获得postgreSQL的当前周
发布时间:2020-12-13 15:49:59 所属栏目:百科 来源:网络整理
导读: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 kno
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
和
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 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |