获得postgreSQL的当前周

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是没有得到任何结果日期/时间.我的任务是将星期日作为本周的开始.

我尝试了与current_date相同但失败了:

select current_week

postgreSQL必须有当前的一周语法.

Answer 1

一种方法是date_trunc()：

select date_trunc('week', current_date) as current_week

Answer 2

知道了extract('dow' from

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

您可以通过减少一天来解决方法:

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

这是样本:

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