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
4 M% F( c3 n  R( N8 @+ p8 x5907  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：
  [% @! D. n5 S  N& d& N; D1 ^  nSELECT 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？
. V. H9 D- t& ?1 I8 n               
- N4 U7 X6 j% G4 y/ S7 U解决方案:
2 _; A& x7 }- U7 _+ P               

6 Q. i3 `$ T1 }5 p
                此条件将返回上周日至周六的记录：
9 i8 Z+ N/ n' }1 x' O3 yWHERE created BETWEEN
6 v5 K8 N* w2 I/ t6 `    NOW():ATE-EXTRACT(DOW FROM NOW())::INTEGER-7
    AND NOW():ATE-EXTRACT(DOW from NOW())::INTEGER
3 C8 [1 Y% N9 B) w有一个例子：
. @' Y5 q/ K4 O9 ^WITH compras AS (
! `2 Q( ~% E- ~* d    SELECT ( NOW() + (s::TEXT || ' day')::INTERVAL )::TIMESTAMP(0) AS created
, }' e' F9 q( X- z# ~    FROM generate_series(-20, 20, 1) AS s
/ q! a5 Q5 m$ K% A7 O% F7 k)
SELECT to_char( created, 'DY'::TEXT), created
FROM compras
1 I' r' O1 ?- ]; Z7 g% v8 OWHERE created BETWEEN
, Q/ {3 z$ n6 n, K3 Y    NOW():ATE-EXTRACT(DOW FROM NOW())::INTEGER-7
    AND NOW():ATE-EXTRACT(DOW from NOW())::INTEGER
在对@ d456的回答中：
' q& f0 D, y" o, ?( R$ a
BETWEEN在间隔的两端不使用星期日的午夜吗？

; c3 k% i4 [4 h' V! o/ `4 b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
2 S& [; F. }9 p. z$ R; e) I* T4 A    FROM generate_series( -- this would produce timestamps with 20 minutes step
0 E+ \- [/ A0 X7 i             (now() - '20 days'::interval)::date,
' E0 r6 f9 y- O3 c( P! P! m             (now() + '20 days'::interval)::date,
             '20 minutes'::interval) AS s
$ a- t# ]1 |5 f( U6 {0 p)
SELECT to_char( created, 'DY'::TEXT), created
FROM compras
WHERE TRUE
5 o  R8 c, D7 O2 x9 ], d    AND created >= NOW():ATE-EXTRACT(DOW FROM NOW())::INTEGER-7
, f6 w/ q- x7 @' b5 F0 x1 C1 ]    AND created <  NOW():ATE-EXTRACT(DOW from NOW())::INTEGER