将给定日期拆分为几天

时间:2015-07-22 10:57:04

标签: sql sql-server sql-server-2008

我正在写一个存储过程。基本上我需要从用户那里获取年份和月份信息并做一些事情。但我的问题是我没有发现给定月份有多少天。让我用例子解释一下;

如我所说,用户给我一年和一个月;

docker run --name cuda_app $(for dev in /dev/nvidia*; do echo -n "--device $dev:$dev "; done) cuda_app

所以我必须创建像

这样的行
@year = 2015

@month = 07

我的计划是将这些行逐个添加到另一个临时表中。

我的SP的最后状态是;

2015-07-01

2015-07-02

.....

2015-07-31

我希望能正确解释。

8 个答案:

答案 0 :(得分:1)

我已将年和月变量更改为int数据类型, 您的日期列的名称为dates_date,并使用convert和带有dateadd函数的while循环来填充@dates表:

ALTER PROCEDURE [dbo].[Usp_CreateStats]
(
 @Year int,
 @Month int
 )
AS
BEGIN

DECLARE @StartTime VARCHAR (10),
        @EndTime VARCHAR (10),
        @Date date
SET @StartTime = '00:00:00'
SET @EndTime = '23:59:59'

SET @Date = CONVERT(date, right('0000' + cast(@year as char(4)), 4) + right('00' + cast(@month as char(2)), 2)+ '01', 112)

DECLARE @Peoples TABLE (name VARCHAR(50))
INSERT INTO @Peoples (name) select distinct name from USERINFO

DECLARE @Dates TABLE (dates_date date)

while month(@date) = @Month    
begin
    insert into @Dates (dates_date) values (@Date)
    set @Date = dateadd(day, 1, @date)
end

END

答案 1 :(得分:1)

要根据@year@month生成日期,您可以使用Tally Table

DECLARE @year   INT = 2015,
        @month  INT = 7

DECLARE @startDate DATE = DATEADD(MONTH, @month - 1, DATEADD(YEAR, @year - 1900, CAST('19000101' AS DATE)))

;WITH Tally (n) AS
(   -- 100 rows
    SELECT TOP(31) 
        ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
    FROM (VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) a(n)
    CROSS JOIN (VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) b(n)
)
SELECT
    DATEADD(DAY, N-1, @startDate)
FROM Tally t
WHERE
    DATEADD(DAY, N-1, @startDate) < DATEADD(MONTH, 1, @startDate)

SQL Fiddle

注意:为变量使用适当的数据类型。在这种情况下,@month@year应为INT

答案 2 :(得分:1)

您可以使用递归CTE。

Declare @Year as varchar(5) = '2015'
Declare @Month as varchar(5) = '05'

Declare @DateOfMonth as DateTime =  @Month + '-01-' + @Year

Declare @startdate datetime, @enddate datetime

SET @startdate = DATEADD(s,0,DATEADD(mm, DATEDIFF(m,0,@DateOfMonth),0))
SET @enddate = DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,@DateOfMonth)+1,0))
SET @enddate = DATEADD(day,-1,@enddate)

;With DateSequence( Date ) as
(
    Select @startdate as Date
        union all
    Select dateadd(day, 1, Date)
        from DateSequence
        where Date < @enddate
)

--select result
Select * from DateSequence option (MaxRecursion 1000)

答案 3 :(得分:0)

DECLARE @Month int= 7 
Declare @Year int = 2015
Declare @MonthStart date = CAST(@Year  as varchar(4)) +  Right('00' +    Cast(@Month as varchar(2)), 2)  + '01'


DECLARE @DaysInMonth int = (SELECT
                    datediff(day, dateadd(day, 1-day(@MonthStart),     @MonthStart),
                    dateadd(month, 1, dateadd(day, 1-day(@MonthStart), @MonthStart))))

Declare @Dates TABLE (date date) 

DECLARE @Counter int = 0 
WHILE @Counter < @DaysInMonth 
BEGIN
INSERT INTO @Dates SELECT DATEADD(Day, @Counter,  @MonthStart)

SET @Counter = @Counter + 1 
END

答案 4 :(得分:0)

您可以使用CTE:

DECLARE @year               INT = 2015,
        @month              INT = 7,
        @first_day_of_month DATETIME,
        @last_day_of_month  DATETIME 

select @first_day_of_month = cast(cast(@year as varchar(4))+'/'+cast(@month as varchar(2))+'/'+'01' as date)
select @last_day_of_month = dateadd(mm,datediff(mm,0,@first_day_of_month)+1,0)-1

;with mycte as 
(
select @first_day_of_month as src
union all
select dateadd(dd,1,src) from mycte
where src < @last_day_of_month

)select * from mycte

答案 5 :(得分:0)

使用递归CTE,

DECLARE @year  VARCHAR(4) = '2015',
        @month VARCHAR(2)= '02',
        @date  DATETIME

SET @date= CONVERT(DATETIME, @year + '-' + @month + '-01');

WITH cte
     AS (SELECT @date AS dt
         UNION ALL
         SELECT Dateadd(dd, 1, dt) AS dt
         FROM   cte
         WHERE  dt < Dateadd(dd, -1, Dateadd(mm, Datediff(mm, 0, @date) + 1, 0)))
SELECT *
FROM   cte 

答案 6 :(得分:0)

这是另一个(更快)的计数解决方案:

DECLARE @year   INT = 2015,
        @month  INT = 7

DECLARE @start datetime = dateadd(m, (@year- 1900) * 12 + 7, -31)

SELECT TOP(datediff(d, @start, dateadd(m, 1, @start))) 
  dateadd(d, ROW_NUMBER() OVER (ORDER BY (SELECT 1)) - 1, @start)
FROM 
  (VALUES(0),(0),(0),(0),(0),(0)) a(n)
   CROSS JOIN (VALUES(0),(0),(0),(0),(0),(0)) b(n)

答案 7 :(得分:-2)

convert(varchar,OrderDate,111)='2015-07-01'