我需要使用 SQL Server 2005 获取日期范围内的所有日期
Harish KV
问问题
32127 次
7 回答
56
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);
于 2008-11-07T12:09:06.587 回答
8
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。
于 2008-11-07T09:23:56.213 回答
5
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
于 2016-12-04T12:48:14.320 回答
1
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。
于 2008-11-07T16:09:43.480 回答
0
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
于 2009-08-13T09:38:52.330 回答
-2
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
于 2008-11-07T12:22:02.097 回答
-3
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
于 2008-11-07T11:07:13.760 回答