
)
SELECT date FROM T OPTION (MAXRECURSION 32767);

Answer 2

如果您在表格中有日期并且只想选择两个日期之间的日期，您可以使用

select * from yourTable where yourDate between date1 and date2

如果您想从无到有生成日期，您可以使用循环来完成，或者您可以使用日期填充临时表，然后从中进行选择。

Answer 3

DECLARE @Date1 DATE='2016-12-21', @Date2 DATE='2016-12-25'
SELECT DATEADD(DAY,number,@Date1) [Date] FROM master..spt_values WHERE type = 'P' AND DATEADD(DAY,number,@Date1) <= @Date2

Answer 4

这是日期生成的 Oracle 版本：

SELECT TO_DATE ('01-OCT-2008') + ROWNUM - 1 g_date
  FROM all_objects
 WHERE ROWNUM <= 15

而不是 all_objects 它可以是任何具有足够行来覆盖所需范围的表。

Answer 5

稍微复杂一点但可能更灵活的方法是使用包含一组连续数字的表格。这允许多个具有不同间隔的日期范围。

/* holds a sequential set of number ie 0 to max */
/* where max is the total number of rows expected */
declare @Numbers table ( Number int  )

declare @max int 
declare @cnt int

set @cnt = 0
/* this value could be limited if you knew the total rows expected */
set @max = 999 

/* we are building the NUMBERS table on the fly */
/* but this could be a proper table in the database */
/* created at the point of first deployment */
while (@cnt <= @max)
begin
      insert into @Numbers select @cnt
      set @cnt = @cnt + 1
end

/* EXAMPLE of creating dates with different intervals */

declare @DateRanges table ( 
   StartDateTime datetime, EndDateTime datetime, Interval int )

/* example set of date ranges */
insert into @DateRanges
select '01 Jan 2009', '10 Jan 2009', 1 /* 1 day interval */
union select '01 Feb 2009', '10 Feb 2009', 2 /* 2 day interval */

/* heres the important bit generate the dates */
select
      StartDateTime
from
(
      select
            d.StartDateTime as RangeStart,
            d.EndDateTime as RangeEnd,
            dateadd(DAY, d.Interval * n.Number, d.StartDateTime) as StartDateTime
      from 
            @DateRanges d, @Numbers n
) as dates
where
      StartDateTime between RangeStart and RangeEnd
order by StartDateTime

我实际上使用它的变体将日期分成时间段（有不同的时间间隔，但通常是 5 分钟长）。我的@numbers 表最多包含 288 个，因为这是您在 24 小时内可以拥有的 5 分钟插槽的总数。

/* EXAMPLE of creating times with different intervals */

delete from @DateRanges 

/* example set of date ranges */
insert into @DateRanges
select '01 Jan 2009 09:00:00', '01 Jan 2009 12:00:00', 30 /* 30 minutes interval */
union select '02 Feb 2009 09:00:00', '02 Feb 2009 10:00:00', 5 /* 5 minutes interval */

/* heres the import bit generate the times */
select
      StartDateTime,
      EndDateTime
from
(
      select
            d.StartDateTime as RangeStart,
            d.EndDateTime as RangeEnd,
            dateadd(MINUTE, d.Interval * n.Number, d.StartDateTime) as StartDateTime,
            dateadd(MINUTE, d.Interval * (n.Number + 1) , StartDateTime) as EndDateTime
      from 
            @DateRanges d, @Numbers n
) as dates
where
      StartDateTime >= RangeStart and EndDateTime <= RangeEnd
order by StartDateTime

Answer 6

如果您想要获取数据库中两个日期之间的所有日期（即客户在 2008 年第三季度下订单的日期），您可以编写如下内容：

select distinct(orderPlacedDate) 
from orders 
where orderPlacedDate between '2008-07-01' and 2008-09-30' 
order by orderPlacedDate

Answer 7

要生成一系列日期，您可以编写一个表值函数。这是一个为数据仓库创建日期维度的函数——您可以通过删除特殊内容来相当容易地调整它。

编辑：这里没有日期维度层次结构。

if object_id ('ods.uf_DateHierarchy') is not null
    drop function ods.uf_DateHierarchy
go

create function ods.uf_DateHierarchy (
       @DateFrom datetime
      ,@DateTo   datetime
) returns @DateHierarchy table (
        DateKey           datetime
) as begin
    declare @today           datetime  
    set @today = @Datefrom

    while @today <= @DateTo begin
        insert @DateHierarchy (DateKey) values (@today)
        set @today = dateadd (dd, 1, @today)
    end

    return
end

go