Question

我有5张桌子，

user, plan, achievement, user_plan, user_achievement

结构

User
id
name

Plan
id
name

achievement
id
name
*planID

user_plan
*planid
*userid

user_achievement
*achievementid
*userid
*createDate

我使用Codeigniter（没关系，我想要的是mysql查询）

$this->db->select('*');
$this->db->from('user');
$query = $this->db->get();

实际上，对于每个计划，都有一个成就列表，如果没有实现，它不会存储在user_achievement中，那么我怎样才能“为每个用户，得到1”他的名字2）计划他加入，3）他取得的成就，而不是实现（需要突出成就）“？

感谢。

Answer 1

使用join

 $this->db->select('i.id,u.name,up.planid,p.name as planname,ua.achievementid,ach.name as ach_name');
 $this->db->from('user as u');
 $this->db->join('user_plan as up','up.userid = u.id','left');
 $this->db->join('plan as p','p.id = up.planid','left');
 $this->db->join('user_achievement as ua','ua.userid = u.id','left');
 $this->db->join('achievement as ach','ach.id = ua.achievementid','left');
 $query = $this->db->get();

Answer 2

您需要在LEFT JOIN，user，plan和achievement之间user_achievement，如下所示

select u.name as username,
p.name as planname,
a.name as achivementname
from User u
left join user_achievement ua on u.id = ua.userid
and ua.createDate is not null
left join achivement a on ua.achievementid = a.id
left join Plan p on a.planid = p.id

SQL Query在不同的表中获取相关数据

2 个答案: