一个数据库查询问题
有2个表
表1
EntryId 
1
2
3
4

表2
id EntryId date ip
1 2 2001-11-12 11:11 61.135.179.155
2 1 2001-11-12 11:23 61.135.179.155
3 3 2001-11-12 13:34 61.135.179.152
4 1 2001-11-12 12:43 61.135.179.152
5 2 2001-11-12 12:34 61.135.179.155

查询成
EntryId date visit ip_visit
1 2001-11-12 2 2
2 2001-11-12 2 1
3 2001-11-12 1 1
4 2001-11-12 0 0


那个date是查询条件 输出的时候可有可无
不用存储过程
sql hql都可

------解决方案--------------------

SQL code

select a.EntryId,format(b.date,'yyyy-mm-dd') as cdate, count(*) as visit,
    (select count(*) from (select distinct ip from 表2 where EntryId=a.EntryId and int(date)=int(b.date))) as ip_visit
from 表1 a inner join 表2 b on a.EntryId=b.EntryId
group by a.EntryId,format(b.date,'yyyy-mm-dd')

------解决方案--------------------

SQL code

select a.entryid,date,count(date) as visit,count(distinct ip) as visit_ip
from 表1 a left join 表2 b on a.EntryId=b.EntryId
group by a.EntryId,date;