距离今天最早8天之内的讯息
select id, addtime,dateadd(day,8,addtime) as plus
from protuct
where dateadd(day,8,addtime)< getdate()
距离今天最早8月之内的讯息
select id, addtime,dateadd(month,8,addtime) as plus
from protuct
where dateadd(month,8,addtime)< getdate()
距离今天最早8年之内的讯息
select id, addtime,dateadd(month,8,addtime) as plus
from protuct
where dateadd(year,8,addtime)< getdate()
例:
create view vTopChitsContent as
select top 10 * from content
where dateadd(day,10,sDatetime) > getdate()
order by chits desc
select id, addtime,dateadd(day,8,addtime) as plus
from protuct
where dateadd(day,8,addtime)< getdate()
距离今天最早8月之内的讯息
select id, addtime,dateadd(month,8,addtime) as plus
from protuct
where dateadd(month,8,addtime)< getdate()
距离今天最早8年之内的讯息
select id, addtime,dateadd(month,8,addtime) as plus
from protuct
where dateadd(year,8,addtime)< getdate()
例:
create view vTopChitsContent as
select top 10 * from content
where dateadd(day,10,sDatetime) > getdate()
order by chits desc
本文介绍如何使用SQL语句查询特定时间范围内的记录,包括日、月、年等不同时间跨度,并提供了一个视图创建示例。
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