Question

我有一张桌子，里面有我购买的所有客户。我想选择上周的所有参赛作品（周日从星期日开始）。

id    value  date
5907  1.20   "2015-06-05 09:08:34-03"
5908  120.00 "2015-06-09 07:58:12-03"

我试过这个：

SELECT id, valor, created, FROM compras WHERE created >= now() - interval '1 week' and parceiro_id= '1'

但我从上周得到的数据包括本周的数据，我只想要上周的数据。

如何仅从上周获取数据？

Answer 1

这种情况将从上周日到周六返回记录：

WHERE created BETWEEN
    NOW()::DATE-EXTRACT(DOW FROM NOW())::INTEGER-7 
    AND NOW()::DATE-EXTRACT(DOW from NOW())::INTEGER

有一个例子：

WITH compras AS (
    SELECT ( NOW() + (s::TEXT || ' day')::INTERVAL )::TIMESTAMP(0) AS created
    FROM generate_series(-20, 20, 1) AS s
)
SELECT to_char( created, 'DY'::TEXT), created
FROM compras
WHERE created BETWEEN
    NOW()::DATE-EXTRACT(DOW FROM NOW())::INTEGER-7 
    AND NOW()::DATE-EXTRACT(DOW from NOW())::INTEGER

回答@ d456：

不会使用BETWEEN包括星期日午夜时间间隔的两端？

这是正确的，BETWEEN包括星期日午夜的间隔两端。要在星期日的间隔结束时排除午夜，必须使用运算符>=和<：

WITH compras AS (
    SELECT s as created
    FROM generate_series( -- this would produce timestamps with 20 minutes step
             (now() - '20 days'::interval)::date,
             (now() + '20 days'::interval)::date,
             '20 minutes'::interval) AS s
)
SELECT to_char( created, 'DY'::TEXT), created
FROM compras
WHERE TRUE
    AND created >= NOW()::DATE-EXTRACT(DOW FROM NOW())::INTEGER-7
    AND created <  NOW()::DATE-EXTRACT(DOW from NOW())::INTEGER

Answer 2

默认情况下，Postgres会在周日开始几周，所以你很幸运。您可以使用date_trunc()来获取上周的开头：

WHERE (created >= date_trunc('week', CURRENT_TIMESTAMP - interval '1 week') and
       created < date_trunc('week', CURRENT_TIMESTAMP)
      )

编辑：

默认情况下，Postgres会在星期一为 date_trunc 开始一周，但周日为dow。所以，你可以通过尼古拉在他的回答中使用的那种逻辑做你想做的事。

PostgreSQL查询从上周选择数据？

2 个答案: