舍入毫秒 SQL Server 2000

Question

在 Microsoft SQL Server 2000 中，我有这个数据。

1900-01-01 00:10:10.830
1900-01-01 00:10:10.430

从上面的列中，我想要 select 日期时间并舍入毫秒，以获得以下输出

1900-01-01 00:10:11
1900-01-01 00:10:10

提前致谢

Answer 1

对于 SQL Server 2008 及更高版本，您可以使用 DATETIME2。 DATETIME2 在 SQL Server 2008 及更高版本中可用 - 有关详细信息，请参阅 here:

SELECT CAST('1900-01-01 00:10:10.830' AS DATETIME2(0));
SELECT CAST('1900-01-01 00:10:10.430' AS DATETIME2(0));

Confirmed Output

对于SQL服务器的早期版本，例如SQL Server 2000。你可以这样做：

SELECT DATEADD(ms, -DATEPART(ms, DATEADD(ms, 500, CAST('1900-01-01 00:10:10.830' AS DATETIME))) , DATEADD(ms, 500, CAST('1900-01-01 00:10:10.830' AS DATETIME)));
SELECT DATEADD(ms, -DATEPART(ms, DATEADD(ms, 500, CAST('1900-01-01 00:10:10.430' AS DATETIME))) , DATEADD(ms, 500, CAST('1900-01-01 00:10:10.430' AS DATETIME)));

Answer 2

SELECT *
     , DateAdd(ss, rounded_second, round_down_seconds) As result
FROM   (
        SELECT *
             , Round(nanoseconds / 1000.0, 0) As rounded_second
        FROM   (
                SELECT the_date
                     , DatePart(ms, the_date) As nanoseconds
                     , DateAdd(ss, DateDiff(ss, 0, the_date), 0) As round_down_seconds
                FROM   (
                        SELECT '1900-01-01 00:10:10.830' As the_date
                        UNION ALL 
                        SELECT '1900-01-01 00:10:10.430'
                       ) As x
               ) As y
       ) As z

为了尽可能清楚，我将每个步骤都分开了。

如果你想要单衬垫：

SELECT the_date
     , DateAdd(ss, Round(DatePart(ms, the_date) / 1000.0, 0), DateAdd(ss, DateDiff(ss, 0, the_date), 0)) As result
FROM   (
        SELECT '1900-01-01 00:10:10.830' As the_date
        UNION ALL 
        SELECT '1900-01-01 00:10:10.430'
       ) As x

舍入毫秒 SQL Server 2000

Rounding milliseconds SQL Server 2000

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