sql-server - SQL 两次中最小的时间

Question

Declare @time1 datetime;
Declare @time2 datetime;

right(@time1,7) = 9:00 Am
right(@time2,7) = 3:10 Pm

case when @time1 > @time2 then 1 else -1 end as small

我需要的答案是-1

Answer 1

SQL 服务器 2005

我将从分别显示每个步骤的详细方法开始。

DECLARE @time1 datetime;
    SET @time1 = Current_Timestamp;
DECLARE @time2 datetime;
    SET @time2 = DateAdd(hh, -29, @time1);

; WITH dateparts AS (
  SELECT @time1 As time1
       , DatePart(hh, @time1) As hh1
       , DatePart(mi, @time1) As mi1
       , DatePart(ss, @time1) As ss1
       , DatePart(ms, @time1) As ms1
       , @time2 As time2
       , DatePart(hh, @time2) As hh2
       , DatePart(mi, @time2) As mi2
       , DatePart(ss, @time2) As ss2
       , DatePart(ms, @time2) As ms2
)
, construct_time_only AS (
  SELECT time1
       , DateAdd(hh, hh1, DateAdd(mi, mi1, DateAdd(ss, ss1, DateAdd(ms, ms1, 0)))) As timeonly1
       , time2
       , DateAdd(hh, hh2, DateAdd(mi, mi2, DateAdd(ss, ss2, DateAdd(ms, ms2, 0)))) As timeonly2
  FROM   dateparts
)
SELECT time1
     , time2
     , timeonly1
     , timeonly2
     , CASE WHEN timeonly1 > timeonly2 THEN 1 ELSE -1 END As small
FROM   construct_time_only
;

这里的想法是我们设置一个共同的基准日期进行比较。在这种情况下，我们使用1900-01-01(ie Cast(0 As datetime)) 作为我们的共同日期。

使用通用日期的原因是为了满足我们起点的日期部分不同的情况。

SQL Server 2008

SQL Server 2008 引入了新的date数据time类型。我们可以简单地Cast()对time结果进行比较

DECLARE @time1 datetime = Current_Timestamp;
DECLARE @time2 datetime = DateAdd(hh, -1, @time1);

SELECT @time1 As time1
     , Cast(@time1 As time) As timeonly1
     , @time2 As time2
     , Cast(@time2 As time) As timeonly2
     , CASE WHEN Cast(@time1 As time) > Cast(@time2 As time) THEN 1 ELSE -1 END As small

Answer 2

您应该将日期作为日期类型进行比较：

只需执行（此处为小提琴演示）：

Declare @date1 datetime = '20131017 09:00:00'
Declare @date2 datetime = '20131017 15:00:00'

--right(@time1,7) = 9:00 Am
--right(@time2,7) = 3:10 Pm

select case when @date1 > @date2 then 1 else -1 end as small

您也可以使用Datediff()函数，例如；

select case when datediff(second, @date1, @date2) > 0 then -1 else 1 end as small