在數據庫中加入03表codeIgniter

Question

與我的表。

person_id serial NOT NULL, 
    firstname character varying(30) NOT NULL, 
    lastname character varying(30), 
    email character varying(50), 
    username character varying(20) NOT NULL, 
    "password" character varying(100) NOT NULL, 
    gender character varying(10), 
    dob date, 
    accesslevel smallint NOT NULL, 
    company_id integer NOT NULL,//Reference to table company 
    position_id integer NOT NULL,//Reference to table position 
company_id serial NOT NULL, 
    company_name character varying(80) NOT NULL, 
    description character varying(255), 
    address character varying(100) NOT NULL, 

在我的控制器

........................ 
// load data 
$persons = $this->person_model->get_paged_list(10,0); 
// generate table data 
$this->load->library('table'); 
$this->table->set_empty(" "); 
$this->table->set_heading('No', 'FirstName', 'LastName','E-mail','Company''Gender', 'Date of Birth', 'Actions'); 
foreach ($persons as $person){ 
    $this->table->add_row(++$i, $person->firstname, 
            $person->lastname, 
            $person->email, 
            $person->company_name, 
            //HOW CAN I GOT THE POSITION TITLE ?, 
          strtoupper($person->gender)=='M'? 'Male':'Female', 
          date('d-m-Y',strtotime($person->dob)), 
} 

我的模型

<?php 
class Person_Model extends Model { 

    private $person= 'person'; 

    function Person(){ 
     parent::Model(); 
    } 

    function list_all(){ 
     $this->db->order_by('person_id','asc'); 
     return $this->db->get($person); 
    } 

    function count_all(){ 
     return $this->db->count_all($this->person); 
    } 

    function get_paged_list($limit = 0, $offset = 0) { 
     $this->db->limit($limit, $offset); 
     $this->db->select("person.*, company.company_name as company"); 
     $this->db->from('person'); 
     $this->db->join('company','person.company_id = company.company_id','left'); 
     //MY QUESTION:? CAN I JOIN MORE WITH TABLE POSITION? 

     $query = $this->db->get();  
     return $query->result(); 
    } 

    function get_by_id($id){ 
     $this->db->where('person_id', $id); 
     return $this->db->get($this->person); 
    } 

    function save($person){ 
     $this->db->insert($this->person, $person); 
     return $this->db->insert_id(); 
    } 

    function update($id, $person){ 
     $this->db->where('person_id', $id); 
     $this->db->update($this->person, $person); 
    } 

    function delete($id){ 
     $this->db->where('person_id', $id); 
     $this->db->delete($this->person); 
    } 
} 
?> 

Answer 1

是的，可以。只需使用盡可能多的$ this-> db-> join（...）即可。