Question

我需要创建一个度量标准，以显示一周中特定日期每小时每小时创建的订单数量。但是，DB中的任何记录都有一段时间。但是我这次仍然需要证明。

select count(id),
       date_trunc('hour', created_at at time zone 'US/Pacific')::time as "time"
from "order"
where to_char(created_at, 'day') like '%wed%'
group by date_trunc('hour', created_at at time zone 'US/Pacific')::time

结果，我有这样的东西

计数，时间
| 6 | 00:00:00 |
+-+ ------------ +
| 3 | 01:00:00 |
+-+ ------------ +
| 4 | 03:00:00 |
+-+ ------------ +
| 5 | 05:00:00 |
+-+ ------------ +

但是需要这个

计数，时间
| 6 | 00:00:00 |
+-+ ------------ +
| 3 | 01:00:00 |
+-+ ------------ +
| 0 | 02:00:00 |
+-+ ------------ +
| 4 | 03:00:00 |
+-+ ------------ +
| 0 | 04:00:00 |
+-+ ------------ +
| 5 | 05:00:00 |
+-+ ------------ +

Answer 1

一种选择是使用包含所有可能时间的日历表，然后加入该日历表：

WITH times AS (
    SELECT '00:00:00'::time AS timeval UNION ALL
    SELECT '01:00:00'::time UNION ALL
    SELECT '02:00:00'::time UNION ALL
    ...
    SELECT '23:00:00'::time
)

SELECT
    COUNT(*) AS count,
    t.timeval
FROM times t
LEFT JOIN "order" o
    ON t.timeval = date_trunc('hour', o.created_at at time zone 'US/Pacific')::time AND
    to_char(o.created_at, 'day') LIKE '%wed%'
GROUP BY
    t.timeval;

如果您不想报告整整24小时，则只需修改times CTE以仅包括您想要的小时数即可。

Answer 2

为什么不这样：

select count(id),
       date_trunc('hour', created_at at time zone 'US/Pacific')::time as "time"
from "order"
where to_char(created_at, 'day') like '%wed%' and count(id)=0 or count(id)!=0
group by date_trunc('hour', created_at at time zone 'US/Pacific')::time

没有时间订单，但需要显示

2 个答案: