将存储在datetime字段中的第一个日期转换为字符串，然后将存储在datetime字段中的时间转换为字符串，附加两个日期并转换回datetime字段，所有这些都使用已知的转换格式。

Convert(datetime, Convert(char(10), MYDATETIMEFIELD, 103) + ' ' + Convert(char(8), MYTIMEFIELD, 108), 103) 

如果你没有使用SQL Server 2008(即你只有一个DateTime数据类型)，你可以使用以下(承认粗糙和准备就绪)TSQL来实现你想要的:

DECLARE @DateOnly AS datetime
DECLARE @TimeOnly AS datetime 

SET @DateOnly = '07 aug 2009 00:00:00'
SET @TimeOnly = '01 jan 1899 10:11:23'


-- Gives Date Only.
SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, @DateOnly))

-- Gives Time Only.
SELECT DATEADD(Day, -DATEDIFF(Day, 0, @TimeOnly), @TimeOnly)

-- Concatenates Date and Time parts.
SELECT
CAST(
    DATEADD(dd, 0, DATEDIFF(dd, 0, @DateOnly)) + ' ' +
    DATEADD(Day, -DATEDIFF(Day, 0, @TimeOnly), @TimeOnly)           
as datetime)

虽然粗糙，但很管用!

这对我很有效

CAST(Tbl.date as DATETIME) + CAST(Tbl.TimeFrom AS TIME)

(适用于SQL 2008 R2)

这是我的解决方案，它忽略了时间列的日期值

CAST(Tbl.date as DATETIME) + CAST(CAST(Tbl.TimeFrom AS TIME) as DATETIME)

希望这能帮助到其他人

结合日期从一个datetime列和时间从另一个datetime列，这是最好的，最快的解决方案为您:

select cast(cast(DateColumn as date) as datetime) + cast(TimeColumn as datetime) from YourTable

简单地将两者连接起来，但首先将它们强制转换，如下所示

select cast(concat(Cast(DateField as varchar), ' ', Cast(TimeField as varchar)) as datetime) as DateWithTime from TableName;