我有这个查询:
Select Trunc(Create_Dtime),Count(Trunc(Create_Dtime)) as Day_0 From Player
Group By Trunc(Create_Dtime)
Order By Trunc(Create_Dtime) Asc
它给了我回溯日期,但如果某个日期没有任何结果,则会被跳过。我想填写从 8-05-12 到现在的所有日期,如果这些日子没有任何值,则结果中只有一个 0。
根据您要如何创建它们,您可以创建日历表connect by
或使用 Oracle语法动态生成行。
with the_dates as (
select max(trunc(Create_Dtime)) as max_date
, min(trunc(Create_Dtime)) as min_date
from player
)
, generator as (
select min_date + level as the_date
from the_dates
connect by level <= max_date
)
select g.the_date, count(trunc(p.Create_Dtime))
from generator g
left outer join player p
on g.the_date = trunc(p.Create_Dtime)
group by g.the_date
order by g.the_date desc
如果您使用日历表选项,它会更干净一点:
with the_dates as (
select max(trunc(Create_Dtime)) as max_date
, min(trunc(Create_Dtime)) as min_date
from player
)
select c.the_date, count(trunc(p.Create_Dtime))
from calender c
join the_dates td
on c.the_date between td.min_date and td.max_date
left outer join join player p
on c.the_date = trunc(p.Create_Dtime)
group by c.the_date
order by c.the_date
或者,刚刚注意到您的日期限制:
with the_dates as (
select to_date('07-05-2012','dd-mm-yyyy') + level as the_date
from dual
connect by level <= trunc(to_date('07-05-2012','dd-mm-yyyy') - sysdate)
)
select td.the_date, count(trunc(p.create_dtime))
from the_dates td
left outer join player p
on td.the_date = trunc(p.create_dtime)
group by td.the_date
order by td.the_date
对于所有这些,我建议trunc(create_dtime)
在你的player
桌子上建立一个索引。