select t.orderdate, count(t.ticketid),sum(t.fare),sum(t.fuel_tax),sum(t.airport_tax),sum(t.paymoney)
from metticket t
where t.status = '2'
and t.orderdate is not null
group by t.orderdate
order by t.orderdate
这样统计每天的金额，就用日期分组了。
20100929 1 1460 0 50 1510
20101019 1 920 0 50 970
20101020 1 800 0 50 850
20101021 2 2500 0 100 2600
20101025 5 3570 0 250 3820
20101027 4 2330 0 200 2530
20101028 3 1710 0 150 1860
20101029 6 3920 0 300 4220
20101102 1 680 0 50 730
20101105 1 870 0 50 920
......
我想实现：没有数据的那天，也能查询到记录。因为没有记录，所有金额都为0
比如20101022没有记录，则为：
20101022 0 0 0 0 0
不知道怎么实现，求帮助
------解决方案--------------------
把下面的365换成你想要从今天开始统计到过去多少天的数据。
- SQL code
withdate_table as (select (sysdate-level) orderdate from dual connect by level < 365),order_table as (select t.orderdate, count(t.ticketid) cnt, sum(t.fare) sumfare, sum(t.fuel_tax) sumfueltax, sum(t.airport_tax) sumairtax, sum(t.paymoney) sumpay from metticket t where t.status = '2' and t.orderdate is not null group by t.orderdate order by t.orderdate)select date_table.orderdate, decode(cnt,null,0,cnt), decode(sumfare,null,0,sumfare), decode(sumfueltax,null,0,sumfueltax), decode(sumairtax,null,0,sumairtax), decode(sumpay,null,0,sumpay) from date_table left join order_table on date_table.orderdate=order_table.orderdate order by 1;