Php 在Codeigniter中连接3个表没有结果

即使数据库中有数据，也没有结果：

$this->db->select('*');
$this->db->from('user');
$this->db->join('userprofile', 'user.userID = userprofile.userID'); 
$this->db->join('classroom', 'user.classroomID = classroom.classroomID'); 
$this->db->where('roleID',"4");
$query = $this->db->get();
return $query->result();

---

在连接时，必须使用where条件定义表名。如下所示

        $this->db->select('*');
        $this->db->from('user');
        $this->db->join('userprofile', 'user.userID = userprofile.userID'); 
        $this->db->join('classroom', 'user.classroomID = classroom.classroomID'); 
        $this->db->where('table_name.roleID',4);//table_name is name of table having column roleID
        $query = $this->db->get();
        return $query->result();

---

是否可以共享需要快照的表数据。。。。。

$this->db->select('*');
$this->db->from('user');
$this->db->join('userprofile', 'user.userID = userprofile.userID'); 
$this->db->join('classroom', 'user.classroomID = classroom.classroomID'); 
$this->db->where('roleID',"4");
$query = $this->db->get();
return $query->result();

your query work with one table data when you join another table you can must define $this->db->where('user.roleID',4) like this.

you can write this query like this for when you get the join table data 

$this->db->select('user.*,userprofile.youdesirename,classroom.class name');
$this->db->from('user');
$this->db->join('userprofile', 'user.userID = userprofile.userID'); 
$this->db->join('classroom', 'user.classroomID = classroom.classroomID'); 
$this->db->where('user.roleID',4)
$query = $this->db->get();
return $query->result();