Question

我一直在网上搜索current_week的正确postgreSQL语法。我搜索了所附的链接，但无法从中获得任何结果Date/Time。我的任务是将星期日作为一周的开始。

我尝试过与current_date相同但失败了：

select current_week

postgreSQL必须有当前的一周语法。

Answer 1

extract('dow' from

的

knowing that

星期日（0）到星期六（6）的星期几

和

根据定义，ISO周从星期一开始

您可以通过减去一天来解决问题：

select date_trunc('week', current_date) - interval '1 day' as current_week
  current_week
------------------------
 2016-12-18 00:00:00+00
(1 row)

以下是样本：

t=# with d as (select generate_series('2016-12-11','2016-12-28','1 day'::interval) t)
select date_trunc('week', d.t)::date  - interval '1 day' as current_week, extract('dow' from d.t), d.t from d
;
    current_week     | date_part |           t
---------------------+-----------+------------------------
 2016-12-04 00:00:00 |         0 | 2016-12-11 00:00:00+00
 2016-12-11 00:00:00 |         1 | 2016-12-12 00:00:00+00
 2016-12-11 00:00:00 |         2 | 2016-12-13 00:00:00+00
 2016-12-11 00:00:00 |         3 | 2016-12-14 00:00:00+00
 2016-12-11 00:00:00 |         4 | 2016-12-15 00:00:00+00
 2016-12-11 00:00:00 |         5 | 2016-12-16 00:00:00+00
 2016-12-11 00:00:00 |         6 | 2016-12-17 00:00:00+00
 2016-12-11 00:00:00 |         0 | 2016-12-18 00:00:00+00
 2016-12-18 00:00:00 |         1 | 2016-12-19 00:00:00+00
 2016-12-18 00:00:00 |         2 | 2016-12-20 00:00:00+00
 2016-12-18 00:00:00 |         3 | 2016-12-21 00:00:00+00
 2016-12-18 00:00:00 |         4 | 2016-12-22 00:00:00+00
 2016-12-18 00:00:00 |         5 | 2016-12-23 00:00:00+00
 2016-12-18 00:00:00 |         6 | 2016-12-24 00:00:00+00
 2016-12-18 00:00:00 |         0 | 2016-12-25 00:00:00+00
 2016-12-25 00:00:00 |         1 | 2016-12-26 00:00:00+00
 2016-12-25 00:00:00 |         2 | 2016-12-27 00:00:00+00
 2016-12-25 00:00:00 |         3 | 2016-12-28 00:00:00+00
(18 rows)

Time: 0.483 ms

Answer 2

一种方法是date_trunc()：

select date_trunc('week', current_date) as current_week

获得postgreSQL的当前周

2 个答案: