我知道OP正在询问SQL 2005的答案，但这个问题已经很老了，所以如果你正在运行SQL 2012或以上版本，你可以使用以下命令:

SELECT DATEADD(DAY, 1, EOMONTH(@somedate, -1))

参考: https://learn.microsoft.com/en-us/sql/t-sql/functions/eomonth-transact-sql?view=sql-server-2017&viewFallbackFrom=sql-server-previousversions

试试这个:

Declare @DayOfMonth TinyInt Set @DayOfMonth = 13
Declare @Month TinyInt Set @Month = 6
Declare @Year Integer Set @Year = 2006
-- ------------------------------------
Select DateAdd(day, @DayOfMonth - 1, 
          DateAdd(month, @Month - 1, 
              DateAdd(Year, @Year-1900, 0)))

It works as well, has added benefit of not doing any string conversions, so it's pure arithmetic processing (very fast) and it's not dependent on any date format This capitalizes on the fact that SQL Server's internal representation for datetime and smalldatetime values is a two part value the first part of which is an integer representing the number of days since 1 Jan 1900, and the second part is a decimal fraction representing the fractional portion of one day (for the time) --- So the integer value 0 (zero) always translates directly into Midnight morning of 1 Jan 1900...

或者，感谢@brinary的建议，

Select DateAdd(yy, @Year-1900,  
       DateAdd(m,  @Month - 1, @DayOfMonth - 1)) 

2014年10月编辑。正如@cade Roux所指出的，SQL 2012现在有一个内置函数: DATEFROMPARTS(年，月，日) 这是一样的。

编辑2016年10月3日，(感谢@bambams注意到这一点，@brinary修复了它)，最后的解决方案，由@brinary提出。除非先执行年份加法，否则似乎对闰年不起作用

select dateadd(month, @Month - 1, 
     dateadd(year, @Year-1900, @DayOfMonth - 1)); 

我个人更喜欢Substring，因为它提供了清理选项，并能够根据需要分割字符串。假设数据的格式是'dd, mm, yyyy'。

--2012 and above
SELECT CONCAT (
        RIGHT(REPLACE(@date, ' ', ''), 4)
        ,'-'
        ,RIGHT(CONCAT('00',SUBSTRING(REPLACE(@date, ' ', ''), CHARINDEX(',', REPLACE(@date, ' ', '')) + 1, LEN(REPLACE(@date, ' ', '')) - CHARINDEX(',', REPLACE(@date, ' ', '')) - 5)),2)
        ,'-'
        ,RIGHT(CONCAT('00',SUBSTRING(REPLACE(@date, ' ', ''), 1, CHARINDEX(',', REPLACE(@date, ' ', '')) - 1)),2)
        )

--2008 and below
SELECT   RIGHT(REPLACE(@date, ' ', ''), 4)
        +'-'
        +RIGHT('00'+SUBSTRING(REPLACE(@date, ' ', ''), CHARINDEX(',', REPLACE(@date, ' ', '')) + 1, LEN(REPLACE(@date, ' ', '')) - CHARINDEX(',', REPLACE(@date, ' ', '')) - 5),2)
        +'-'
        +RIGHT('00'+SUBSTRING(REPLACE(@date, ' ', ''), 1, CHARINDEX(',', REPLACE(@date, ' ', '')) - 1),2)

下面演示了如果数据存储在列中，如何使用它。不用说，在应用到列之前检查结果集是最理想的

DECLARE @Table TABLE (ID INT IDENTITY(1000,1), DateString VARCHAR(50), DateColumn DATE)

INSERT INTO @Table
SELECT'12, 1, 2007',NULL
UNION
SELECT'15,3, 2007',NULL
UNION
SELECT'18, 11 , 2007',NULL
UNION
SELECT'22 , 11, 2007',NULL
UNION
SELECT'30, 12, 2007  ',NULL

UPDATE @Table
SET DateColumn = CONCAT (
        RIGHT(REPLACE(DateString, ' ', ''), 4)
        ,'-'
        ,RIGHT(CONCAT('00',SUBSTRING(REPLACE(DateString, ' ', ''), CHARINDEX(',', REPLACE(DateString, ' ', '')) + 1, LEN(REPLACE(DateString, ' ', '')) - CHARINDEX(',', REPLACE(DateString, ' ', '')) - 5)),2)
        ,'-'
        ,RIGHT(CONCAT('00',SUBSTRING(REPLACE(DateString, ' ', ''), 1, CHARINDEX(',', REPLACE(DateString, ' ', '')) - 1)),2)
        ) 

SELECT ID,DateString,DateColumn
FROM @Table

Try

CAST(STR(DATEPART(year, DATE))+'-'+ STR(DATEPART(month, DATE)) +'-'+ STR(DATEPART(day, DATE)) AS DATETIME)

Sql Server 2012有一个函数，它将根据部件(DATEFROMPARTS)创建日期。对于我们其他人来说，这是我创建的一个db函数，它将确定部件的日期(谢谢@Charles)…

IF EXISTS (SELECT * FROM dbo.sysobjects WHERE id = object_id(N'[dbo].[func_DateFromParts]'))
    DROP FUNCTION [dbo].[func_DateFromParts]
GO

CREATE FUNCTION [dbo].[func_DateFromParts]
(
    @Year INT,
    @Month INT,
    @DayOfMonth INT,
    @Hour INT = 0,  -- based on 24 hour clock (add 12 for PM :)
    @Min INT = 0,
    @Sec INT = 0
)
RETURNS DATETIME
AS
BEGIN

    RETURN DATEADD(second, @Sec, 
            DATEADD(minute, @Min, 
            DATEADD(hour, @Hour,
            DATEADD(day, @DayOfMonth - 1, 
            DATEADD(month, @Month - 1, 
            DATEADD(Year, @Year-1900, 0))))))

END

GO

你可以这样称呼它……

SELECT dbo.func_DateFromParts(2013, 10, 4, 15, 50, DEFAULT)

返回……

2013-10-04 15:50:00.000

对于低于12的SQL Server版本，我可以建议使用SET DATEFORMAT结合使用CAST

-- 26 February 2015
SET DATEFORMAT dmy
SELECT CAST('26-2-2015' AS DATE)

SET DATEFORMAT ymd
SELECT CAST('2015-2-26' AS DATE)

如何创建这些字符串取决于你