在我正在处理的提取中,我有2个datetime列。一列存储日期,另一列存储如下所示的时间。

如何查询表,将这两个字段组合成类型为datetime的1列?

日期

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

1899-12-30 12:30:00.000
1899-12-30 10:00:00.000
1899-12-30 10:00:00.000

当前回答

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!

其他回答

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) 

SELECT (CAST(@TimeField As Date) As DateTime) + CAST(CAST(@TimeField As Time) As DateTime)

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

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!

SELECT CAST(your_date_column AS date) + CAST(your_time_column AS datetime) FROM your_table

效果非常好