如何实现分组统计？

[kulg]

表如下
id bookdate bookname booknumber
1 2004-12-19 00:00:00.000 计算机组装 2
2 2004-12-19 00:00:00.000 powerbuilder9.0 5
3 2004-12-18 00:00:00.000 internet 4
4 2004-12-18 00:00:00.000 计算机组装 8
5 2004-12-18 00:00:00.000 中华美食 4
6 2004-12-17 00:00:00.000 局域网组装 5
7 2004-12-17 00:00:00.000 汇编程序设计 3

想用SQL语句实现

日期 购书数量
2004-12-19 00:00:00.000 7
2004-12-18 00:00:00.000 16
2004-12-17 00:00:00.000 8

请问如何实现分组统计?

[子陌红尘]

select
bookdate as 日期，
sum(booknumber) as 购书数量
from
表
group by
日期
order by
日期 desc

[dulei115]

select bookdate as 日期, sum(booknumber) as 购书数量
from 表
group by bookdate
order by bookdate desc

[NinGoo]

你确认你的日期后面都是00:00:00的格式？

select bookdate as 日期, sum(booknumber) as 购书数量
from table1
group by bookdate

[vinsonshen]

select 日期=bookdate,购书数量=sum(booknumber) from 表 group by bookdate

[vinsonshen]

若要按每天来统计的话，这样显示的结果更好：
select 日期=convert(varchar(10),bookdate,120),购书数量=sum(booknumber) from 表 group by
convert(varchar(10),bookdate,120)

[wuzhouhome]

select bookdate as 日期,
sum(booknumber) as 购书数量
from
表
group by
日期
order by
日期 desc

[gimy007]

同以上；
我是来噌分的~~~
^O^

[kulg]

谢谢！搞定！