按日期分组,当count()不产生任何行时为0

Question

我正在使用Postgresql 9,当没有计算行数时,我正在计算和分组.

我们假设以下架构:

create table views {
 date_event timestamp with time zone ;
 event_id integer;
}

让我们想象以下内容:

2012-01-01 00:00:05    2
2012-01-01 01:00:05    5
2012-01-01 03:00:05    8
2012-01-01 03:00:15    20

我想按小时分组,并计算行数.我希望我能找回以下内容:

2012-01-01 00:00:00    1
2012-01-01 01:00:00    1
2012-01-01 02:00:00    0
2012-01-01 03:00:00    2
2012-01-01 04:00:00    0
2012-01-01 05:00:00    0
.
.

2012-01-07 23:00:00    0

我的意思是,对于每个时间范围槽,我计算我的表中日期对应的行数,否则,我返回一个计数为零的行.

以下肯定不会起作用(只会将计数行数大于0的行).

SELECT  extract ( hour from date_event ),count(*)
FROM views
where date_event > '2012-01-01' and date_event <'2012-01-07'
GROUP BY extract ( hour from date_event );

请注意,我可能还需要按分钟,按小时,按天,按月或按年分组(当然可以进行多次查询).

我只能使用普通的旧sql,因为我的视图表可能非常大(> 100M记录),所以我会尽量记住性能.

怎么能实现这一目标？

谢谢 !

Answer 1

鉴于您没有表中的日期,您需要一种方法来生成它们.你可以使用这个generate_series功能:

SELECT * FROM generate_series('2012-01-01'::timestamp, '2012-01-07 23:00', '1 hour') AS ts;

这将产生如下结果:

         ts          
---------------------
 2012-01-01 00:00:00
 2012-01-01 01:00:00
 2012-01-01 02:00:00
 2012-01-01 03:00:00
...
 2012-01-07 21:00:00
 2012-01-07 22:00:00
 2012-01-07 23:00:00
(168 rows)

剩下的任务是使用外连接加入这两个选择,如下所示:

select extract ( day from ts ) as day, extract ( hour from ts ) as hour,coalesce(count,0) as count from 
(
    SELECT  extract ( day from date ) as day , extract ( hour from date ) as hr ,count(*)
    FROM    sr
    where date>'2012-01-01' and date <'2012-01-07'
    GROUP BY   extract ( day from date ) , extract ( hour from date )
) AS cnt 
 right outer join ( SELECT * FROM generate_series ( '2012-01-01'::timestamp, '2012-01-07 23:00', '1 hour') AS ts ) as dtetable on extract ( hour from ts ) = cnt.hr and extract ( day from ts ) = cnt.day 
 order by day,hour asc;