日期 时间
0-1点 1-2点 2-3点 3-4点 4-5点..... 23-24点
2007-08-27 20 24 28 13 8 33
2007-08-28 12 54 45 14 8 43
2007-08-29 32 23 28 23 32 34
2007-08-30 12 34 43 14 42 23
2007-08-31 22 23 28 13 8 33
...
按日期/时间 显示不同时刻,登陆某网站的人数
表中有 记录着 登陆人的 id 和 登陆时间
------解决方案--------------------
select count(id) from table group by 时间(取小时)
------解决方案--------------------
select t1.id,t2.num1,t3.num2,t4.num3 from tablename t1
left join (select id,count(id) as num1 from tableNmae where substring(date,12,2) > =0 and substring(date,12,2) <1) t2 on t1.id=t2.id
left join (select id,count(id) as num2 from tableNmae where substring(date,12,2) > =1 and substring(date,12,2) <2 ) t3 on t1.id=t3.id
left join (select id,count(id) as num3 from tableNmae where substring(date,12,2)> =2 and substring(date,12,2) <3 ) t4 on t1.id=t4.id
group by substring(date,0,10) 试试能不能通过编译,我的没有Sql server