postgresql - 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，我正在与计数和分组作斗争。

让我们假设以下架构:

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，统计行数。我希望我能检索到以下内容:

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，我计算表中日期对应的行数，否则，我返回计数为零的行。

以下肯定行不通(只会产生计数行数 > 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、小时、天、月或年进行分组(当然可以进行多个查询)。

我只能使用普通的旧 sql，并且由于我的 View 表可能非常大(>100M 记录)，所以我尽量将性能放在心上。

如何实现？

谢谢!

Answer 1

鉴于表格中没有日期，您需要一种生成日期的方法。您可以使用 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)

剩下的任务是像这样使用外部连接连接两个选择:

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;