如何从没有时区列的时间戳中提取时间？

Question

我想计算每天每个时间间隔的订单数量。

日期的格式是没有时区的时间戳。 我似乎不能只提取时间。 我每天都使用这个查询，但是有没有办法在一个表中包含一个月中每一天的时间间隔？

CASE WHEN date_created_utc >= timestamp '2020-09-01 08:00:00' AND date_created_utc <= timestamp '2020-09-01 11:00:00' THEN 'Q1'    
WHEN date_created_utc >= timestamp '2020-09-01 11:00:01' AND date_created_utc <= timestamp '2020-09-01 14:00:00' THEN 'Q2'    
WHEN date_created_utc >= timestamp '2020-09-01 14:00:01' AND date_created_utc <= timestamp '2020-09-01 16:00:00' THEN 'Q3'    
WHEN date_created_utc >= timestamp '2020-09-01 16:00:01' AND date_created_utc <= timestamp '2020-09-01 20:00:00' THEN 'Q4'    
WHEN date_created_utc >= timestamp '2020-09-01 20:00:01' AND date_created_utc <= timestamp '2020-09-01 23:59:00' THEN 'Q5'    
END AS interval,    
COUNT(id) as cnt    
FROM order_processing    
GROUP BY 1;

所需的输出表：

Day Q1  Q2  Q3  Q4  Q5
1   28  57  50  65  27
2   23  50  60  90  66
3   58  60  80  70  67

Answer 1

您只需要小时部分来实现逻辑：您可以使用extract() ：

select
    date_created_utc::date day,
    count(*) filter(where extract(hour from date_created_utc) between 8 and 10) q1,
    count(*) filter(where extract(hour from date_created_utc) between 11 and 14) q2,
    ...
from order_processing
group by date_created_utc::date

Answer 2

您可以转换为time ，然后使用比较。 对于聚合：

COUNT(*) FILTER (WHERE date_created_utc::time >= '08:00:00' and date_created_utc::time < '11:00:00') as cnt_1