******** * *********
* GROUPS
-------------
users * uid(Primary KEy)
gname
------ * gdesc
uid *
username *
password *
FULL_NAME *
date_of_birth *
phone_number *
valid *
manu_gps
editor_map
operator
gid (Foreign Key) reference to GROUPS(uid)
您可以帮助SQL查询吗?我应该加入这个表格。换句话说,我应该把关联gid和uid,而不是gid pun适当的gname
答案 0 :(得分:0)
尝试类似
的内容SELECT u.uid,
u.username,
g.uid,
g.gname
FROM users u
LEFT JOIN groups g
ON g.uid = u.gid;