PostgreSQL查询以选择上周的数据？

风随心旨

我有一张桌子，上面有我所有顾客购买的东西。我想选择上周（从周日开始的一周）中的所有条目。
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？

解决方案:


            此条件将返回上周日至周六的记录：
WHERE created BETWEEN
NOW():

ATE-EXTRACT(DOW FROM NOW())::INTEGER-7
AND NOW():

ATE-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():

ATE-EXTRACT(DOW FROM NOW())::INTEGER-7
AND NOW():

ATE-EXTRACT(DOW from NOW())::INTEGER
在对@ d456的回答中：

BETWEEN在间隔的两端不使用星期日的午夜吗？

没错，BETWEEN包括间隔两端的周日午夜。要在间隔结束时排除周日的午夜，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():

ATE-EXTRACT(DOW FROM NOW())::INTEGER-7
AND created < NOW():

ATE-EXTRACT(DOW from NOW())::INTEGER

PostgreSQL查询以选择上周的数据？

风随心旨 LV1