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請教時間處理問題,该怎么解决

热度:200   发布时间:2016-04-27 21:58:12.0
請教時間處理問題

表tb1結構:
        f1   int     自增
        f2   datetime                

f2的值大致如下:
2006-12-20   22:22:32.000
2006-12-20   23:32:38.000
2006-12-20   00:24:49.000

declare   t1   char(5),
                t2   char(5),
                t3   datetime

set   t1= '23:30 '
set   t1= '00:30 '     第二天的l凌晨時間

現在請教如何在表tb1中查詢出字段f2時間部分在t1至t2范圍的紀錄




------解决方案--------------------
select * from tablename where f2 between cast(( '2006-12-20 '+ '23:30 '+ ':000 ') as datetime) and cast(( '2006-12-21 '+ '00:30 '+ ':000 ') as datetime)
------解决方案--------------------
select * from tablename
where convert(varchar(10),f2,108) between '23:30 ' and '00:30 '
------解决方案--------------------
create table T(f1 int, f2 datetime)
insert T select 1, '2006-12-20 22:22:32.000 '
union all select 2, '2006-12-20 23:32:38.000 '
union all select 3, '2006-12-20 00:24:49.000 '

declare @t1 char(5), @t2 char(5)
set @t1= '23:30 '
set @t2= '00:30 '

select * from T
where f2 between (convert(char(10), f2, 120)+ ' '[email protected]) and (convert(char(10), f2+1, 120)+ ' '[email protected])
--result
f1 f2
----------- ------------------------------------------------------
2 2006-12-20 23:32:38.000

(1 row(s) affected)

------解决方案--------------------


create table T(f1 int, f2 datetime)
insert T select 1, '2006-12-20 22:22:32.000 '
union all select 2, '2006-12-20 23:32:38.000 '
union all select 3, '2006-12-20 00:24:49.000 '

select f2 from T where
DATEDIFF(Hour, Convert(char(8),f2,112),f2) in (23,0)
and
(DATEDIFF(Minute, Convert(char(10),f2,112),f2)-DATEDIFF(Hour, Convert(char(8),f2,112),f2)*60)
between 0 and 59
------解决方案--------------------

create table T(f1 int, f2 datetime)
insert T select 1, '2006-12-20 22:22:32.000 '
union all select 2, '2006-12-20 23:32:38.000 '
union all select 3, '2006-12-20 00:24:49.000 '
union all select 4, '2006/12/28 00:34:11.000 '

select f2 from T where
(DATEDIFF(Hour, Convert(char(8),f2,112),f2)=23
and
(DATEDIFF(Minute, Convert(char(10),f2,112),f2)-DATEDIFF(Hour, Convert(char(8),f2,112),f2)*60)
between 0 and 59)
or
(DATEDIFF(Hour, Convert(char(8),f2,112),f2)=0
and
(DATEDIFF(Minute, Convert(char(10),f2,112),f2)-DATEDIFF(Hour, Convert(char(8),f2,112),f2)*60)
between 0 and 30)
------解决方案--------------------
字段f2時間部分在t1至t2范圍的紀錄?LZ 你表達的不是很清楚.
2006-12-20 00:24:49.000
2006-12-20 23:34:49.000
2006-12-19 23:34:49.000
2006-12-19 00:24:49.000
時間在 00:30 和 23:30 中的記錄是?
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