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
次
1899-12-30 12:30:00.000
1899-12-30 10:00:00.000
1899-12-30 10:00:00.000
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
次
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
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
其他回答
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!
select s.SalesID from SalesTbl s
where cast(cast(s.SaleDate as date) as datetime) + cast(cast(s.SaleCreatedDate as time) as datetime) between @FromDate and @ToDate
我遇到了类似的情况,我必须将日期和时间字段合并到DateTime字段。上面提到的解决方案都不行,特别是添加两个字段作为添加这两个字段的数据类型是不一样的。
我创建了下面的解决方案,其中我添加了小时和分钟部分的日期。这对我来说非常有效。请查看一下,如果遇到任何问题请告诉我。
与资源; 作为 ( select StatusTime = '12/30/1899 5:17:00 PM', StatusDate = '7/24/2019 12:00:00 AM' ) select DATEADD(MI, DATEPART(MINUTE,CAST(tbl))StatusTime AS TIME)),DATEADD(HH, DATEPART(小时,CAST(tbl))。StatusTime AS TIME)), CAST(tbl。StatusDate为DATETIME)) 从台
执行结果:2019-07-24 17:17:00.000
SELECT (CAST(@TimeField As Date) As DateTime) + CAST(CAST(@TimeField As Time) 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