Codeigniter, using URL as parameter for query - php

Question

i am using codeigniter, i am trying to create a page where if the url entered example.com/mobil/bekas/toyota/avanza it shows all used car that has toyota as brand and avanza as model, and if the url entered example.com/mobil/bekas/toyota it shows all used car that has toyota as brand.
Here is my controller :
public function bekas($brand_nama,$model_nama='NULL')
{
$this->load->model('listing_model');
$data['cars'] = $this->listing_model->viewListingByBrandAndModel($brand_nama, $model_nama);
$this->load->view('product_listing.php', $data);
}
Here is the model :
function viewListingByBrandAndModel($brand_nama, $model_nama)
{
$this->load->library('pagination');
$this->load->library('table');
$config['base_url'] = 'http://example.com/mobil/bekas/'.$brand_nama.'/'.$model_nama;
$config['total_rows'] = $this->db->select('*')
->join('car_list_tbl','car_list_tbl.car_list_ID = user_listing_tbl.car_list_ID')
->join('member_tbl','member_tbl.mID = user_listing_tbl.mID')
->join('model_tbl','model_tbl.model_ID = car_list_tbl.model_ID')
->join('series_tbl','series_tbl.series_ID = car_list_tbl.series_ID')
->join('body_type_tbl','body_type_tbl.body_type_nama = car_list_tbl.body_type_nama')
->join('brand_tbl','brand_tbl.brand_name = car_list_tbl.brand_name')
->where('car_list_tbl.brand_name',$brand_nama)
->like('model_tbl.model_nama', $model_nama)
->where('user_listing_tbl.listing_type','BEKAS')
->get('user_listing_tbl')->num_rows();
$config['per_page'] = 20;
$config['num_links'] = 10;
$config['display_pages'] = TRUE;
$config['full_tag_open'] = '<ul class="pagination">';
$config['full_tag_close'] = '</ul>';
$config['cur_tag_open'] = '<li class="active"><a href="#">';
$config['cur_tag_close'] = '</a></li>';
$config['num_tag_open'] = '<li>';
$config['num_tag_close'] = '</li>';
$config['first_link'] = FALSE;
$config['last_link'] = FALSE;
$config['prev_link'] = false;
$config['next_link'] = false;
$config['next_tag_open'] = '<li><a href="#"><i class="fa fa-chevron-left">';
$config['next_tag_close'] = '</i></a></li>';
$config['prev_tag_open'] = '<li><a href="#"><i class="fa fa-chevron-right">';
$config['prev_tag_close'] = '</i></a></li>';
$this->pagination->initialize($config);
//Pagination End
$sql = $this->db->select('*')
->join('car_list_tbl','car_list_tbl.car_list_ID = user_listing_tbl.car_list_ID')
->join('member_tbl','member_tbl.mID = user_listing_tbl.mID')
->join('brand_tbl','brand_tbl.brand_name = car_list_tbl.brand_name')
->join('model_tbl','model_tbl.model_ID = car_list_tbl.model_ID')
->join('series_tbl','series_tbl.series_ID = car_list_tbl.series_ID')
->where('car_list_tbl.brand_name',$brand_nama)
->like('model_tbl.model_nama', $model_nama)
->where('user_listing_tbl.listing_type','BEKAS')
->get('user_listing_tbl', $config['per_page'], $this->uri->segment(5));
return $sql->result();
I am still newbie at web programming, can i have inputs on which part i am lacking ? Because it works when i type example.com/mobil/bekas/toyota/avanza but it wont show anything when i type example.com/mobil/bekas/toyota

Answer 1

1 ) You are passing NULL as a string in parameter
2) Please use the if condition base on the $model_name at the time of the database query. Do not pass the extra conditions in query with like model_name LIKE '';
$this->db->select('*')
->join('car_list_tbl','car_list_tbl.car_list_ID = user_listing_tbl.car_list_ID')
->join('member_tbl','member_tbl.mID = user_listing_tbl.mID')
->join('brand_tbl','brand_tbl.brand_name = car_list_tbl.brand_name')
->join('model_tbl','model_tbl.model_ID = car_list_tbl.model_ID')
->join('series_tbl','series_tbl.series_ID = car_list_tbl.series_ID')
->where('car_list_tbl.brand_name',$brand_nama);
if($model_nama){
$this->db->like('model_tbl.model_nama', $model_nama);
}
$this->db->where('user_listing_tbl.listing_type','BEKAS');
->get('user_listing_tbl', $config['per_page'], $this->uri->segment(5));
return $this->db->result();

Answer 2

Your function declaration has a small syntax error. The value NULL is a special value in PHP and as such, should not be surrounded in quotes. For this reason, you were querying the database for the type 'toyota' with the make 'NULL' rather than NULL.
Changing your controller code should fix the issue:
public function bekas($brand_nama,$model_nama=NULL)
{
//...
}