How obtain directly the model from a query? -> CodeIgniter - php

Question

Can I obtain the model directly from a query in Code Igniter?
Model
<?php
class User_Model extends CI_Model {
public $user_id;
public $name;
public $team_id
public function getTeamName() {
$this->db->select('name');
$this->db->from('team');
$this->db->where('team_id',$this->team_id);
$query = $this->db->get();
$team = $query->row();
if($empresa == null){
return "";
} else {
return $team->name;
}
}
public function getUser($id) {
$query = $this->db->get_where('users', array('user_id' => $id));
$user = $query->row();
return $user;
}
I would like getUser($id) function returns a user object like 'new User_Model()' to be able to use getTeamName() function in the view. Can I do it?
Now returns an object, but not a User_Model() object.

Answer 1

is wrong call a method in the model from the view.
in the model use
function getUser($id){
$query = $this-db->select('*')
->from('users')
->where('user_id', $id)
->get();
return $query->result();
}
this return a object with all data, then you call the method in the controller like this
$data['user_model'] = $this->model_name->getUser($id);
$this->load->view('Your_View',$data);
then in the view $user_model is the object
in the view
print_r($user_model);

Answer 2

As long as the model is loaded, in your view you can access any public function or property like:
$this->user_model->getUser() or $this->user_model->team_id
View:
<?php
echo $this->user_model->getUser(1)->username;
This is against MVC in some respects but I can understand the logic behind doing so based on the following scenario where you want to say use num_rows and result without doing count:
class User_Model extends CI_Model {
public function getUsers() {
return $this->db->get('users');
}
}
class Some_Controller extends CI_Controller {
public function index() {
$this->load->model('user_model');
$data['users'] = $this->user_model->getUsers();
$this->load->view('some_view', $data);
}
}
View:
<?php
if ($users->num_rows() > 0) {
foreach ($users->result() as $row) {
echo $row->username;
}
} else {
echo 'No users';
}

Answer 3

Actually there is another possibility you can use
create a Team_model and an User_model
The Team_model could look like
class Team_model extends CI_Model
{
$arrTeamNames = [];
public function getTeamName($team_id)
{
if (!isset($this->arrTeamNames[$team_id]))
{
$query = $this->db
->select('name')
->from('team')
->where('team_id',$this->team_id)
->get();
if ($query->num_rows() == 1)
{
$objTeam = $query->row();
$this->arrTeamNames[$team_id] = $objTeam->name;
return $objTeam->name;
}
return '';
}
return $this->arrTeamNames[$team_id];
}
}
and the User_model
class User_Model extends CI_Model
{
public function getUser($id)
{
$query = $this->db->get_where('users', array('user_id' => $id));
if ($query->num_rows() == 1)
{
$user = $query->row(0, 'User_Object');
return $user;
}
}
}
class User_Object
{
public function getTeamName()
{
$ci = get_instance();
$ci->load->model('Team_model');
return $ci->team_model->getTeamName($this->team_id);
}
}
Note in the User_model you have an additional User_Object which
gets mapped by the row() function which is well documented
here.
Now if you call this model in one of your controller and pass it to your view you actually can access to a model function in a proper way e.g.
class User extends CI_Controller
{
public function detail($user_id)
{
$this->load->model('User_model');
$objUser = $this->User_model->getUser($user_id);
$this->load->view('user-detail', ['objUser' => $objUser]);
}
}
//user-detail view
echo $objUser->getTeamName();