求SQL代码 思路
id a b c1401 144 137 2009-11-17 08:37:38.000
901 142 140 2009-11-17 09:37:38.000
1110 128 125 2009-11-17 19:19:45.000
610 126 128 2009-11-17 20:19:45.000
我想取出0-12点 12点--24点 各自A最大值，结果是
1401 144 137 2009-11-17 08:37:38.000
1110 128 125 2009-11-17 19:19:45.000 --------------------编程问答-------------------- 0-12的
select max(a) from table where Convert(varchar(2),datepart(hour,[时间])) <=12
12-24的
select max(a) from table where Convert(varchar(2),datepart(hour,[时间])) >=12
--------------------编程问答-------------------- 如果不取12点的，把等于去掉 --------------------编程问答--------------------
select max(a) from table where Convert(varchar(2),datepart(hour,[时间])) <=12--------------------编程问答--------------------
union all
select max(a) from table where Convert(varchar(2),datepart(hour,[时间])) >=12 and Convert(varchar(2),datepart(hour,[时间])) <=24
--------------------编程问答-------------------- --------------------编程问答--------------------
declare @t table ([id] int, a int, b int, c datetime)
insert into @t
select 1401,144,137,'2009-11-17 08:37:38.000' union all
select 901,142,140,'2009-11-17 09:37:38.000' union all
select 1110,128,125,'2009-11-17 19:19:45.000' union all
select 610,126,128,'2009-11-17 20:19:45.000'
select * from @t t
where exists(select 1 from
(select max(id) as id from @t
group by case when convert(char(10), c, 114)<= '12:00:00:0000' then 'a' else 'b' end
) t1 where t.id=t1.id)
+1 --------------------编程问答--------------------
select convert(varchar(12),c,23),max(a) as zd,px from (--------------------编程问答-------------------- 4楼代码 正解 --------------------编程问答-------------------- 解决了啊。我来看下有份接吗。。 --------------------编程问答-------------------- 4楼正解！~~~~~ --------------------编程问答-------------------- 1楼的也可以吧
select *,
case when datepart(hour,c)<12 then '上午' else '下午' end as px
from t) as s group by convert(varchar(12),s.c,23),s.px
主要用max和DataPart
补充：.NET技术 , ASP.NET
