【发布时间】:2018-02-15 17:41:11
【问题描述】:
我有一个类似推特的网络,用户在其中关注其他用户,我需要向他们展示关注新人的建议
TABLE USERS
user id_user
A 1
B 2
C 3
D 4
E 5
F 6
TABLE COMMUNITY
id_follower id_followed
3 4
3 5
3 6
3 (C) 跟随到 4,5,6 (D,E,F)
我得到了显示关注用户的声明,4,5,6(D,E,F):
SELECT user,id_followed
FROM users,community
WHERE users.id_user=community.id_followed AND community.id_follower=3
GROUP BY user
如何显示未跟随 C(3) 的用户,即 1,2 (A,B)
我需要EXCEPT 吗? LEFT OUTER JOIN?
SELECT id_followed
FROM community
EXCEPT
(
SELECT user,id_followed
FROM users,community
WHERE users.id_user=community.id_followed AND community.id_follower=3
GROUP BY user
)
RETRIEVE ERROR.
【问题讨论】:
标签: sql left-join outer-join except