如果两个字段都是datetime，那么简单地添加它们就可以了。 例如: 声明@d datetime， @t datetime Set @d = '2009-03-12 00:00:00.000'; Set @t = ' 189-12-30 12:30:00.000'; 选择@d + @t 如果您使用Date & Time数据类型，则只需将时间转换为datetime 例如: 声明@d日期，@t时间 Set @d = '2009-03-12'; Set @t = '12:30:00.000'; 选择@d + cast(@t as datetime)

DECLARE @Dates table ([Date] datetime);
DECLARE @Times table ([Time] datetime);

INSERT INTO @Dates VALUES('2009-03-12 00:00:00.000');
INSERT INTO @Dates VALUES('2009-03-26 00:00:00.000');
INSERT INTO @Dates VALUES('2009-03-30 00:00:00.000');

INSERT INTO @Times VALUES('1899-12-30 12:30:00.000');
INSERT INTO @Times VALUES('1899-12-30 10:00:00.000');
INSERT INTO @Times VALUES('1899-12-30 10:00:00.000');

WITH Dates (ID, [Date])
AS (
    SELECT ROW_NUMBER() OVER (ORDER BY [Date]), [Date] FROM @Dates
), Times (ID, [Time])
AS (
    SELECT ROW_NUMBER() OVER (ORDER BY [Time]), [Time] FROM @Times
)
SELECT Dates.[Date] + Times.[Time] FROM Dates
    JOIN Times ON Times.ID = Dates.ID

打印:

2009-03-12 10:00:00.000
2009-03-26 10:00:00.000
2009-03-30 12:30:00.000

这对我很有效

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

(适用于SQL 2008 R2)

将两个字段转换为DATETIME:

SELECT CAST(@DateField as DATETIME) + CAST(@TimeField AS DATETIME)

如果你正在使用Getdate()，首先使用这个:

DECLARE @FechaActual DATETIME = CONVERT(DATE, GETDATE());
SELECT CAST(@FechaActual as DATETIME) + CAST(@HoraInicioTurno AS DATETIME)

另一种方法是使用CONCATand CAST，请注意，需要使用DATETIME2(x)才能使其工作。你可以将x设置为0- 7,7之间的任何值，这意味着没有精度损失。

DECLARE @date date = '2018-03-12'
DECLARE @time time = '07:00:00.0000000'
SELECT CAST(CONCAT(@date, ' ', @time) AS DATETIME2(7))

返回2018-03-12 07:00:00.0000000

在SQL Server 14上测试

如上所述，我有很多错误，所以我这样做

try_parse(concat(convert(date,Arrival_date),' ',arrival_time) as datetime) AS ArrivalDateTime

这对我很管用。