获得postgreSQL的当前周

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却无法从中获得任何结果 Date/Time.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.

解决方法

knowing that提取物(‘dow’来自

The day of the week as Sunday (0) to Saturday (6)

和

By deFinition,ISO weeks start on Mondays

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