Question

我想计算每天每个时间间隔的订单数。

日期的格式是没有时区的时间戳。我似乎无法仅提取时间。我每天都使用此查询，但是有没有一种方法可以将一个月中每一天的时间间隔放在一个表中？

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

您只需要小时部分即可实现逻辑：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

如何从没有时区的时间戳列中提取时间？

2 个答案: