求教一条SQL语句（关于Group By）
大家好！
        想请教一个问题，我有一个表
ID       Productcode       num       createdate     .......
1           PSR0W001             1           2007-01-02   09:30:01
2           PSR0W002             3           2007-01-02   01:30:12
3           PSR0W001             4           2007-01-02   10:10:10
4           PSR0W001             2           2007-01-03   03:00:15  
5           PSR0W001             2           2007-01-03   08:01:00
..........
每天的08：00：00到第二天早上的07：59：59为一天（用以统计每天的产量）
如产品代码为PSR0W001的产品2007-01-02这天生产了7个。
请问怎么样用SQL语句实现啊？


------解决方案--------------------
--如：
Select Productcode,sum(num) as nums,
to_char(Createdate, 'yyyy-mm-dd ') as Createdate
from (
Select Productcode,num,Createdate+(
case when to_char(Createdate, 'hh24 ') < '08 '
then -1 else 0 end) as Createdate from 表名 ) t
group by Productcode,to_char(Createdate, 'yyyy-mm-dd ')
------解决方案--------------------

Select Productcode,to_char(Createdate-1/3, 'yyyy-mm-dd ') as Createdate,sum(num) as nums
from table_name
group by Productcode,to_char(Createdate-1/3, 'yyyy-mm-dd ')
------解决方案--------------------
SQL> select zz.Productcode,sum(zz.num) sum_num
2 from (select tt.*,
3 decode(sign(to_char(tt.createdate, 'HH24 ')- '08 '),-1,to_char(tt.createdate-1, 'yyyy-mm-dd '),to_char(tt.createdate, 'yyyy-mm-dd ')) as sumdate
4 from ( select 1 as id, 'PSR0W001 ' as Productcode,1 as num,to_date( '2007-01-02 09:30:01 ', 'yyyy-mm-dd hh24:mi:ss ') as createdate from dual
5 union all
6 select 2 as id, 'PSR0W002 ' as Productcode,3 as num,to_date( '2007-01-02 13:30:12 ', 'yyyy-mm-dd hh24:mi:ss ') as createdate from dual
7 union all
8 select 3 as id, 'PSR0W001 ' as Productcode,4 as num,to_date( '2007-01-02 23:10:10 ', 'yyyy-mm-dd hh24:mi:ss ') as createdate from dual
9 union all
10 select 4 as id, 'PSR0W001 ' as Productcode,2 as num,to_date( '2007-01-03 03:00:15 ', 'yyyy-mm-dd hh24:mi:ss ') as createdate from dual
11 union all
12 select 5 as id, 'PSR0W001 ' as Productcode,2 as num,to_date( '2007-01-03 08:01:00 ', 'yyyy-mm-dd hh24:mi:ss ') as createdate from dual
13 )tt
14 )zz
15 group by zz.Productcode,zz.sumdate;

PRODUCTCODE SUM_NUM
----------- ----------
PSR0W001 7
PSR0W001 2
PSR0W002 3

------解决方案--------------------
trunc(sysdate-1/3)
把当前时间-8小时后转换成整数(无小数位)
假设当前时间是2007-08-31 08:00:00
转换出来的时间就是2007-08-31


和trunc(sysdate)-1/3有什么区别呢？
把当前时间转换成整数(无小数位)后减8小时
假设当前时间是2007-08-31 08:00:00