SQL有关问题,求解决

SQL问题,求解决
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。

20100929114600501510
201010191920050970
201010201800050850
201010212250001002600
201010255357002503820
201010274233002002530
201010283171001501860
201010296392003004220
201011021680050730
201011051870050920
  ......

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：
2010102200000
不知道怎么实现，求帮助

[解决办法]
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;