sql - PostgreSQL 查询选择上周的数据？

Question

我有一张表，上面有我的顾客的所有购买。我想选择上周的所有条目(一周从星期日开始)。

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

我已经试过了:

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

但是我得到了上周的数据包括本周的数据，我只想要上周的数据。

如何只获取上周的数据？

Answer 1

此条件将返回上周星期日到星期六的记录:

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

对@d456 的回答:

Wouldn't using BETWEEN include midnight on Sunday at both ends of the interval?

没错，BETWEEN 在间隔的两端都包含星期日的午夜。要在间隔结束时排除周日午夜，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